Courses

Courses for Kids

Free study material

Store

Talk to our experts

1800-120-456-456

NCERT Solutions

Class 11

Maths

Chapter 12 Limits And Derivatives

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

Latest Updates

Book Free Demo :

Maximise Every Mark with One-to-One Learning

Solve Maths Class 11 Limits And Derivatives With Vedantu's Expert Guidance

NCERT Solution for Limits And Derivatives Class 11 Chapter 12 marks the beginning of calculus, a significant branch of mathematics that deals with change and motion. Understanding Limits And Derivatives Class 11 Solutions is crucial as these concepts form the foundation for more advanced topics in calculus and have wide-ranging applications in various fields such as physics, engineering, economics, and beyond. Access the NCERT Solutions for Class 11 Maths here.

Table of Content

Access Exercise wise NCERT Solutions for Chapter 12 Maths Class 11

Current Syllabus Exercises of Class 11 Maths Chapter 12

NCERT Solutions of Class 11 Maths Limits and Derivatives Exercise 12.1

NCERT Solutions of Class 11 Maths Limits and Derivatives Exercise 12.2

NCERT Solutions of Class 11 Maths Limits and Derivatives Miscellaneous Exercise

Popular Vedantu Learning Centres Near You

22 No Phatak, Patiala

#2, Guhman Road, Near Punjabi Bagh, 22 No Phatak-Patiala

Visit Centre

Chhoti Baradari, Patiala

Vedantu Learning Centre, SCO-144 -145, 1st & 2nd Floor, Chotti Baradari Scheme Improvement Trust, Patiala-147001

Visit Centre

Janakpuri, Delhi

Vedantu Learning Centre, A-1/173A, Najafgarh Road, Opposite Metro Pillar 613, Block A1, Janakpuri, New Delhi 110058

Visit Centre

Mithanpura, Muzaffarpur

Vedantu Learning Centre, 2nd Floor, Ugra Tara Complex, Club Rd, opposite Grand Mall, Mahammadpur Kazi, Mithanpura, Muzaffarpur, Bihar 842002

Visit Centre

Boring Road, Patna

Vedantu Learning Centre , 4th & 5th Floor , SKU Centre Boring Road Patna - 800001

Visit Centre

Jankipuram, Lucknow

Vedantu Learning Center Plot No CP-3 , sector E jankipuram Sitapur Road Yojna lucknow U.P 226021

Visit Centre

Courses

Master Class 11 Limits And Derivatives With Vedantu's Expert Guidance

Exercise 12.1: This exercise introduces the concept of limits of a function, their properties, and different types of limits. Students will learn how to find the limit of a function using algebraic techniques and the Squeeze rule. They will also learn about left and right-hand limits and the existence of a limit.
Exercise 12.2: In this exercise, students will learn about the concept of derivatives of a function, their geometrical interpretation, and the rules of differentiation. They will also practice finding the derivative of a function using various differentiation rules.
Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of limits and derivatives to solve various problems and answer questions. They will also practice finding the limit of a function using algebraic techniques and the Squeeze rule and finding the derivative of a function using various differentiation rules.

Access NCERT Solutions for Class 11 Maths Chapter 12 – Limits and Derivatives

Exercise 12.1

1: Evaluate the Given limit: $lim_{x \to 3} x + 3$

Ans: $lim_{x \to 3} x + 3 = 3 + 3 = 6$

2: Evaluate the Given limit: $lim_{x \to z} (x - \frac{22}{7})$

Ans: $lim_{x \to z} (x - \frac{22}{7}) = (π - \frac{22}{7})$

3: Evaluate the Given limit: $lim_{r \to 1} π r^{2}$

Ans: $lim π r^{2} = π (1^{2}) = π$

4: Evaluate the Given limit: $lim_{x \to 1} \frac{4 x + 3}{x - 2}$

Ans: $lim_{x \to 1} \frac{4 x + 3}{x - 2} = \frac{4 (4) + 3}{4 - 2} = \frac{16 + 3}{2} = \frac{19}{2}$

5: Evaluate the Given limit: $lim_{x \to - 1} \frac{x^{10} + x^{5} + 1}{x - 1}$

Ans: $lim_{x \to - 1} \frac{x^{10} + x^{5} + 1}{x - 1} = \frac{(- 1)^{10} + (- 1)^{5} + 1}{- 1 - 1} = \frac{1 - 1 + 1}{- 2} = - \frac{1}{2}$

6: Evaluate the Given limit: $lim_{x \to 0} \frac{(x + 1)^{5} - 1}{x}$

Ans: $lim_{x \to 0} \frac{(x + 1)^{5} - 1}{x}$

Put $x + 1 = y s o$ that $y \to 1$ as $x \to 0$

Accordingly, $lim_{x \to 0} \frac{(x + 1)^{5} - 1}{x} = lim_{x \to 1} \frac{(y)^{5} - 1}{y - 1}$

5. $1^{5 - 1} [lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n d^{n - 1}]$

$= 5$

$∴ lim_{x \to 0} \frac{(x + 1)^{5} - 1}{x} = 5$

7: Evaluate the Given limit: $lim_{x \to 2} \frac{3 x^{2} - x - 10}{x^{2} - 4}$

Ans: At $x - 2$ , the value of the given rational function takes the form $\frac{0}{0}$

$lim_{x \to 2} \frac{3 x^{2} - x - 10}{x^{2} - 4} = lim_{x \to 2} \frac{(x - 2) (3 x + 5)}{(x - 2) (x + 2)}$

$lim_{x \to 2} \frac{3 x + 5}{x + 2}$

$= \frac{3 (2) + 5}{2 + 2}$

$- \frac{11}{4}$

8: Evaluate the Given limit: $lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3}$

Ans: At $x = 2$ , the value of the given rational function takes the form $\frac{0}{0}$

$lim_{x \to 3} \frac{x^{4} - 81}{2 x^{2} - 5 x - 3} = lim_{x \to 3} \frac{(x - 3) (x + 3) (x^{2} + 9)}{(x - 3) (2 x + 1)}$

$lim_{x \to 3} \frac{(x + 3) (x^{2} + 9)}{(2 x + 1)}$

$= \frac{(3 + 3) (3^{2} + 9)}{2 (3) + 1}$

$= \frac{6 \times 18}{7}$

$= \frac{108}{7}$

9: Evaluate the Given limit: $lim_{x \to 0} \frac{a x + b}{c x + 1}$

Ans:

$lim_{x \to 0} \frac{a x + b}{c x + 1} = \frac{a (0) + b}{d (0) + 1} = b$

10: Evaluate the Given limit: $lim_{z \to 1} \frac{\frac{1}{3^{3}} - 1}{\frac{1}{2^{6}} - 1}$

Ans: $lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{\frac{1}{z^{6}} - 1}$

At $z = 1$ , the value of the given function takes the form $\frac{0}{0}$ Put $z^{\frac{1}{6}} = x$ so that $z \to 1$ as $x \to 1$ .

Accordingly, $lim_{x \to 1} \frac{\frac{1}{\vec{e}} - 1}{z^{\frac{1}{t}} - 1} = lim_{x \to 1} \frac{x^{2} - 1}{x - 1}$

$= lim_{x \to 1} \frac{x^{2} - 1}{x - 1}$

$= {2.1}^{2 \cdot 1} [lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n d^{n - 1}]$

$= 2$

$lim_{x \to 1} \frac{z^{\frac{1}{3}} - 1}{\frac{1}{t^{6}} - 1} = 2$

11: Evaluate the Given limit: $lim_{x \to 1} \frac{a x^{2} + b x + c}{c x^{2} + b x + a}, a + b + c \neq 0$

Ans: $lim_{x \to 1} \frac{a x^{2} + b x + c}{c x^{2} + b x + a} = \frac{a (1)^{2} + b (1) + c}{c (1)^{2} + b (1) + a}$

$= \frac{a + b + c}{a + b + c}$

$= 1$

$[a + b + c \neq 0]$

12: Evaluate the given limit: $lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

Ans: $lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

At $x = - 2$ , the value of the given function takes the form $\frac{0}{0}$

Now, $lim_{x \to - 2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} = lim_{x \to - 2} \frac{(\frac{2 + x}{2 x})}{x + 2}$

$lim_{x \to - 2} \frac{1}{2 x}$

$\frac{1}{2 (- 2)} = \frac{- 1}{4}$

13: Evaluate the given limit: $lim_{x \to 0} \frac{\sin a x}{b x}$

Ans: $lim_{x \to 0} \frac{\sin a x}{b x}$

At $x = 0$ , the value of the given function takes the form $\frac{0}{0}$

Now, $lim_{x \to 0} \frac{\sin a x}{b x} = lim_{x \to 0} \frac{\sin a x}{a x} \times \frac{a x}{b x}$

$\begin{array}{l} = lim_{x \to 0} (\frac{\sin a x}{a x}) \times \frac{a}{b} \\ \frac{a}{b} lim_{ax \to 0} (\frac{\sin a x}{a x}) [x \to 0 \Rightarrow a x \to 0] \\ = \frac{a}{b} \times 1 \\ = [lim_{x \to 0} (\frac{\sin y}{y})] \\ = \frac{a}{b} \end{array}$

14: Evaluate the given limit: $lim_{x \to 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

Ans: $lim_{x \to 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

At $x = 0$ , the value of the given function takes the form $\frac{0}{0}$

Now, $lim_{x \to 0} \frac{\sin a x}{\sin b x} = lim_{x \to 0} (\frac{(\frac{\sin a x}{a x}) \times a x}{(\frac{\sin b x}{a x}) \times b x})$

$\begin{aligned} = \frac{a}{b} \times \frac{lim_{a x \to 0} (\frac{\sin a x}{a x})}{lim_{b x \to 0} (\frac{\sin b x}{a x})} \\ x \to 0 \Rightarrow a x \to 0 \end{aligned}$

$and x \to 0 \Rightarrow b x \to 0$

and $x \to 0 \Rightarrow b x \to 0$

$\frac{a}{b} \times \frac{1}{1}$

$[lim_{x \to 0} (\frac{\sin y}{y}) = 1]$

15: Evaluate the given limit: $lim_{x \to z} \frac{\sin (π - x)}{π (π - x)}$

Ans: $lim x \to z \frac{\sin (π - x)}{π (π - x)}$

It is seen that $x \to π \Rightarrow (π . x) \to 0$

$lim_{x \to 0} (\frac{\sin a x}{a x}) \times \frac{a}{b}$

$\frac{a}{b} lim_{a x \to 0} (\frac{\sin a x}{a x}) [x \to 0 \Rightarrow a x \to 0]$

$= \frac{a}{b} \times 1$

$[lim_{x \to 0} (\frac{\sin y}{y})]$

$= \frac{a}{b}$

16: Evaluate the given limit: $lim_{x \to 0} \frac{\cos x}{π - x}$

Ans: $lim_{x \to 0} \frac{\cos x}{π - x} = \frac{\cos 0}{π - 0} = \frac{1}{π}$

17: Evaluate the given limit: $lim_{x \to 0} \frac{\cos 2 x - 1}{\cos x - 1}$

Ans:

$lim_{x \to 0} \frac{\cos 2 x - 1}{\cos x - 1}$

At, x =0 the value of given function takes a form $\frac{0}{0}$

Now, $lim_{x \to 0} \frac{\cos 2 x - 1}{\cos x - 1} = lim_{x \to 0} \frac{1 - \sin^{2} x - 1}{1 - 2 \sin^{2} \frac{x}{2} - 1}$

$[\cos x = 1 - 2 \sin^{2} \frac{x}{2}]$

$- lim_{x \to 0} \frac{\sin^{2} x}{\sin^{2} \frac{x}{2}} = lim_{x \to 0} \frac{(\frac{\sin^{2} x}{x^{2}}) \times x^{2}}{(\frac{\sin^{2} \frac{x}{2}}{(\frac{x}{2})^{2}}) \times \frac{x^{2}}{4}}$

$\begin{aligned} = 4 \frac{lim_{x \to 0} (\frac{\sin^{2} x}{x^{2}})}{lim_{x \to 0} (\frac{\sin^{2} \frac{x}{2}}{{(\frac{x}{2})}^{2}})} \\ = 4 \frac{{(lim_{x \to 0} \frac{\sin^{2} x}{x^{2}})}^{2}}{{(lim_{x \to 0} \frac{\sin \frac{x}{2}}{{(\frac{x}{2})}^{2}})}^{2}} \\ [x \to 0 \Rightarrow \frac{x}{2} \to 0] \\ = 4 \frac{1^{2}}{1^{2}} [lim_{y \to 0} \frac{\sin y}{y} = 1] \end{aligned}$

18: Evaluate the given limit: $lim_{x \to 0} \frac{a x + x \cos x}{b \sin x}$

Ans: $lim_{x \to 0} \frac{a x + x \cos x}{b \sin x}$

At $x = 0$ , the value of the given function takes the form $\frac{0}{0}$

Now, $lim_{x \to 0} \frac{a x + x \cos x}{b \sin x} = \frac{1}{b} lim_{x \to 0} \frac{x (a + \cos x)}{\sin x}$

$lim_{b \to 0} (\frac{x}{\sin x}) \times lim_{x \to 0} (a + \cos x)$

$\frac{1}{b} (\frac{1}{lim_{x \to 0} (\frac{\sin x}{x})}) \times lim_{x \to 0} (a + \cos x)$

$\frac{1}{b} \times (a + \cos 0) [lim_{y \to 0} \frac{\sin x}{x} = 1]$

$- \frac{a + 1}{b}$

19: Evaluate the given limit: $lim_{x \to 0} x \sec x$

Ans: $lim_{x \to 0} x \sec x = lim_{x \to 0} \frac{x}{\cos x} = \frac{0}{\cos 0} = \frac{0}{1} = 0$

20: Evaluate the given limit: $lim_{x \to 0} \frac{\sin a x + b x}{a x + \sin b x} a, b, a + b \neq 0$

Ans: At $x = 0$ , the value of the given function takes the form $\frac{0}{0}$ Now, $lim_{x \to 0} \frac{\sin a x + b x}{a x + \sin b x}$

$= lim_{x \to 0} \frac{(\frac{\sin a x}{a x}) a x + b x}{a x + b x (\frac{\sin b x}{b x})}$

$= \frac{(lim_{x \to 0} \frac{\sin a x}{a x}) \times lim_{x \to 0} (a x) + lim_{x \to 0} (b x)}{lim_{x \to 0} a x + lim_{x \to 0} b x (lim_{x \to 0} \frac{\sin b x}{b x})} [$ As $x \to 0 \Rightarrow ax \to 0$ and $b x \to 0]$

$= \frac{lim_{x \to 0} (a x) + lim_{x \to 0} b x}{lim_{x \to 0} a x + lim_{x \to 0} b x}$

$[lim_{y \to 0} \frac{\sin x}{x} = 1]$

$= \frac{lim_{x \to 0} (a x + b x)}{lim_{x \to 0} (a x + b x)}$

$= lim_{x \to 0} (1)$

$= 1$

21: Evaluate the given limit: $lim_{x \to 0} (cosec x - \cot x)$

Ans: At $x = 0$ , the value of the given function takes the form $\infty - \infty$ Now, $lim_{x \to 0} (cosec x - \cot x)$

$lim_{x \to 0} (\frac{1}{\sin x} - \frac{\cos x}{\sin x})$

$lim_{x \to 0} (\frac{1 - \cos x}{\sin x})$

$= lim_{x \to 0} \frac{(\frac{1 - \cos x}{x})}{(\frac{\sin x}{x})}$

$= \frac{lim_{x \to 0} \frac{1 - \cos x}{x}}{lim_{x \to 0} \frac{\sin x}{x}}$

$- \frac{0}{1}$

$[lim_{y \to 0} \frac{1 - \cos x}{x} = 0$ and $lim_{y \to 0} \frac{\sin x}{x} = 1]$

$= 0$

22: Evaluate the given limit: $lim_{x \to \frac{x}{2}} \frac{\tan 2 x}{x - \frac{π}{2}}$

Ans: $lim_{x \to \frac{z}{2}} \frac{\tan 2 x}{x - \frac{π}{2}}$

At $x = \frac{π}{2}$ , the value of the given function takes the form $\frac{0}{0}$ Now, put So that $x - \frac{π}{2} - y$ so that $x \to \frac{π}{2}, y \to 0$

$∴ lim_{x \to \frac{π}{2}} \frac{\tan 2 x}{x - \frac{π}{2}} = lim_{y \to 0} \frac{\tan 2 (y + \frac{π}{2})}{y}$

$lim_{y \to 0} \frac{\tan (π + 2 y)}{y}$

$- lim_{y \to 0} \frac{\tan 2 y}{y}$

$[\tan (π + 2 y) = \tan 2 y]$

$- lim_{y \to 0} \frac{\sin 2 y}{y \cos 2 y}$

$lim_{y \to 0} (\frac{\sin 2 y}{2 y} \times \frac{2}{\cos 2 y})$

$- (lim_{y \to 0} \frac{\sin 2 y}{2 y}) \times lim_{y \to 0} (\times \frac{2}{\cos 2 y})$

$[y \to 0 \Rightarrow 2 y \to 0]$

$- 1 \times \frac{2}{\cos 0} [lim_{\leftrightarrow \infty 0} \frac{\sin x}{x} = 1]$

$= 1 \times \frac{2}{1}$

$- 2$

23: Find $lim_{x \to 0} f (x)$ and $lim_{x \to 1} f (x)$ , where $f (x) = {\begin{cases} 2 x + 3, & x \leq 0 \\ 3 (x + 1), & x > 0 \end{cases}$

Ans: The given function is

$f (x) = {\begin{cases} 2 x + 3, & x \leq 0 \\ 3 (x + 1), & x > 0 \end{cases}$

$lim_{x \to 0} f (x) = lim_{x \to 0} [2 x + 3] = 2 (0) + 3 - 3$

$lim_{v \to 0^{+}} f (x) = lim_{x \to 0} 3 (x + 1) - 3 (0 + 1) - 3$

$∴ lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) = lim_{x \to 0} f (x) = 3$

$lim_{x \to π} f (x) = lim_{x \to 1} (x + 1) = 3 (1 + 1) = 6$

$∴ lim_{x \to +} f (x) = lim_{x \to 1^{-}} f (x) = lim_{x \to 1} f (x) = 6$

24: Find $lim_{x \to 1} f (x)$ , when $f (x) = {\begin{cases} x^{2} - 1, & x \leq 1 \\ - x - 1, & x > 1 \end{cases}$

Ans:

The given function is

$f (x) = {\begin{cases} x^{2} - 1, & x \leq 1 \\ - x - 1, & x > 1 \end{cases}$

$∴ lim_{x \to T} f (x) = lim_{x \to 1} [x^{2} - 1] - 1^{2} - 1 - 1 - 1 = 0$

It is observed that $lim_{x \to π} f (x) \neq lim_{x \to 1^{+}} f (x)$ .

Hence, lim $f (x)$ does not exist.

25: Evaluate $lim_{x \to 0} f (x)$ , where $f (x) = {\begin{cases} \frac{| x |}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

Ans: The given function is $f (x) = {\begin{cases} \frac{| x |}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} [\frac{| x |}{x}]$

$- lim_{x \to 0} (\frac{- x}{x})$

(When $x$ is negative, $| x | = \cdot x$ )

$- lim_{x \to 0} (- 1)$

$- \cdot 1$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} [\frac{| x |}{x}]$

$- lim_{x \to 0} (\frac{x}{x})$

(When $x$ is positive, $| x | - x$ )

$= lim_{x \to 0} (1)$

$- 1$

It is observed that $lim_{x \to σ} f (x) \neq lim_{x \to 0^{+}} f (x)$ .

Hence, $lim_{x \to 0} f (x)$ does not exist.

26: Find $lim_{x \to 0} f (x) - {\begin{cases} \frac{x}{| x |}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

Ans: The given function is

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0} [\frac{x}{| x |}]$

$- lim_{x \to 0} (\frac{x}{- x})$

[When $x < 0, | x | - \cdot x]$

$- lim_{x \to 0} (- 1)$

$= \cdot 1$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} [\frac{x}{| x |}]$

$- lim_{x \to 0} (\frac{x}{x})$

$[$ When $x > 0, | x | = x]$

$- lim_{x \to 0} (1)$

$- 1$

It is observed that $lim_{x \to σ} f (x) \neq lim_{x \to 0^{+}} f (x)$ .

Hence, $lim_{x \to 0} f (x)$ does not exist.

27: Find $lim_{x \to 5} f (x)$ , where $f (x) = | x | \cdot 5$

Ans: The given function is $f (x) = | x | \cdot 5$

$lim_{y \to 5^{-}} f (x) = lim_{x \to 5} (| x | - 5)$

$= lim_{x \to 5} (x - 5)$

[When $x > 0, | x | - x]$

$- 5 \cdot 5$

$- 0$

$lim_{x \to^{+}} f (x) = lim_{x \to 5} (| x | - 5)$

$= lim_{x \to 5} (x - 5)$

( When $x > 0, | x | - x]$ )

$- 5 - 5$

$= 0$

$lim_{x \to 5^{-}} f (x) = lim_{x \to 5^{+}} f (x) = 0$

Hence, $lim_{x \to 5} f (x) = 0$

28: Suppose $f (x) = {\begin{cases} a + b x, & x < 0 \\ 4, & x = 1 \\ b - a x, & x > 1 \end{cases}$ and if $lim_{x \to 1} f (x) = f (1)$ what are possible values of a and b?

Ans: The given function is

$f (x) = {\begin{cases} a + b x, & x < 0 \\ 4, & x = 1 \\ b - a x, & x > 1 \end{cases}$

$lim_{- π} f (x) = lim_{x \to 1} (a + b x) = a + b$

$lim_{x, 1 -} f (x) = lim_{x \to 1} (b - a x) = b - a$

$f (1) = 4$

It is given that $lim_{x \to 1} f (x) = f (1)$ .

$∴ lim_{x \to π} f (x) = lim_{x \to 1^{+}} f (x) = lim_{x \to 1} f (x) = f (1)$

$\Rightarrow a + b - 4$ and $b - a = 4$

On solving these two equations, we obtain $a = 0$ and $b = 4$ . Thus, the respective possible values of a and $b$ are 0 and 4 .

29: Let $a_{1}, a_{2}, \dots, a_{n}$ be fixed real numbers and define a function

$f (x) = (x - a_{1}) (x - a_{2}) \dots . . (x - a_{n})$

What is $lim f (x) ?$ For some $a \neq a_{1}, a_{2}, \dots, a_{n}$ . Compute $lim_{x \to a} f (x)$ .

Ans: The given function is $f (x) = (x - a_{1}) (x - a_{2}) \dots . . (x - a)$ $lim_{x \to + 3} f (x) = lim_{x \to a} [(x - a_{1}) (x - a_{2}) \dots . . (x - a_{n})]$

$= (a_{1} - a_{1}) (a_{1} - a_{2}) \dots . . (a_{1} - a_{n}) = 0$

$∴ lim_{x \to 3} f (x) = 0$

Now, $lim_{x \to a} f (x) = lim_{x \to a} [(x - a_{1}) (x - a_{2}) \dots . . (x - a_{n})]$

$- (a - a_{1}) (a - a_{2}) \dots . . (a - a)$

$∴ lim f (x) = (a - a_{1}) (a - a_{2}) \dots \dots (a - a)$

30. If $f (x) = {\begin{aligned} | x | + 1 \\ 0 \\ | x | - 1 \end{aligned}$ $\begin{aligned} x < 0 \\ x = 0 \\ x > 1 \end{aligned}$

For what value (s) does $lim_{x \to a} f (x)$ exists?

Ans: The given function is

If $f (x) = {\begin{aligned} | x | + 1 \\ 0 \\ | x | - 1 \end{aligned}$ $\begin{aligned} x < 0 \\ x = 0 \\ x > 1 \end{aligned}$

When $a = 0$

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (| x | + 1)$

$= lim_{x \to 0} (- x + 1)$

If $x < 0, | x | = - x$

$= 0 + 1$

$= 1$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} (| x | + 1)$

$= lim_{x \to 0} (x - 1)$

If $x > 0, | x | = - x$

$= 0 - 1$

$= - 1$

Here, it is observed that $lim_{x \to 0^{-}} f (x) \neq lim_{x \to 0^{+}} f (x)$ .

$∴ lim_{x \to 0} f (x)$ does not exist.

When $a < 0$ $lim_{x \to a^{-}} f (x) = lim_{x \to a^{-}} (| x | + 1)$

$= lim_{x \to a} (- x + 1)$

$[x < a < 0 \Rightarrow | x | - - x]$

$= - a + 1$

$lim_{x \to \vec{t}} f (x) = lim_{x \to Δ} (| x | + 1)$

$= lim_{x \to a} (- x + 1)$

$[a < x < 0 \Rightarrow | x | - - x]$

$= - a$

$+ 1$

$∴ lim_{x \to a^{+}} f (x) = lim_{x \to a^{+}} f (x) = - a + 1$

Thus, limit of $f (x)$ exists at $x - a$ , where $a < 0$ .

When $a > 0$

$lim_{x \to a^{-}} f (x) = lim_{x \to a^{-}} (| x | + 1)$

$= lim_{x \to a} (- x - 1)$

$[0 < x < a \Rightarrow | x | - x]$

$= a - 1$

$lim_{x \to^{2}} f (x) = lim_{x \to Δ} (| x | - 1)$

$= lim_{x \to a} (- x - 1)$

$[0 < x < a \Rightarrow | x | = x]$

$= a - 1$

$∴ lim_{x \to π} f (x) = lim_{x \to^{+}} f (x) = a - 1$

Thus, limit of $f (x)$ exists at $x = a$ , where $a > 0$ Thus, $lim_{x \to a} f (x)$ exists for all $a \neq 0$ .

31: If the function f(x) satisfies, $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$ , evaluate $lim_{x \to 1} f (x)$

Ans: $lim_{x \to 1} \frac{f (x) - 2}{x^{2} - 1} = π$

$\Rightarrow \frac{lim_{x \to 1} (f (x) - 2)}{lim_{x \to 1} (x^{2} - 1)} = π$

$\Rightarrow lim_{x \to 1} (f (x) - 2) = π lim_{x \to 1} (x^{2} - 1)$

$\Rightarrow lim_{x \to 1} (f (x) - 2) = π (1^{2} - 1)$

$\Rightarrow lim_{x \to 1} (f (x) - 2) = 0$

$\Rightarrow lim_{x \to 1} f (x) - lim_{x \to 1} 2 = 0$

$\Rightarrow lim_{x \to 1} f (x) - 2 = 0$

$∴ lim_{x \to 1} f (x) = 2$

32: If $f (x) = {\begin{cases} m x^{2} + n, & x < 0 \\ n x + m, & 0 \leq x \leq 1 \\ n x^{3} + m, & x > 1 \end{cases}$

For what integers m and n does $lim_{x \to 0} f (x)$ and $lim_{x \to 1} f (x)$ exist?

Ans: $f (x) = {\begin{cases} m x^{2} + n, & x < 0 \\ n x + m, & 0 \leq x \leq 1 . \\ n x^{3} + m, & x > 1 \end{cases}$

$lim_{x \to 0} f (x) = lim_{x \to 0} (m x^{2} + n)$

$= m (0)^{2} + n$

$= n$

$= n (0) + m$

$= m$

Thus, $lim_{x \to 0^{+}} f (x)$ exists if $m = n$ .

$lim_{x \to \sqrt{-}} f (x) = lim_{x \to 1} (n x + m)$

$= n (1) + m$

$= m + n$

$lim_{x \to 1^{+}} f (x) = lim_{x \to 1} (n x^{3} + m)$

$= n (1)^{3} + m$

$= m + n$

$∴ lim_{x \to \sqrt{}} f (x) = lim_{x \to +} f (x) = lim_{x \to 1} f (x) .$

Thus, $lim_{u \to 1} f (x)$ exists for any internal value of $m$ and $n$ .

Exercise 12.2

1: Find the derivative of $x^{2} - 2$ at $x = 10$ .

Ans: Let $f (x) = x^{2} - 2 .$ Accordingly. $f^{'} (10) = lim_{h \to 0} \frac{f (10 + h) - f (10)}{h}$

$= lim_{h \to 0} \frac{[(10 + h)^{2} - 2] - (10^{2} - 2)}{h}$

$= lim_{h \to 0} \frac{10^{2} + 2 \cdot 10 \cdot h + h^{2} - 2 - 10^{2} + 2}{h}$

$= lim_{h \to 0} \frac{20 h + h^{2}}{h}$

$= lim_{h \to 0} (20 + h) = 20 + 0 = 20$

Thus, the derivative of $x^{2} - 2$ at $x - 10$ is 20 .

2. Find the derivative of x at x=1.

Ans: Let $f (x) = x .$ Accordingly.

$f^{'} (1) = lim_{n \to 0} \frac{f (1 + h) - f (1)}{h}$

$= lim_{h \to 0} \frac{(1 + h) - 1}{h}$

$= lim_{h \to 0} \frac{h}{h}$

$= lim_{h \to 0} (1) = 1$

Thus, the derivative of $x$ at $x = 1$ is 1 .

3: Find the derivative of $99 x$ at $x$ -100.

Ans: Let $f (x) = 99 x$ . Accordingly,

$f^{'} (100) = lim_{h \to 0} \frac{f (100 + h) - f (100)}{h}$

$= lim_{n \to 0} \frac{99 (100 + h) - 99 (100)}{h}$

$= lim_{h \to 0} \frac{99 \times 100 + 99 h - 99 \times 100}{h}$

$= lim_{h \to 0} \frac{99 h}{h}$

$= lim_{h \to 0} (99) = 99$

Thus, the derivative of $99 x$ at $x = 100$ is 99 .

4: Find the derivative of the following functions from first principles.

(i) $x^{3} - 27$

(ii) $(x - 1) (x - 2)$

(iii) $\frac{1}{x^{2}}$

(iv) $\frac{x + 1}{x - 1}$

Ans: (i) Let $f (x) = x^{3} - 27$ . Aocordingly, from the first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{[(x + h)^{3} - 27] - (x^{3} - 27)}{h}$

$= lim_{h \to 0} \frac{x^{3} + h^{3} + 3 x^{2} h + 3 x t^{2} - x^{3}}{h}$

$= lim_{h \to 0} \frac{h^{3} + 3 x^{2} h + 3 x h^{2}}{h}$

$= lim_{h \to 0} (h^{3} + 3 x^{2} h + 3 x h^{2})$

$= 0 + 3 x^{2} + 0 = 3 x^{2}$

(ii) Let $f (x) = (x - 1) (x - 2)$ . Accordingly, from the first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{(x + h - 1) (x + h - 2) - (x - 1) (x - 2)}{h}$

$= lim_{n \to 0} \frac{(x^{2} + h x - 2 x + h x + t^{2} - 2 h - x - h + 2) - (x^{2} - 2 x - x + 2)}{h}$

$= lim_{h \to 0} \frac{(h x + h x + h^{2} - 2 h - h)}{h}$

$= lim_{h \to 0} \frac{2 h x + h^{2} - 3 h}{h}$

$= lim_{n \to 0} (2 x + h - 3)$

$- 2 x - 3$

(iii) Let $f (x) = \frac{1}{x^{2}}$ . Accordingly, from the first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{\frac{1}{(x + h)^{2}} - \frac{1}{x^{2}}}{h}$

$= lim_{h \to 0} \frac{1}{h} [\frac{x^{2} - (x + h)^{2}}{x^{2} (x + h)^{2}}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{x^{2} - x^{2} - 2 h x - h^{2}}{x^{2} (x + h)^{2}}]$

$= lim_{n \to 0} \frac{1}{h} [\frac{- A - 2 h x}{x^{2} (x + h)^{2}}]$

$= lim_{h \to 0} [\frac{- h^{2} - 2 x}{x^{2} (x + h)^{2}}]$

$= \frac{0 - 2 x}{x^{2} (x + 0)^{2}} = \frac{- 2}{x^{3}}$

(iv) Let $f (x) = \frac{x + 1}{x - 1}$ . Accordingly, from the frst principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{(\frac{x + h + 1}{x + h - 1} - \frac{x + 1}{x - 1})}{h}$

$= lim_{n \to 0} \frac{1}{h} [\frac{(x - 1) (x + h + 1) - (x + 1) (x + h - 1)}{(x - 1) (x + h - 1)}]$

$= lim_{n \to 0} \frac{1}{h} [\frac{(x^{2} + h x + x - x - h - 1) - (x^{2} + h x - x + x + h - 1)}{(x - 1) (x + h - 1)}]$

$= lim_{n \to 0} \frac{1}{h} [\frac{- 2 h}{(x - 1) (x + h - 1)}]$

$= lim_{n \to 0} [\frac{- 2}{(x - 1) (x + h - 1)}]$

$= \frac{- 2}{(x - 1) (x - 1)} = \frac{- 2}{(x - 1)^{2}}$

5: For the function $F (x) = \frac{x^{10}}{100} + \frac{x^{59}}{99} + \dots + \frac{x^{2}}{2} + x + 1$

Prove that $f (1) = 100 f^{'} (0)$

Ans: The given function is

$F (x) = \frac{x^{m}}{100} + \frac{x^{59}}{99} + \dots + \frac{x^{2}}{2} + x + 1$

$\frac{d}{d x} f (x) = \frac{d}{d x} [\frac{x^{1 m}}{100} + \frac{x^{m}}{99} + \dots + \frac{x^{2}}{2} + x + 1]$

$\frac{d}{d x} f (x) = \frac{d}{d x} (\frac{x^{100}}{100}) + \frac{d}{d x} (\frac{x^{99}}{99}) + \dots + \frac{d}{d x} (\frac{x^{2}}{2}) + \frac{d}{d x} (x) + \frac{d}{d x} (1)$

On using theorem $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ , we obtain

$\frac{d}{d x} f (x) = \frac{100 x^{99}}{100} + \frac{99 x^{88}}{99} + \dots + \frac{2 x}{2} + 1 + 0$

$= x^{99} + x^{88} + \dots + x + 1$

$∴ f^{'} (x) = x^{99} + x^{18} + \dots + x + 1$

At $x = 0$

$f^{'} (0) = 1$

At $x = 1$ ,

Thus, $f (1) = 100 f (0)$

6: Find the derivative of $x^{n} + a x^{n - 1} + a^{2} x^{n - 2} + \dots + a^{n - 1} χ + a^{n}$ for some fixed real number a.

Ans: Let $f (x) = x^{n} + a x^{n - 1} + a^{2} x^{n - 2} + \dots + a^{n - 1} x + a^{n}$

$\frac{d}{d x} f (x) = \frac{d}{d x} (x^{n} + a x^{n - 1} + a^{2} x^{n - 2} + \dots + a^{n - 1} x + d^{n})$

$= \frac{d}{d x} (x^{n}) + a \frac{d}{d x} (x^{n - 1}) + a^{2} \frac{d}{d x} (x^{n - 2}) + \dots + a^{n - 1} \frac{d}{d x} (x) + a^{n} \frac{d}{d x} (1)$

On using theorem $\frac{d}{d x} (x^{n}) = n^{n - 1}$ , we obtain

$f^{'} (x) = n x^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + \dots + a^{n - 1} + a^{n} (0)$

$∴ f^{'} (x) = n x^{n - 1} + a (n - 1) x^{n - 2} + a^{2} (n - 2) x^{n - 3} + \dots + a^{n - 1}$

7: For some constants a and $b$ , find the derivative of

(i) $(x - a) (x - b)$

(ii) ${(a x^{2} + b)}^{2}$

(iii) $\frac{x - a}{x - b}$

Ans: (i) Let $f (x) = (x - a) (x - b)$

$\Rightarrow f (x) = x^{2} - (a + b) x + a b$

$∴ f^{'} (x) = \frac{d}{d x} (x^{2} - (a + b) x + a b)$

$= \frac{d}{d x} (x^{2}) - (a + b) \frac{d}{d x} (x) + \frac{d}{d x} (a b)$

On using theorem $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ , we obtain

$f^{'} (x) = 2 x - (a + b) + 0$

$= 2 x - a - b$

(ii) Let $f (x) - {(a x^{2} + b)}^{2}$

$\Rightarrow f (x) = a^{2} x^{4} + 2 a b x^{2} + b^{2}$

$∴ f (x) = \frac{d}{d x} (a^{2} x^{4} + 2 a b x^{2} + b^{2})$

$= a^{2} \frac{d}{d x} (x^{4}) + 2 a b \frac{d}{d x} (x^{2}) + \frac{d}{d x} b^{2}$

On using theorem $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ , we obtain

$f^{'} (x) = a^{2} (4 x^{3}) + 2 a b (2 x) + b^{2} (0)$

$= 4 a^{2} x^{3} + 4 a b x$

$= 4 a x (a x^{2} + b)$

(iii) Let $f (x) = \frac{x - a}{x - b}$

$\Rightarrow f^{'} (x) = \frac{d}{d x} (\frac{x - a}{x - b})$

By quotient rule,

$f^{'} (x) = \frac{(x - b) \frac{d}{d x} (x - a) - (x - a) \frac{d}{d x} (x - b)}{(x - b)^{2}}$

$= \frac{(x - b) (1) - (x - a) (1)}{(x - b)^{2}}$

$= \frac{x - b - x + a}{(x - b)^{2}}$

$= \frac{a - b}{(x - b)^{2}}$

8: Find the derivative of $\frac{x^{n} - a^{n}}{x - a}$ for some constant a.

Ans: Let $f (x) = \frac{x^{n} - d^{n}}{x - a}$

$\Rightarrow f^{'} (x) = \frac{d}{d x} (\frac{x^{n} - a^{n}}{x - a})$

By quotient rule, $f^{'} (x) = \frac{(x - a) \frac{d}{d x} (x^{n} - a^{n}) - (x^{n} - a^{n}) \frac{d}{d x} (x - a)}{(x - a)^{2}}$

$= \frac{(x - a) (n x^{n - 1} - 0) - (x^{n} - a^{n})}{(x - a)^{2}}$

$= \frac{n x^{n} - a n x^{n - 1} - x^{n} + a^{n}}{(x - a)^{2}}$

9: Find the derivative of

(i) $2 x - \frac{3}{4}$

(ii) $(5 x^{3} + 3 x - 1) (x - 1)$

(iii) $x^{- 3} (5 + 3 x)$

(iv) $x^{5} (3 - 6 x^{- 9})$

(v) $x^{- 4} (3 - 4 x^{- 5})$

(vi) $\frac{2}{x + 1} - \frac{x^{2}}{3 x - 1}$

Ans: (i) Let $f (x) = 2 x - \frac{3}{4}$

$f^{'} (x) = \frac{d}{d x} (2 x - \frac{3}{4})$

$= 2 \frac{d}{d x} (x) - \frac{d}{d x} (\frac{3}{4})$

$- 2 - 0$

(ii) Let $f (x) = (5 x^{3} + 3 x - 1) (x - 1)$

By Leibnitz product rule,

$f^{'} (x) = (5 x^{3} + 3 x - 1) \frac{d}{d x} (x - 1) + (x - 1) \frac{d}{d x} (5 x^{3} + 3 x - 1)$

$- (5 x^{3} + 3 x - 1) (1) + (x - 1) (5.3 x^{2} + 3 - 0)$

$- (5 x^{3} + 3 x - 1) + (x - 1) (15 x^{2} + 3)$

$- 5 x^{3} + 3 x - 1 + 15 x^{3} + 3 x - 15 x^{2} - 3$

$= 20 x^{3} - 15 x^{2} + 6 x - 4$

(iii) Let $f (x) = x^{- 3} (5 + 3 x)$

By Leibnitz product rule,

$f^{'} (x) = x^{- 3} \frac{d}{d x} (5 + 3 x) + (5 + 3 x) \frac{d}{d x} (x^{- 3})$

$= x^{- 3} (0 + 3) + (5 + 3 x) (3 x^{- 3 - 1})$

$= x^{- 3} (3) + (5 + 3 x) (3 x^{- 4})$

$= 3 x^{- 3} - 15 x^{- 4} - 9 x^{- 3}$

$= - 6 x^{3} - 15 x^{4}$

$= - 3 x^{- 3} (2 + \frac{5}{x})$

$= \frac{- 3 x^{- 3}}{x} (2 x + 5)$

$= \frac{- 3}{x^{4}} (5 + 2 x)$

(iv) Let $f (x) = x^{5} (3 - 6 x^{- 9})$

By Leibnitz product rule,

$f^{'} (x) = x^{5} \frac{d}{d x} (3 - 6 x^{9}) + (3 - 6 x^{- 9}) \frac{d}{d x} (x^{5})$

$= x^{5} {0 - 6 (- 9) x^{- 2 - 1}} + (3 - 6 x^{9}) (5 x^{4})$

$= x^{5} (54 x^{- 10}) + 15 x^{4} - 30 x^{- 5}$

$= 54 x^{5} + 15 x^{4} - 30 x^{5}$

$= 24 x^{5} + 15 x^{4}$

$= 15 x^{4} + \frac{24}{x^{5}}$

(v) Let $f (x) = x^{- 4} (3 - 4 x^{5})$

By Leibnitz product rule,

$f^{'} (x) = x^{- 4} \frac{d}{d x} (3 - 4 x^{- 5}) + (3 - 4 x^{- 5}) \frac{d}{d x} (x^{- 4})$

$= x^{- 4} {0 - 4 (- 5) x^{- 5 - 1}} + (3 - 4 x^{5}) (- 4) x^{- 4 - 1}$

$= x^{- 1} (20 x^{6}) + (3 - 4 x^{5}) (- 4 x^{- 5})$

$= 20 x^{10} - 12 x^{- 5} + 16 x^{- 0}$

$= 36 x^{- 10} - 12 x^{- 6}$

$= \frac{12}{x^{5}} + \frac{36}{x^{20}}$

(vi) Let $f (x) = \frac{2}{x + 1} - \frac{x^{2}}{3 x - 1}$

$f^{'} (x) = \frac{d}{d x} (\frac{2}{x + 1}) - \frac{d}{d x} (\frac{x^{2}}{3 x - 1})$

By quotient rule,

$f^{'} (x) = [\frac{(x + 1) \frac{d}{d x} (2) - 2 \frac{d}{d x} (x + 1)}{(x + 1)^{2}}] - [\frac{(3 x - 1) \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (3 x - 1)}{(3 x - 1)^{2}}]$

$= [\frac{(x + 1) (0) - 2 (0)}{(x + 1)^{2}}] - [\frac{(3 x - 1) (2 x) - x^{2} (3)}{(3 x - 1)^{2}}]$

$= \frac{- 2}{(x + 1)^{2}} - [\frac{6 x^{2} - 2 x - 3 x^{2}}{(3 x - 1)^{2}}]$

$= \frac{- 2}{(x + 1)^{2}} - [\frac{3 x^{2} - 2 x^{2}}{(3 x - 1)^{2}}]$

$= \frac{- 2}{(x + 1)^{2}} - \frac{x (3 x - 2)}{(3 x - 1)^{2}}$

10: Find the derivative of cos $x$ from first principle.

Ans: Let $f (x) - \cos x$ . Accordingly, from the first principle, $f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{n \to 0} [\frac{\cos (x + h) - \cos (x)}{h}]$

$= lim_{n \to 0} [\frac{\cos x \cos h - \sin x \sin h - \cos x}{h}]$

$= lim_{n \to 0} [\frac{- \cos x (1 - \cos h)}{h} \frac{- \sin x \sin h}{h}]$

$= - \cos x [lim_{n \to 0} (\frac{1 - \cos h}{h})] - \sin x [lim_{n \to 0} (\frac{\sin h}{h})]$

$= - \cos x (0) - \sin x (1) [lim_{n \to 0} \frac{1 - \cos h}{h} = 0$ and $lim_{n \to 0} \frac{\sin h}{h} = 1]$

$∴ f^{'} (x) = - \sin x$

11: Find the derivative of the following functions:

(i) $\sin x \cos x$

(ii) $\sec x$

(iii) $5 \sec x + 4 \cos x$

(iv) $cosec x$

(v) $3 \cot x + 5 cosec x$

(vi) $5 \sin x - 6 \cos x + 7$

(vii) $2 \tan x - 7 \sec x$

Ans: (i) Let $f (x) = \sin x \cos x .$ Accordingly, from the first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{n \to 0} \frac{\sin (x + h) \cos (x + h) - \sin x \cos x}{h}$

$= lim_{h \to 0} \frac{1}{2 h} [2 \sin (x + h) \cos (x + h) - 2 \sin x \cos x]$

$= lim_{n \to 0} \frac{1}{2 h} [\sin 2 (x + h) - \sin 2 x]$

$= lim_{h \to 0} \frac{1}{2 h} [2 \cos \frac{2 x + 2 h + 2 x}{2} \cdot \sin \frac{2 x + 2 h - 2 x}{2}]$

$= lim_{h \to 0} \frac{1}{2 h} [2 \cos \frac{4 x + 2 h}{2} \cdot \sin \frac{2 h}{2}]$

$= lim_{n \to 0} \frac{1}{2 h} [\cos (2 x + h) \sin h]$

$= lim_{n \to 0} \cos (2 x + h) \cdot lim_{n \to 0} \frac{\sin h}{h}$

$= \cos (2 x + h) \cdot 1$

$= \cos 2 x$

(ii) Let $f (x) = \sec x .$ Accordingly, from the first principle,

$\begin{array}{l} f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{n \to 0} \frac{\sec (x + h) - \sec x}{h} \\ = lim_{h \to 0} \frac{1}{h} [\frac{1}{\cos (x + h)} - \frac{1}{\cos x}] \\ = lim_{n \to 0} \frac{1}{h} [\frac{\cos x - \cos (x + h)}{\cos x \cos (x + h)}] \\ = \frac{1}{\cos x} \cdot lim_{n \to 0} \frac{1}{h} [\frac{- 2 \sin (\frac{x + x + h}{2}) \sin (\frac{x - x - h}{2})}{\cos (x + h)}] \\ = \frac{1}{\cos x} \cdot lim_{h \to 0} \frac{1}{h} [\frac{- 2 \sin (\frac{2 x + h}{2}) \sin (\frac{- h}{2})}{\cos (x + h)}] \\ = \frac{1}{\cos x} \cdot lim_{h \to 0} \frac{1}{2 h} \frac{- 2 \sin (\frac{2 x + h}{2}) \frac{\sin (\frac{- h}{2})}{(\frac{h}{2})}]}{\cos (x + h)} \\ = \frac{1}{\cos x} \cdot lim_{i \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \cdot lim_{i \to 0} \frac{\sin (\frac{2 x + h}{2})}{\cos (x + h)} \\ = \frac{1}{\cos x} \cdot 1 \cdot \frac{\sin x}{\cos x} \\ = \sec x \tan x \\ (iii) Let f (x) = 5 \sec x + 4 \cos x . Accordingly, from the first principle, \\ f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{n \to 0} \frac{5 \sec (x + h) + 4 \cos (x + h) - [5 \sec x + 4 \cos x]}{h} \end{array}$

$\begin{array}{l} = 5 lim_{n \to 0} \frac{[\sec (x + h) - \sec x]}{h} + 4 lim_{n \to 0} \frac{[\cos (x + h) - \cos x]}{h} \\ = 5 lim_{n \to 0} \frac{1}{h} [\frac{1}{\cos (x + h)} - \frac{1}{\cos x}] + 4 lim_{n \to 0} \frac{1}{h} [\cos (x + h) - \cos x] \\ = 5 lim_{n \to 0} \frac{1}{h} [\frac{\cos x - \cos (x + h)}{\cos x \cos (x + h)}] + 4 lim_{n \to 0} \frac{1}{h} [\cos x \cos h - \sin x \sin h - \cos x] \\ = \frac{5}{\cos x} \cdot lim_{h \to 0} \frac{1}{h} [\frac{- 2 \sin (\frac{2 x + h}{2}) \sin (\frac{- h}{2})}{\cos (x + h)}] + 4 [- \cos x lim_{h \to 0} \frac{(1 - \cos x)}{h} - \sin x lim_{h \to 0} \frac{\sin h}{h}] \\ = \frac{5}{\cos x} \cdot lim_{h \to 0} \frac{\sin (\frac{2 x + h}{2}) \frac{\sin (\frac{- h}{2})}{(\frac{h}{2})}]}{\cos (x + h)} + 4 [- \cos x (0) - \sin x (1)] \\ = \frac{5}{\cos x} \cdot [lim_{n \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \cdot lim_{n \to 0} \frac{\sin (\frac{2 x + h}{2})}{\cos (x + h)}] - 4 \sin x \\ = \frac{5}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot 1 - 4 \sin x \\ = 5 \sec x \tan x - 4 \sin x \\ (iv) Let f (x) = cosec x . Accordingly, from the first principle. \\ f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{n \to 0} \frac{1}{h} [cosec (x + h) - cosec x] \\ = lim_{n \to 0} \frac{1}{h} [\frac{1}{\sin (x + h)} - \frac{1}{\sin x}] \\ = lim_{n \to 0} \frac{1}{h} [\frac{\sin x - \sin (x + h)}{\sin x \sin (x + h)}] \end{array}$

$= lim_{h \to 0} \frac{1}{h} [\frac{2 \cos (\frac{x + x + h}{2}) \cdot \sin (\frac{x - x - h}{2})}{\sin x \sin (x + h)}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{2 \cos (\frac{2 x + h}{2}) \cdot \sin (\frac{- h}{2})}{\sin x \sin (x + h)}]$

$= lim_{h \to 0} \frac{- \cos (\frac{2 x + h}{2}) \frac{\sin (\frac{- h}{2})}{(\frac{h}{2})}]}{\sin x \sin (x + h)}$

$= lim_{h \to 0} (\frac{- \cos (\frac{2 x + h}{2})}{\sin x \sin (x + h)}) \cdot lim_{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}$

$= (\frac{- \cos x}{\sin x \sin x}) \cdot 1$

$= - cosec x \cot x$

(v) Let $f (x) = 3 \cot x + 5 cosec x$ . Accordingly, from the first principle, $f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{n \to 0} \frac{1}{h} [3 \cot (x + h) + 5 cosec (x + h) - 3 \cot x - 5 cosec x]$

$= 3 lim_{n \to 0} \frac{1}{h} [\cot (x + h) - \cot x] + 5 lim_{n \to 0} \frac{1}{h} [cosec (x + h) - cosec x]$

$\dots .$

Now, $lim_{n \to 0} \frac{1}{h} [\cot (x + h) - \cot x] = lim_{n \to 0} \frac{1}{h} [\frac{\cos (x + h)}{\sin (x + h)} - \frac{\cos x}{\sin x}]$

$= lim_{n \to 0} \frac{1}{h} [\frac{\cos (x + h) \sin x - \cos x \sin (x + h)}{\sin x \sin (x + h)}]$

$= lim_{n \to 0} \frac{1}{h} [\frac{\sin (x - x - h)}{\sin x \sin (x + h)}]$

$\begin{array}{l} = lim_{n \to 0} \frac{1}{h} [\frac{\sin (- h)}{\sin x \sin (x + h)}] \\ = lim_{h \to 0} \frac{\sin h}{h} \cdot lim_{n \to 0} [\frac{1}{\sin x \sin (x + h)}] \\ = - 1 \cdot \frac{1}{\sin x \sin (x + h)} = \frac{- 1}{\sin^{2} x} = - {cosec}^{2} x \dots . (2) \\ lim_{n \to 0} \frac{1}{h} [cosec (x + h) - cosec x] = lim_{n \to 0} \frac{1}{h} [\frac{1}{\sin (x + h)} - \frac{1}{\sin x}] \\ = lim_{n \to 0} \frac{1}{h} [\frac{\sin x - \sin (x + h)}{\sin x \sin (x + h)}] \\ = lim_{h \to 0} \frac{1}{h} [\frac{2 \cos (\frac{x + x + h}{2}) \cdot \sin (\frac{x - x - h}{2})}{\sin x \sin (x + h)}] \\ = lim_{h \to 0} \frac{1}{h} [\frac{2 \cos (\frac{2 x + h}{2}) \cdot \sin (\frac{- h}{2})}{\sin x \sin (x + h)}] \\ = lim_{h \to 0} \frac{- \cos (\frac{2 x + h}{2}) \cdot \frac{\sin (\frac{- h}{2})}{(\frac{h}{2})}}{\sin x \sin (x + h)} \\ = lim_{h \to 0} (\frac{- \cos (\frac{2 x + h}{2})}{\sin x \sin (x + h)}) \cdot lim_{n \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \\ = (\frac{- \cos x}{\sin x \sin x}) \cdot 1 \\ = - cosec x \cot x \\ From (1), (2), and (3), we obtain \\ f^{'} (x) = - 3 {cosec}^{2} x - 5 cosec x \cot x \end{array}$

(vi) Let $f (x) = 5 \sin x - 6 \cos x + 7$ . Accordingly, from the first principle, $f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{1}{h} [5 \sin (x + h) - 6 \cos (x + h) + 7 - 5 \sin x + 6 \cos x - 7]$

$= 5 lim_{n \to 0} \frac{1}{h} [\sin (x + h) - \sin x] - 6 lim_{n \to 0} \frac{1}{h} [\cos (x + h) - \cos x]$

$= 5 lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{x + h + x}{2}) \cdot \sin (\frac{x + h - x}{2})] - 6 lim_{n \to 0} \frac{\cos x \cos h - \sin x \sinh - \cos x}{h}$

$= 5 lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{2 x + h}{2}) \cdot \sin (\frac{h}{2})] - 6 lim_{n \to 0} [\frac{- \cos x (1 - \cos h) - \sin x \sin h}{h}]$

$= 5 lim_{n \to 0} \frac{1}{h} [\cos (\frac{2 x + h}{2}) \cdot \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}] - 6 lim_{n \to 0} [\frac{- \cos x (1 - \cos h)}{h} - \frac{\sin x \sin h}{h}]$

$= 5 [lim_{n \to 0} \cos (\frac{2 x + h}{2})] [lim_{h \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}] - 6 [- \cos x (lim_{n \to 0} \frac{1 - \cos h}{h}) - \sin x (lim_{n \to 0} \frac{\sin h}{h})]$

$= 5 \cos x \cdot 1 - 6 [(- \cos x) \cdot (0) - \sin x \cdot 1]$

$= 5 \cos x + 6 \sin x$

(vii) Let $f (x) = 2 \tan x - 7 \sec x .$ Accordingly, from the first principle, $f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{n \to 0} \frac{1}{h} [2 \tan (x + h) - 7 \sec (x + h) - 2 \tan x + 7 \sec x]$

$= 2 lim_{n \to 0} \frac{1}{h} [\tan (x + h) - \tan x] - 7 lim_{n \to 0} \frac{1}{h} [\sec (x + h) - \sec x]$

$= 2 lim_{h \to 0} \frac{1}{h} [\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}] - 7 lim_{n \to 0} \frac{1}{h} [\frac{1}{cosec (x + h)} - \frac{1}{cosec x}]$

$= 2 lim_{n \to 0} \frac{1}{h} [\frac{\cos x \sin (x + h) - \sin x \cos (x + h)}{\cos x \cos (x + h)}] - 7 lim_{n \to 0} \frac{1}{h} [\frac{\cos x - \cos (x + h)}{\cos x \cos (x + h)}]$

$= 2 lim_{h \to 0} \frac{1}{h} [\frac{\sin x + h - x}{\cos x \cos (x + h)}] - 7 lim_{n \to 0} \frac{1}{h} [\frac{- 2 \sin (\frac{x + x + h}{2}) \sin (\frac{x - x - h}{2})}{\cos x \cos (x + h)}]$

$= 2 [lim_{n \to 0} (\frac{\sin h}{h}) \frac{1}{\cos x \cos (x + h)}] - 7 lim_{n \to 0} \frac{1}{h} [\frac{- 2 \sin (\frac{2 x + h}{2}) \sin (\frac{- h}{2})}{\cos x \cos (x + h)}]$

$= 2 (lim_{h \to 0} \frac{\sin h}{h}) [lim_{h \to 0} \frac{1}{\cos x \cos (x + h)}] - 7 (lim_{h \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}}) (lim_{h \to 0} \frac{\sin (\frac{2 x + h}{2})}{\cos x \cos (x + h)})$

$= 2 \cdot 1 \cdot 1 \frac{1}{\cos x \cos x} - 7 \cdot 1 (\frac{\sin x}{\cos x \cos x})$

$= 2 \sec^{2} x - 7 \sec x \tan x$

Miscellaneous Exercise

1: Find the derivative of the following functions from first principle:

(i) -x

(ii) $(- x)^{- 1}$

(iii) $\sin (x + 1)$

(iv) $\cos (x - \frac{π}{8})$

Ans: (i) Let $f (x) = - x$ . Accordingly, $f (x + h) = - (x + h)$

By first principle, $f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{- (x + h) - (- x)}{h}$

$= lim_{n \to 0} \frac{- x - h + x}{h}$

$= lim_{n \to 0} \frac{- h}{h}$

$= lim_{h \to 0} (- 1) = - 1$

(ii) Let $f (x) = (- x)^{- 1} = \frac{1}{- x} = \frac{- 1}{x} .$ Accordingly, $f (x + h) = \frac{- 1}{(x + h)}$

By first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{1}{h} [\frac{- 1}{(x + h)} - (\frac{- 1}{x})]$

$= lim_{n \to 0} \frac{1}{h} [\frac{- x + (x + h)}{x (x + h)}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{h}{x (x + h)}]$

$= lim_{n \to 0} \frac{1}{x (x + h)}$

$= \frac{1}{x \cdot X} = \frac{1}{x^{2}}$

(iii) Let $f (x) = \sin (x + 1)$ . Accordingly, $f (x + h) = \sin (x + h + 1)$

By first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{n \to 0} \frac{1}{h} [\sin (x + h + 1) - \sin (x + 1)]$

$= lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{x + h + 1 + x + 1}{2}) \sin (\frac{x + h + 1 - x - 1}{2})]$

$= lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{2 x + h + 2}{2}) \sin (\frac{h}{2})]$

$= lim_{n \to 0} [\cos (\frac{2 x + h + 2}{2}) \cdot \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}]$

$\begin{array}{l} = lim_{n \to 0} \frac{1}{h} \cos (\frac{2 x + h + 2}{2}) \cdot lim_{\frac{b}{2} \to} \frac{\sin (\frac{h}{2})}{h} \frac{h}{\frac{h}{2})} [As h \to 0 \Rightarrow \frac{h}{2} \to 0] \\ = \cos (\frac{2 x + 0 + 2}{2}) \cdot 1 [lim_{h \to 0} \frac{\sin x}{x} = 1] \\ = \cos (x + 1) \\ (iv) Let f (x) = \cos (x - \frac{π}{8}) . Accordingly, f (x + h) - \cos (x + h - \frac{π}{8}) \\ By first principle, \\ f^{'} (x) = lim_{n \to 0} \frac{f (x + f) - f (x)}{h} \\ = lim_{h \to 0} \frac{1}{h} [\cos (x + h - \frac{π}{8}) - \cos (x - \frac{π}{8})] \\ = lim_{n \to 0} \frac{1}{h} [- 2 \sin \frac{(x + h - \frac{π}{8} + x - \frac{π}{8})}{2} \sin (\frac{x + h - \frac{π}{8} - x + \frac{π}{8}}{2})] \\ = lim_{n \to 0} \frac{1}{h} [- 2 \sin (\frac{2 x + h - \frac{π}{4}}{2}) \sin (\frac{h}{2})] \\ = lim_{n \to 0} [- \sin (\frac{2 x + h - \frac{π}{4}}{2}) \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}] \\ = lim_{n \to 0} [- \sin (\frac{2 x + h - \frac{π}{4}}{2})] \cdot lim_{\frac{π}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} [As h \to 0 \Rightarrow \frac{h}{2} \to 0] \\ = - \sin (\frac{2 x + 0 - \frac{π}{4}}{2}) \cdot 1 \end{array}$

$= - \sin (x - \frac{π}{8})$

2: Find the derivative of the following functions (it is to be understood that a, b, c, d. p, q, r and s are fixed non-zero constants and $m$ and $n$ are integers): $(x + a)$

Ans: Let $f (x) = x + a$ . Accordingly. $f (x + h) - x + h + a$ By first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$lim_{n \to 0} \frac{x + h + a - x - a}{h}$

$lim_{n \to 0} (\frac{h}{h})$

$- lim_{n \to 0} (1)$

$= 1$

3: Find the derivative of the following functions (it is to be understood that $a, b, c$ . d, $p, q, r$ and s are fixed non- zero constants and $m$ and $n$ are integers): $(p x + q) (\frac{r}{x} + s)$

Ans: Let $f (x) = (p x + q) (\frac{r}{x} + s)$

By Leibnitz product rule.

$f^{'} (x) = (p x + q) {(\frac{r}{x} + s)}^{'} + (\frac{r}{x} + s) (p x + q)^{'}$

$- (p x + q) {(r x^{- 1} + s)}^{'} + (\frac{r}{x} + s) (p)$

$- (p x + q) (- n x^{2}) + (\frac{r}{x} + s) p$

$= (p x + q) (\frac{- r}{x^{2}}) + (\frac{r}{x} + s) p$

$= \frac{- p x}{x} - \frac{q r}{x^{2}} + \frac{p r}{x} + p s$

$p s - \frac{q r}{x^{2}}$

4: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $(a x + b) (c x + d)^{2}$

Ans: Let $f^{'} (x) = (a x + b) (c x + d)^{2}$

By Leibnitz product rule,

$f^{'} (x) = (a x + b) \frac{d}{d x} (c x + d)^{2} \frac{d}{d x} (a x + b)$

$(a x + b) \frac{d}{d x} (c^{2} x^{2} + 2 c d x^{2}) + (c x + d)^{2} \frac{d}{d x} (a x + b)$

$(a x + b) [\frac{d}{d x} (c^{2} x^{2}) + \frac{d}{d x} (2 c d x) + \frac{d}{d x} d^{2}] + (c x + d)^{2} [\frac{d}{d x} a x + \frac{d}{d x} b]$

$= (a x + b) (2 c^{2} x + 2 c d) + (c x + d)^{2} a$

$- 2 c (a x + b) (c x + d) + a (c x + d)^{2}$

5: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non zero constants and $m$ and $n$ are integers): $\frac{a x + b}{c x + d}$

Ans: Let $f (x) = \frac{a x + b}{c x + d}$

By quotient rule,

$f (x) = \frac{(c x + d) \frac{d}{d x} (a x + b) - (a x + b) \frac{d}{d x} (c x + d)}{(c x + d)^{2}}$

$= \frac{(c x + d) (a) - (a x + d) (c)}{(c x + d)^{2}}$

$\frac{a c x + a d - a c x - b c}{(c x + d)^{2}}$

$\frac{a d - b c}{(c x + d)^{2}}$

6: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$

Ans: Let $f (x) = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}} = \frac{\frac{x + 1}{x}}{\frac{x - 1}{x}} = \frac{x + 1}{x - 1}$ , where $x \neq 0$

By quotient rule, $f^{'} (x) = \frac{(x - 1) \frac{d}{d x} (x - 1) - (x + 1) \frac{d}{d x} (x - 1)}{(x - 1)^{2}}, x \neq 0, 1$

$= \frac{(x - 1) (1) - (x + 1) (1)}{(x - 1)^{2}}, x \neq 0, 1$

$\frac{x - 1 - x - 1}{(x - 1)^{2}}, x \neq 0, 1$

$\frac{- 2}{(x - 1)^{2}}, x \neq 0, 1$

7: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers) $: \frac{1}{a x^{2} + b x + c}$

Ans: Let $f (x) = \frac{1}{a x^{2} + b x + c}$

By quotient rule,

$f^{'} (x) = \frac{(a x^{2} + b x + c) \frac{d}{d x} (1) - \frac{d}{d x} (a x^{2} + b x + c)}{{(a x^{2} + b x + c)}^{2}}$

$\frac{(a x^{2} + b x + c) (0) - (2 a x + b)}{{(a x^{2} + b x + c)}^{2}}$

$\frac{- (2 a x + b)}{{(a x^{2} + b x + c)}^{2}}$

8: Find the derivative of the following functions (it is to be understood that $a, b, c$ d, p, q, rand s are fixed non-zero constants and m and $n$ are integers): $\frac{a x + b}{p x^{2} + q x + r}$

Ans: Let $f (x) = \frac{a x + b}{p x^{2} + q x + r}$

By quotient rule,

$f^{'} (x) = \frac{(p x^{2} + q x + r) \frac{d}{d x} (a x + b) - (a x + b) \frac{d}{d x} (p x^{2} + q x + r)}{{(p x^{2} + q x + r)}^{2}}$

$= \frac{(p x^{2} + q x + r) (a) - (a x + b) (2 p x + q)}{{(p x^{2} + q x + r)}^{2}}$

$= \frac{a p x^{2} + a q x + a r - a q x + 2 n p x + b q}{{(p x^{2} + q x + r)}^{2}}$

$\frac{- a p x^{2} + 2 b p x + a r - b q}{{(p x^{2} + q x + r)}^{2}}$

9: Find the derivative of the following functions (it is to be understood that $a, b, c$ . d, $p$ , q, $r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\frac{p x^{2} + q x + r}{a x + b}$

Ans: Let $f (x) = \frac{p x^{2} + q x + r}{a x + b}$

By quotient rule,

$\dot{f} (x) = \frac{(a x + b) \frac{d}{d x} (p x^{2} + q x + η) - (p x^{2} + q x + r) \frac{d}{d x} (a x + b)}{(a x + b)^{2}}$

$\frac{(a x + b) (2 p x + q) - (p x^{2} + q x + r) (a)}{(a x + b)^{2}}$

$= \frac{2 a p x^{2} + a q x + 2 b p x + b q - a q x^{2} - a q x - a r}{(a x + b)^{2}}$

$\frac{a p x^{2} + 2 b p x + b q - a r}{(a x + b)^{2}}$

10: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p$ , q, $r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$

Ans: Let $f (x) = \frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$

$f^{'} (x) = \frac{d}{d x} (\frac{a}{x^{4}}) - \frac{d}{d x} (\frac{a}{x^{2}}) + \frac{d}{d x} (\cos x)$

$a \frac{d}{d x} (x^{- 4}) - b \frac{d}{d x} (x^{2}) + \frac{d}{d x} (\cos x)$

$- a (- 4 x^{- 5}) - b (- 2 x^{3}) + (- \sin x) [\frac{d}{d x} (x^{n}) = n x^{n - 1}$ and $\frac{d}{d x} (\cos x) = - \sin x]$

$\frac{- 4 a}{x^{5}} + \frac{2 b}{x^{3}} - \sin x$

11: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed nonzero constants and $m$ and $n$ are integers): $4 \sqrt{x} - 2$

Ans: Let $f (x) = 4 \sqrt{x} - 2$

$f^{'} (x) = \frac{d}{d x} (4 \sqrt{x} - 2) = \frac{d}{d x} (4 \sqrt{x}) - \frac{d}{d x} (2)$

$= 4 \frac{d}{d x} (x^{\frac{1}{2}}) - 0 = 4 (\frac{1}{2} x^{\frac{1}{2}})$

$= (2 x^{- \frac{1}{2}}) = \frac{2}{\sqrt{x}}$

12: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $(a x + b)^{n}$

Ans: Let $f (x) = (a x + b)^{n} .$ Accordingly, $f (x + h) - {a (x + h) + b}^{n} - (a x + a h + b)^{n}$

By first principle,

$f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{(a x + a h + b) - (a x + b)^{n}}{h}$

$= lim_{h \to 0} \frac{(a x + b)^{n} {(1 + \frac{a h}{a x + b})}^{n} - (a x + b)^{n}}{h}$

$= (a x + b)^{n} lim_{n \to 0} \frac{1}{h} [{1 + n (\frac{a h}{a x + b}) + \frac{n (n - 1)}{2} {(\frac{a h}{a x + b})}^{2} + \dots} - 1]$ (using binomial theorem)

$= (a x + b)^{n} lim_{n \to 0} \frac{1}{h} [(\frac{a h}{a x + b}) + \frac{n (n - 1) a^{2} h^{2}}{2 (a x + b)^{2}} + \dots$ (Terms cortaining higher degrees of $h)]$

$= (a x + b)^{n} lim_{n \to 0} [\frac{n a}{(a x + b)} + \frac{n (n - 1) ∄^{7} h^{2}}{2 (a x + b)^{2}} + \dots]$

$= (a x + b)^{n} [\frac{n a}{(a x + b)} + 0]$

$= n a \frac{(a x + b)^{n}}{a x + b}$

$- n a (a x + b)^{n - 1}$

13: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $(a x + b)^{n} (c x + d)^{m}$

Ans: Let $f (x) = (a x + b)^{n} (c x + d)^{m}$

By Leibnitz product rule,

$f^{'} (x) = (a x + b)^{n} \frac{d}{d x} (c x + d)^{m} + (c x + d)^{m} \frac{d}{d x} (a x + b)^{n}$

Now let $f_{1} (x) = (c x + d)^{m}$

$f_{1} (x + h) = (c x + c h + d)^{m}$

$f_{1}^{'} (x) = lim_{n \to 0} \frac{f_{1} (x + h) - f_{1} (x)}{h}$

$= lim_{n \to 0} \frac{(c x + c h + d)^{m} - (c x + d)^{m}}{h}$

$= (c x + d)^{m} lim_{n \to 0} \frac{1}{h} [{(1 + \frac{c h}{c x + d})}^{m} - 1]$

$= (c x + d)^{m} lim_{h \to 0} \frac{1}{h} [{(1 + \frac{m c h}{(c x + d)} + \frac{m (m - 1)}{2} \frac{c^{2} h^{2}}{(c x + d)^{2}} + \dots)}^{m} - 1]$

$= (c x + d)^{m} lim_{n \to 0} \frac{1}{h} [\frac{m c h}{(c x + d)} + \frac{m (m - 1) c^{2} h^{2}}{2 (c x + d)^{2}} + \dots$ (Terms containing higher degree oh $h)]$

$= (c x + d)^{m} lim_{h \to 0} [\frac{m c}{(c x + d)} + \frac{m (m - 1) c^{2} h^{2}}{2 (c x + d)^{2}} + \dots]$

$= (C x + a)^{m} [\frac{m c h}{(c x + d)} + 0]$

$= \frac{m c (c x + d)^{m}}{(c x + d)}$

$= m c (c x + d)^{m - 1}$

$\frac{d}{d x} (c x + d)^{m} = m d (x + d)^{m - 1}$

Similarly, $\frac{d}{d x} (a x + b)^{n} = n a (a x + b)^{n - 1}$

... (3)

Therefore, from (1), (2), and (3), we obtain

$f^{'} (x) = (a x + b)^{n} {m c (c x + d)^{m - 1}} + (c + d)^{m} {n a (a x + b)^{n - 1}}$

$= (a x + b)^{n - 1} (c x + d)^{m - 1} [m c (a x + b) + n a (c x + d)]$

14: Find the derivative of the following functions (it is to be understood that $a, b, c$ . d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\sin (x + a)$

Ans: Let, $f (x) = \sin (x + a)$

$f (x + h) = \sin (x + h + a)$

By first principle, $f^{'} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{n \to 0} \frac{\sin (x + h + a) - \sin (x + a)}{h}$

$= lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{x + h + a + x + a}{2}) \sin (\frac{x + h + a - x - a}{2})]$

$= lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{2 x + 2 a + h}{2}) \sin (\frac{h}{2})]$

$= lim_{h \to 0} [\cos (\frac{2 x + 2 a + h}{2}) [\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}]]$

$= lim_{h \to 0} \cos (\frac{2 x + 2 a + h}{2}) \cdot lim_{\frac{h}{2} \to 0} [\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}] [$ As $h \to 0 \Rightarrow \frac{h}{2} \to 0]$

$= \cos (\frac{2 x + 2 a}{2}) \times 1 [lim_{n \to 0} \frac{\sin x}{x} = 1]$

$= \cos (x + a)$

15: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $cosec x \cot x$

Ans: Let $f (x) = cosec x \cot x$

By Leibnitz product rule,

$f^{'} (x) = cosec x (\cot x)^{'} + \cot x (cosec x)^{'} \dots . (1)$

Let $f_{1} (x) = \cot x .$ Accordingly, $f_{1} (x + h) = \cot (x + h)$

By first principle,

$f^{''} (x) = lim_{n \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{\cot (x + h) - \cot (x)}{h}$

$= lim_{h \to 0} \frac{1}{h} (\frac{\cos (x + h)}{\sin (x + h)} - \frac{\cos (x)}{\sin x})$

$= lim_{n \to 0} \frac{1}{h} (\frac{\sin x \cos (x + h) - \cos x \sin (x + h)}{\sin x \sin (x + h)})$

$= lim_{h \to 0} \frac{1}{h} (\frac{\sin (x - x + h)}{\sin x \sin (x + h)})$

$= \frac{1}{\sin x^{n \to 0}} lim_{h} \frac{1}{h} [\frac{\sin (- h)}{\sin (x + h)}]$

$= \frac{- 1}{\sin x} (lim_{n \to 0} \frac{\sin h}{h}) (lim_{n \to 0} \frac{1}{\sin (x + h)})$

$= \frac{- 1}{\sin x} \cdot 1 \cdot (lim_{n \to 0} \frac{1}{\sin (x + 0)})$

$= \frac{- 1}{\sin^{2} x}$

$= - {cosec}^{2} x$

$∴ (\cot x)^{'} = - {cosec}^{2} x \dots$ (2)

Now, let $f_{2} (x) = cosec x .$ Accordingly, $f_{2} (x + h) = cosec (x + h)$

By first principle, $f_{2} (x) = lim_{n \to 0} \frac{f_{2} (x + h) - f_{2} (x)}{h}$

$= lim_{h \to 0} \frac{1}{h} [cosec (x + h) - cosec (x)]$

$= lim_{n \to 0} \frac{1}{h} (\frac{1}{\sin (x + h)} - \frac{1}{\sin x})$

$= lim_{n \to 0} \frac{1}{h} (\frac{\sin x - \sin (x + h)}{\sin x \sin (x + h)})$

$= \frac{1}{\sin x} \cdot lim_{h \to 0} \frac{1}{h} [\frac{2 \cos (\frac{x + x + h}{2}) \sin (\frac{x - x - h}{2})}{\sin (x + h)}]$

$= \frac{1}{\sin x} \cdot lim_{n \to 0} \frac{1}{h} [\frac{2 \cos (\frac{2 x + h}{2}) \sin (\frac{- h}{2})}{\sin (x + h)}]$

$= \frac{1}{\sin x} \cdot lim_{h \to 0} [\frac{- \sin (\frac{h}{2})}{(\frac{h}{2})} \cdot \frac{\cos (\frac{2 x + h}{2})}{\sin (x + h)}]$

$= \frac{- 1}{\sin x} \cdot lim_{h \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \cdot lim_{n \to 0} \frac{\cos (\frac{2 x + h}{2})}{\sin (x + h)}$

$= \frac{- 1}{\sin x} \cdot 1 \cdot \frac{\cos (\frac{2 x + h}{2})}{\sin (x + 0)}$

$= \frac{- 1}{\sin x} \cdot \frac{\cos x}{\sin x}$

$= - cosec x \cdot \cot x$

$∴ (cosec x)^{'} = - cosec x \cdot \cot x$

From (1), (2), and (3), we obtain

$f^{'} (x) = cosec x (- {cosec}^{2} x) + \cot x (- cosec x \cot x)$

$= - {cosec}^{3} x - \cot^{2} x cosec x$

16: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{\cos x}{1 + \sin x}$

Ans: Let $f (x) = \frac{\cos x}{1 + \sin x}$

By quotient rule,

$f^{'} (x) = \frac{(1 + \sin x) \frac{d}{d x} (\cos x) - (\cos x) \frac{d}{d x} (1 + \sin x)}{(1 + \sin x)^{2}}$

$= \frac{(1 + \sin x) (- \sin x) - (\cos x) (\cos x)}{(1 + \sin x)^{2}}$

$= \frac{- \sin x - \sin^{2} x - \cos^{2} x}{(1 + \sin x)^{2}}$

$= \frac{- \sin x - (\sin^{2} x + \cos^{2} x)}{(1 + \sin x)^{2}}$

$= \frac{- \sin x - 1}{(1 + \sin x)^{2}}$

$= \frac{- (1 - \sin x)}{(1 + \sin x)^{2}}$

$= \frac{- 1}{(1 + \sin x)^{2}}$

17: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non zero constants and m and n are integers): $\frac{\sin x + \cos x}{\sin x - \cos x}$

Ans:17: Let $f (x) = \frac{\sin x + \cos x}{\sin x - \cos x}$

By quotient rule,

$f^{''} (x) = \frac{(\sin x - \cos x) \frac{d}{d x} (\sin x + \cos x) - (\sin x + \cos x) \frac{d}{d x} (\sin x - \cos x)}{(\sin x + \cos x)^{2}}$

$= \frac{(\sin x - \cos x) (\cos x - \sin x) - (\sin x + \cos x) (\cos x + \sin x)}{(\sin x + \cos x)^{2}}$

$= \frac{- (\sin x - \cos x)^{2} - (\sin x + \cos x)^{2}}{(\sin x + \cos x)^{2}}$

$= \frac{- [\sin^{2} x + \cos^{2} x - 2 \sin x \cos x + \sin^{2} x + \cos^{2} x + 2 \sin x \cos x]}{(\sin x + \cos x)^{2}}$

$= \frac{- [1 + 1]}{(\sin x - \cos x)^{2}}$

$= \frac{- 2}{(\sin x - \cos x)^{2}}$

18: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{\sec x - 1}{\sec x + 1}$

Ans: Let $f (x) = \frac{\sec x - 1}{\sec x + 1}$

$f (x) = \frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1} = \frac{1 - \cos x}{1 + \cos x}$

By quotient rule,

$f^{'} (x) = \frac{(1 + \cos x) \frac{d}{d x} (1 - \cos x) - (1 - \cos x) \frac{d}{d x} (1 + \cos x)}{(1 + \cos x)^{2}}$

$= \frac{(1 + \cos x) (\sin x) - (1 - \cos x) (- \sin x)}{(1 + \cos x)^{2}}$

$= \frac{\sin x + \cos x \sin x + \sin x - \sin x \cos x}{(1 + \cos x)^{2}}$

$= \frac{2 \sin x}{(1 + \cos x)^{2}}$

$= \frac{2 \sin x}{{(1 + \frac{1}{\sec x})}^{2}} = \frac{2 \sin x}{\frac{(\sec x + 1)^{2}}{\sec^{2} x}}$

$= \frac{2 \sin x \sec^{2} x}{(\sec x + 1)^{2}}$

$= \frac{\frac{2 \sin x}{\cos x} \sec x}{(\sec x + 1)^{2}}$

$= \frac{2 \sec x \tan x}{(\sec x + 1)^{2}}$

19: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d. $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\sin^{n} x$

Ans: Let $y = \sin^{n} x$

Accordingly, for $n = 1, y = \sin x$

$∴ \frac{d y}{d x} = \cos x$ , i.e., $\frac{d}{d x} \sin x = \cos x$

For $n = 2, y = \sin^{2} x$ .

$∴ \frac{d y}{d x} = \frac{d}{d x} (\sin x \sin x)$

$= (\sin x)^{'} (\sin x + \sin x (\sin x)^{'}$ (By Leibnitz product rule)

$= \cos x \sin x + \sin x \cos x$

$= 2 \sin x \cos x$

$\dots . (1)$

For $n = 3, y = \sin^{3} x$

$∴ \frac{d y}{d x} = \frac{d}{d x} (\sin x \sin^{2} x)$

$= (\sin x)^{'} \sin^{2} x + \sin x (\sin x)^{'}$

(By Leibnitz product rule)

$- \cos x \sin^{2} x + \sin x (2 \sin x \cos x) [$ Using $(1)]$

$= \cos x \sin^{2} x + \sin^{2} x \cos x$

$= 3 \sin^{2} x \cos x$

We assert that $\frac{d}{d x} (\sin^{n} x) = n \sin^{(n - 1)} x \cos x$

Let our assertion be true for $n = k$ .

i.e., $\frac{d}{d x} (\sin^{k} x) = k \sin^{(k - 1)} x \cos x \dots .$ (2)

Corsider

$\frac{d}{d x} (\sin^{k + 1} x) = \frac{d}{d x} (\sin x \sin^{(k)} x)$

$= (\sin x)^{'} \sin^{k} x + \sin x {(\sin^{k} x)}^{n}$

(By Leibnitz product rule)

$= \cos x \sin^{k} x + \sin x (k \sin^{k - 1} \cos x) [$ Using $(2)]$

$= \cos x \sin^{k} x + 2 \sin^{k} x \cos x$

$- (k + 1) \sin^{k} x \cos x$

Thus, our assertion is true for $n = k + 1$ .

Hence, by mathematical induction, $\frac{d}{d x} (\sin^{n} x) = n \sin^{(n - 1)} x \cos x$

20: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and s are fixed non-zero constants and $m$ and n are integers): $\frac{a + b \sin x}{c + d \cos x}$

Ans: Let $f (x) = \frac{a + b \sin x}{c + d \cos x}$

By quotient rule,

$f^{'} (x) = \frac{(c + d \cos x) \frac{d}{d x} (a + b \sin x) - (a + b \sin x) \frac{d}{d x} (c + d \cos x)}{(c + d \cos x)^{2}}$

$= \frac{(c + d \cos x) (b \cos x) - (a + b \sin x) (- d \sin x)}{(c + d \cos x)^{2}}$

$= \frac{c b \cos x + b d \cos^{2} x + a d \sin x + b d \sin^{2} x}{(c + d \cos x)^{2}}$

$= \frac{b c \cos x + a d \sin x + b d (\cos^{2} x + \sin^{2} x)}{(C + d \cos x)^{2}}$

$= \frac{b c \cos x + a d \sin x + b d}{(c + d \cos x)^{2}}$

21: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and s are fixed non-zero constants and m and $n$ are integers): $\frac{\sin (x + a)}{\cos x}$

Ans: Let $f (x) = \frac{\sin (x + a)}{\cos x}$

By quotient rule,

$f^{'} (x) = \frac{\cos x \frac{d}{d x} [\sin (x + a)] - \sin (x + a) \frac{d}{d x} \cos x}{\cos^{2} x}$

$f^{'} (x) = \frac{\cos x \frac{d}{d x} [\sin (x + a)] - \sin (x + a) \frac{d}{d x} (- \sin x)}{\cos^{2} x}$

Let $g (x) - \sin (x + a) .$ Accordingly, $g (x + h) = \sin (x + h + a)$

By first principle,

$g^{'} (x) = lim_{h \to 0} \frac{g (x + h) - g (x)}{h}$

$= lim_{n \to 0} \frac{1}{h} [\sin (x + h + a) - \sin (x + a)]$

$= lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{x + h + a + x + a}{2}) \sin (\frac{x + h + a - x - a}{2})]$

$= lim_{n \to 0} \frac{1}{h} [2 \cos (\frac{2 x + 2 a + h}{2}) \sin (\frac{h}{2})]$

$= lim_{n \to 0} [\cos (\frac{2 x + 2 a + h}{h}) {\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}}]$

$= lim_{n \to 0} \cos (\frac{2 x + 2 a + h}{h}) \cdot lim_{n \to 0} {\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}} [$ As $h \to 0 \Rightarrow \frac{h}{2} \to 0]$

$= (\cos \frac{2 x + 2 a}{2}) \times 1 [lim_{n \to 0} \frac{\sin h}{h} = 1]$

$= \cos (x + a) \dots$ (ii)

From (i) and (ii), we obtain $f^{'} (x) = \frac{\cos x \cos (x + a) + \sin x \sin (x + a)}{\cos^{2} x}$

$= \frac{\cos (x + a - x)}{\cos^{2} x}$

$= \frac{\cos a}{\cos^{2} x}$

22: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) $: x^{4} (5 \sin x - 3 \cos x)$

Ans: Let $f (x) = x^{4} (5 \sin x - 3 \cos x)$

Byproduct rule.

$f^{'} (x) = x^{4} \frac{d}{d x} (5 \sin x - 3 \cos x) + (5 \sin x - 3 \cos x) \frac{d}{d x} (x^{4})$

$= x^{4} [5 \frac{d}{d x} (\sin x) - 3 \frac{d}{d x} (\cos x)] + (5 \sin x - 3 \cos x) \frac{d}{d x} (x^{4})$

$= x^{4} [5 \cos x - 3 (- \sin x)] + (5 \sin x - 3 \cos x) (4 x^{3})$

$= x^{3} [5 x \cos x + 3 x \sin x + 20 \sin x - 12 \cos x]$

23: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, p, q, r and s are fixed non-zero constants and m and n are integers): $(x^{2} + 1) \cos x$

Ans: Let $f (x) = (x^{2} + 1) \cos x$

By product rule.

$f^{'} (x) = (x^{2} + 1) \frac{d}{d x} (\cos x) + \cos x \frac{d}{d x} (x^{2} + 1)$

$= (x^{2} + 1) (- \sin x) + \cos x (2 x)$

$= - x^{2} \sin x - \sin x + 2 x \cos x$

24: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $(a x^{2} + \sin x) (p + q) \cos x)$

Ans: Let $f (x) = (a x^{2} + \sin x) (p + q \cos x)$

By product rule.

$f^{'} (x) = (a x^{2} + \sin x) \frac{d}{d x} (p + q \cos x) + (p + q \cos x) \frac{d}{d x} (a x^{2} + \sin x)$

$= (a x^{2} + \sin x) (- q \sin x) + (p + q \cos x) (2 a x + \cos x)$

$= - q \sin x (a x^{2} + \sin x) + (p + q \cos x) (2 a x + \cos x)$

25: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $(x + \cos x) (x - \tan x)$

Ans: Let $f (x) = (x + \cos x) (x - \tan x)$

By product rule,

$f^{'} (x) = (x + \cos x) \frac{d}{d x} (x - \tan x) + (x - \tan x) \frac{d}{d x} (x + \cos x)$

$= (x + \cos x) [\frac{d}{d x} (x) - \frac{d}{d x} (\tan x)] + (x - \tan x) (1 - \sin x)$

$= (x + \cos x) [1 - \frac{d}{d x} (\tan x)] + (x - \tan x) (1 - \sin x)$

Let $g (x) = \tan x .$ Accordingly, $g (x + h) = \tan (x + h)$

By first principle,

$g^{''} (x) = lim_{n \to 0} \frac{g (x + h) - g (x)}{h}$

$= lim_{h \to 0} \frac{\tan (x + h) - \tan (x)}{h}$

$= lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}]$

$= lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h) \cos x - \sin x \cos (x + h)}{\cos x \cos (x + h)}]$

$= \frac{1}{\cos x^{n \to 0}} lim_{h} \frac{1}{h} [\frac{\sin (x + h - x)}{\cos (x + h)}]$

$= \frac{1}{\cos x^{h \to 0}} lim_{h} \frac{1}{h} [\frac{\sin h}{\cos (x + h)}]$

$= \frac{1}{\cos x} (lim_{n \to 0} \frac{\sin h}{h}) (lim_{n \to 0} \frac{1}{\cos (x + h)})$

$= \frac{1}{\cos x} \cdot \cdot (\frac{1}{\cos (x + 0)})$

$= \frac{1}{\cos^{2} x}$

$= \sec^{2} x$ ... (ii)

Therefore, from (i) and (ii). We obtain

$f^{'} (x) = (x + \cos x) (1 - \sec^{2} x) + (x - \tan x) (1 - \sin x)$

$= (x + \cos x) (- \tan^{2} x) + (x - \tan x) (1 - \sin x)$

$= - \tan^{2} x (x + \cos x) + (x - \tan x) (1 - \sin x)$

26: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and s are fixed non-zero constants and m and $n$ are integers): $\frac{4 x + 5 \sin x}{3 x + 7 \cos x}$

Ans: Let $f (x) = \frac{4 x + 5 \sin x}{3 x + 7 \cos x}$

Quotient rule,

$f^{'} (x) = \frac{(3 x + 7 \cos x) \frac{d}{d x} (4 x + 5 \sin x) - (4 x + 5 \sin x) \frac{d}{d x} (3 x + 7 \cos x)}{(3 x + 7 \cos x)^{2}}$

$= \frac{(3 x + 7 \cos x) [4 \frac{d}{d x} (x) + 5 \frac{d}{d x} (\sin x)] - (4 x + 5 \sin x) [3 \frac{d}{d x} (x) + 7 \frac{d}{d x} (\cos x)]}{(3 x + 7 \cos x)^{2}}$

$= \frac{(3 x + 7 \cos x) [4 x + 5 \cos x] - (4 x + 5 \sin x) [3 - 7 \sin x]}{(3 x + 7 \cos x)^{2}}$

$= \frac{12 x + 15 x \cos x + 28 x \cos x + 35 \cos^{2} x - 12 x + 28 x \sin x - 15 \sin x + 35 (\cos^{2} x + \sin^{2} x)}{(3 x + 7 \cos x)^{2}}$

$= \frac{15 x \cos x + 28 \cos x + 28 x \sin x - 15 \sin x + 35 (\cos^{2} x + \sin^{2} x)}{(3 x + 7 \cos x)^{2}}$

$= \frac{35 + 15 x \cos x + 28 \cos x + 28 x \sin x - 15 \sin x}{(3 x + 7 \cos x)^{2}}$

27: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{x^{2} \cos (\frac{π}{4})}{\sin x}$

Ans: Let $f (x) = \frac{x^{2} \cos (\frac{π}{4})}{\sin x}$

By quotient rule, $f^{'} (x) = \cos (\frac{π}{4}) [\frac{\sin x \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (\sin x)}{\sin^{2} x}]$

$= \cos (\frac{π}{4}) [\frac{\sin x (2 x) - x^{2} (\cos x)}{\sin^{2} x}]$

$= \frac{x \cos \frac{π}{4} [2 \sin x - x \cos x]}{\sin^{2} x}$

28: Find the derivative of the following functions (it is to be understood that $a, b, c$ . d, p, q, r and s are fixed non-zero constants and $m$ and $n$ are integers): $\frac{x}{1 + \tan x}$

Ans: Let $f (x) = \frac{x}{1 + \tan x}$

$f (x) = \frac{(1 + \tan x) \frac{d}{d x} (x) - (x) \frac{d}{d x} (1 + \tan x)}{(1 + \tan x)^{2}}$

$f^{'} (x) = \frac{(1 + \tan x) - x \frac{d}{d x} (1 + \tan x)}{(1 + \tan x)^{2}}$

Let $g (x) = 1 + \tan x .$ Accordingly $g (x + h) = 1 + \tan (x + h)$ .

By first principle, $\dot{g} (x) = lim_{n \to 0} \frac{g (x + h) - g (x)}{h}$

$lim_{h \to 0} [\frac{1 + \tan (x + h) - 1 - \tan (x)}{h}]$

$lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}]$

$lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h) \cos x - \sin x \cos x (x + h)}{\cos (x + h) \cos x}]$

$lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h - x)}{\cos (x + h) \cos x}]$

$lim_{h \to 0} \frac{1}{h} [\frac{\sinh}{\cos (x + h) \cos x}]$

$- (lim_{n \to 0} \frac{\sinh}{h}) \cdot (lim_{n \to 0} \frac{1}{\cos (x + h) \cos x})$

$- 1 \times \frac{1}{\cos^{2}} = \sec^{2} x$

$\Rightarrow \frac{d}{d x} (1 + \tan^{2} x) = \sec^{2} x$

From (i) and (ii), we obtain

$\dot{f} (x) = \frac{1 + \tan x - x \sec^{2} x}{(1 + \tan x)^{2}}$

29: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, p, q, r and s are fixed non-zero constants and m and n are integers): $(x + \sec x) (x - \tan x)$

Ans: Let $f (x) = (x + \sec x) (x - \tan x)$

By product rule.

$f (x) = (x + \sec x) \frac{d}{d x} (x - \tan x) + (x - \tan x) \frac{d}{d x} (x + \sec x)$

$- (x + \sec x) [\frac{d}{d x} (x) - \frac{d}{d x} \tan x] + (x - \tan x) [\frac{d}{d x} (x) - \frac{d}{d x} \sec x]$

$- f (x + \sec x) [1 - \frac{d}{d x} \tan x)] + (x - \tan x) [1 + \frac{d}{d x} \sec x]$

$\dots (i)$

Let $f_{1} (x) = \tan x, f_{2} (x) = \sec x$

Accordingly, $f_{1} (x + h) \cdot \tan (x + h)$ and $f_{2} (x + h) - \sec (x + h)$

$f_{1}^{'} (x) = lim_{n \to 0} (\frac{f_{1} (x + h) - f_{1} (x)}{h})$

$= lim_{h \to 0} [\frac{\tan (x + h) - \tan (x)}{h}]$

$\begin{array}{l} - lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h)}{\cos (x + h)} - \frac{\sin x}{\cos x}] \\ - lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h) \cos x - \sin x \cos x (x + h)}{\cos (x + h) \cos x}] \\ - lim_{n \to 0} \frac{1}{h} [\frac{\sin (x + h - x)}{\cos (x + h) \cos x}] \\ - lim_{n \to 0} \frac{1}{h} [\frac{\sinh}{\cos (x + h) \cos x}] \\ - (lim_{n \to 0} \frac{\sinh}{h}) \cdot (lim_{n \to 0} \frac{1}{\cos (x + h) \cos x}) \\ - 1 \times \frac{1}{\cos^{2}} = \sec^{2} x \\ \Rightarrow \frac{d}{d x} (1 + \tan^{2} x) = \sec^{2} x \\ f_{2}^{'} (x) = lim_{n \to 0} (\frac{f_{2} + (x + h) - f_{2} (x)}{h}) \\ = lim_{h \to 0} (\frac{\sec (x + h) - \sec (x)}{h}) \\ = lim_{n \to 0} \frac{1}{h} (\frac{1}{\cos (x + h)} - \frac{1}{\cos x}) \\ = lim_{n \to 0} \frac{1}{h} (\frac{\cos x - \cos (x + h)}{\cos (x + h) \cos x}) \\ = \frac{1}{\cos x^{' \to 0}} lim_{h} \frac{1}{h} [\frac{- 2 \sin (\frac{x + x + h}{2}) \cdot \sin (\frac{x - x - h}{2})}{\cos (x + h)}] \\ = \frac{1}{\cos x} lim_{h \to 0} \frac{1}{h} [\frac{- 2 \sin (\frac{2 x + h}{2}) \cdot \sin (\frac{- h}{2})}{\cos (x + h)}] \end{array}$

$= \frac{1}{\cos x} lim_{h \to 0} \frac{1}{h} [\frac{\sin (\frac{2 x + h}{2}) {\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}}}{\cos (x + h)}]$

$= \sec x \frac{{lim_{n \to 0} \sin (\frac{2 x + h}{2})} {lim_{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}}}{lim_{n \to 0} \cos (x + h)}$

$= \sec x \cdot \frac{\sin x \cdot 1}{\cos x}$

$\Rightarrow \frac{d}{d x} \sec x = \sec x \tan x$

From (i). (ii), and (iii), we obtain

$f^{'} (x) = (x + \sec x) (1 - \sec^{2} x) + (x - \tan x) (1 + \sec x \tan x)$

30: Find the derivative of the following functions (it is to be understood that $a, b, c$ , d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\frac{x}{\sin^{n} x}$

Ans: Let $f (x) = \frac{x}{\sin^{n} x}$

By quotient rule, $f^{'} (x) = \frac{\sin^{n} x \frac{d}{d x} x - x \frac{d}{d x} \sin^{n} x}{\sin^{2 n} x}$

It can be easily shown that $\frac{d}{d x} \sin^{n} x = n \sin^{n - 1} x \cos x$

Therefore,

$f^{'} (x) = \frac{\sin^{n} \times \frac{d}{d x} x - x \frac{d}{d x} \sin^{n} x}{\sin^{2 n} x}$

$= \frac{\sin^{n} x \cdot 1 - x ({nin}^{n - 1} x \cos x)}{\sin^{2 n} x}$

$= \frac{\sin^{n - 1} x (\sin x - n x \cos x)}{\sin^{2 n} x}$

$= \frac{\sin x - n x \cos x}{\sin^{n + 1} x}$

Class 11 Maths Chapter 12: Exercises Breakdown

Exercise	Number of Questions
Exercise 12.1	32 Questions and solutions
Exercise 12.2	11 Questions and solutions
Miscellaneous Exercise	30 Questions and Solutions

Conclusion

The Class 11 Limits and Derivatives Solutions provided by Vedantu, is a valuable tool for 11th-grade students. It helps introduce mathematical concepts in an accessible manner. The provided solutions and explanations simplify complex ideas, making it easier for 11th-grade students to understand the material. By using Vedantu's resources, students can develop a deeper understanding of NCERT concepts. Class 11 Maths Ch Limits And Derivatives are a helpful aid for 11th-grade students, empowering them to excel in their studies and develop a genuine appreciation for Limits And Derivatives. In previous years' exams, this chapter typically features around 5-7 questions.

Other Study Material for CBSE Class 11 Maths Chapter 12

S. No	Important Links for Chapter 12 Limits And Derivatives
1	Class 11 Limits and Derivatives Important Questions
2	Class 11 Limits and Derivatives Revision Notes
3	Class 11 Limits and Derivatives Important Formulas
4	Class 11 Limits and Derivatives NCERT Exemplar Solution

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

S. No	NCERT Solutions Class 11 Maths All Chapters
1	Chapter 1 - Sets Solutions
2	Chapter 2 - Relations and Functions Solutions
3	Chapter 3 - Trigonometric Functions Solutions
4	Chapter 4 - Complex Numbers and Quadratic Equations Solutions
5	Chapter 5 - Linear Inequalities Solutions
6	Chapter 6 - Permutations and Combinations Solutions
7	Chapter 7 - Binomial Theorem Solutions
8	Chapter 8 - Sequences and Series Solutions
9	Chapter 9 - Straight Lines Solutions
10	Chapter 10 - Conic Sections Solutions
11	Chapter 11 - Introduction to Three Dimensional Geometry Solutions
12	Chapter 12 - Limits and Derivatives Solutions
13	Chapter 13 - Statistics Solutions
14	Chapter 14 - Probability Solutions

Important Related Links for CBSE Class 11 Maths

S.No.	Important Study Material for Maths Class 11
1	CBSE Class 11 Maths NCERT Books
2	CBSE Class 11 Maths Revision Notes
3	CBSE Class 11 Maths Important Questions
4	CBSE Class 11 Maths NCERT Solutions
5	CBSE Class 11 Maths NCERT Exemplar Solution
6	CBSE Class 11 Maths RS Aggarwal Solutions
7	CBSE Class 11 Maths RD Sharma Solutions
8	CBSE Class 11 Maths Formulas
9	CBSE Class 11 Maths Syllabus
10	CBSE Class 11 Maths Sample Papers

FAQs on NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

1. Where can I find the stepwise and CBSE-approved NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives?

You can access the latest NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives in a stepwise, CBSE-approved format on trusted educational platforms such as Vedantu. These solutions follow the official NCERT answer pattern and cover all exercise and miscellaneous questions as per the 2025–26 CBSE syllabus.

2. How to solve Exercise 12.1 of NCERT Class 11 Maths Chapter 12 using the official NCERT answer format?

To solve Exercise 12.1 of NCERT Class 11 Maths Chapter 12, begin by identifying the limit or derivative asked in each question. Apply NCERT-prescribed theorems and formulas stepwise, ensuring each calculation and justification is clearly shown, just like the official NCERT answer key. Always show substitution or limit-taking steps as outlined by the CBSE textbook.

3. Which method should be used to solve limits questions in Exercise 12.2 as per NCERT Solutions for Class 11 Maths?

For solving limits questions in Exercise 12.2, always follow the NCERT stepwise approach: first, directly substitute the variable; if indeterminate, simplify the expression using algebraic techniques like factorization or rationalization. If needed, apply standard limit formulas and always adhere to the format shown in the NCERT textbook for correct marks. Each solution must justify the method used.

4. Can I download the NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives as a PDF?

Yes, you can download the complete NCERT Solutions for Class 11 Maths Chapter 12 as a PDF from Vedantu and other CBSE-aligned platforms. The PDF contains clear, chapter-wise solved answers to all exercises and miscellaneous questions, formatted as per the latest NCERT/CBSE guidelines for 2025–26.

5. Are the NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives available in Hindi medium?

Yes, for Hindi medium students, NCERT Solutions for Class 11 Maths Chapter 12 (सीमाएँ और अवकलज) are available in Hindi. These answers follow the official NCERT Hindi textbook pattern and use correct mathematical terminology and methods as per the CBSE format.

6. What stepwise method is recommended for the Miscellaneous Exercise of Chapter 12 Limits and Derivatives?

For the Miscellaneous Exercise, start by carefully reading each question to determine whether it relates to limits or derivatives. Write each step, including substitutions, simplifications, and theorems used, exactly as shown in the NCERT textbook. Justify each step and highlight which formula or property is being applied. Use the CBSE solution structure for all answers.

7. How do I ensure my answers for Chapter 12 Limits and Derivatives are correct according to the NCERT Solutions?

To ensure correctness, follow every calculation and theorem application stepwise, aligning your working with the NCERT solutions. Double-check the final answer with the NCERT answer key and ensure all justifications and explanation steps match the official CBSE format for full marks.

8. Why is the concept of limits important in Class 11 Maths Chapter 12, and how is it applied in NCERT exercises?

The concept of limits is crucial as it forms the foundation of calculus, including derivatives and continuity. In NCERT exercises, limit is applied to evaluate the value of expressions as variables approach specific points. Every solution must show the limit-taking procedure as per NCERT pattern to earn full marks in exams.

9. Do the NCERT Solutions for Limits and Derivatives include reasons and justifications for each answer as required in CBSE board exams?

Yes, CBSE-approved NCERT solutions for Limits and Derivatives always include reasons, justifications, and explanation for each step, especially in questions involving limit operations and differentiation. This ensures your answer is comprehensive, as per the CBSE answer sheet format for 2025–26 exams.

10. What do I do if my answers to derivatives questions do not match with the NCERT Solutions for Chapter 12?

If your answers differ from the official NCERT Solutions, review your calculation steps, ensure you have used the correct rules of differentiation, and check if the question was solved using the precise CBSE method. Refer back to the NCERT stepwise answer, verify each justification, and understand the logic before attempting again.

11. Are NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives sufficient for board exam preparations?

Yes, NCERT Solutions for Class 11 Maths Chapter 12 provide thorough practice for CBSE board exams as they cover all types of questions from the textbook, follow latest CBSE marking scheme, and use stepwise solutions with proper reasoning, making them fully sufficient for strong board exam preparation.

12. How should I attempt application-based or tricky questions in Chapter 12 using the NCERT method?

For application-based or tricky questions, begin by clearly writing the problem, then outline all possible solution pathways using NCERT formulae and properties. Carefully follow the stepwise answer structure, explicitly state any limit or derivative property used, and clearly show all intermediate steps as required in the NCERT answer format.