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Continuity and Differentiability Class 12 Notes CBSE Maths Chapter 5 (Free PDF Download)

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Revision Notes for CBSE Class 12 Maths Chapter 5 (Continuity and Differentiability) - Free PDF Download

Class 12 students often face pressure as board exams approach. As the exam gets closer, it is not the subject matter that is most important, but the final revision notes that can be a great help. Revision is not just about referring to the notes you made in class, but it should also include all the important topics of the chapter.


Vedantu's Continuity and Differentiability notes can help you attempt questions based on continuity and differentiation with ease, whether they are theoretical or numerical. Students often do not prepare revision notes while going through the chapter, and as a result, they tend to miss out on various important topics when they revise. To make your preparation easier, we at Vedantu have created Continuity and Differentiability Class 12 notes that include some practical guidelines.


Therefore, it is advisable to refer to Continuity and Differentiability notes, as they include all the important concepts and formulas that will help you solve all the numerical questions given in Class 12 Chapter 5 easily.


List of Topics Covered in Class 12 Maths Chapter 5

Take a look at the following table to know the topics and subtopics that students of CBSE Class 12 will study and revise while covering Chapter 5.  Elaborated discussion on each of these topics is provided to help students grasp them better and prepare to score high marks.


  • Introduction

  • Continuity

    • Algebra of Continuous Functions

  • Differentiability

    • Derivatives of Composite Functions

    • Derivatives of Implicit Functions

    • Derivatives of Inverse Trigonometric Functions

    • Second Order Derivatives

  • Logarithmic Differentiation


Here are some of the key concepts that are introduced in this chapter:


  • The concept of continuity and its different properties

  • The concept of differentiability and its different properties

  • The derivatives of composite functions, implicit functions, and inverse trigonometric functions

  • The use of logarithmic differentiation to solve differential equations


This chapter is an important foundation for many other topics in calculus, such as applications of derivatives, integral calculus, and differential equations. It is therefore important to have a good understanding of the concepts and techniques introduced in this chapter.


CBSE Class 12 Maths Revision Notes 2023-24 - Chapter Wise PDF Notes for Free

In the table below we have provided the PDF links of all the chapters of CBSE Class 12 Maths whereby the students are required to revise the chapters by downloading the PDF. 


Continuity and Differentiability Related Important Links

It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.


Class 12 Study Materials Important Links

Find a curated selection of study resources for Class 12 subjects, helping students prepare effectively and excel in their academic pursuits.

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Continuity and Differentiability Class 12 Notes Maths - Basic Subjective Questions


Section–A (1 Mark Questions)

1. Find the derivative of $log(tan\;x)$ w.r.t. X.

Ans. $\dfrac{d}{dx}\left \{ log\left ( tan\;x \right ) \right \}=\dfrac{1}{tan\;x}\cdot \dfrac{d}{dx}\left ( tan\;x \right )=\dfrac{sec^{2}x}{tan\;x}$ 


2. If $f(x)=|\cos x|$, then find $f{}'(\dfrac{\pi }{4})$.

Ans. For $0<x<\dfrac{\pi }{2},\;cos\;x>0$

$\Rightarrow f{}'(x)=-sin\;x$

$\Rightarrow f{}'\left ( \dfrac{\pi }{4} \right )=-sin\dfrac{\pi }{4}=-\dfrac{1}{\sqrt{2}}$


3. If $f(x)=\cos x-\sin x$, then

Ans. We have $f(x)=\cos x-\sin x$

$\Rightarrow f{}'(x)=-sin\;x-cos\;x$

$\Rightarrow f{}'\left ( \dfrac{\pi }{3} \right )=-sin(\dfrac{\pi }{3})-cos(\dfrac{\pi }{3})$

$=-\frac{\sqrt{3}}{2}-\dfrac{1}{2}$


4. For the function $\sqrt{x}+\sqrt{y}=1$, find $\dfrac{d y}{d x}$ at $\left(\frac{1}{4}, \frac{1}{4}\right)$

Ans. We have, $\sqrt{x}+\sqrt{y}=1$

Differentiating w.r.t $x$, we get

$\dfrac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$

$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{\sqrt{x}}$

$\left ( \dfrac{dy}{dx} \right )_{\left ( \frac{1}{4},\frac{1}{4} \right )}=-\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}}=-1$


5. Discuss the continuity of the function f (x) = sin x. cos x.

Ans. Since sin x and cos x are continuous functions and product of two continuous function is continuous function. Therefore, f(x) = sin x. cos x is a continuous function.


Section–B (2 Marks Questions)

6. Find the derivative of $\cos ^{-1}\left(2 x^2-1\right)$ w.r.t. $X$.

Ans. Let $\mathrm{u}=\cos ^{-1}\left(2 x^2-1\right)$

$\begin{aligned}& \text { Differentiating both sides w. r. t. } x, \\& \begin{aligned}\frac{d u}{d x} & =-\frac{1}{\sqrt{1-\left(2 x^2-1\right)^2}} \cdot \frac{d}{d x}\left(2 x^2-1\right) \\& =-\frac{4 x}{\sqrt{1-4 x^4-1+4 x^2}}=-\frac{2}{\sqrt{1-x^2}} \end{aligned}\end{aligned}$


7. Find the points of discontinuity of the function $f(x)=\frac{4-x^2}{4 x-x^3}$.

Ans. We have, $f(x)=\frac{4-x^2}{4 x-x^3}$,

which is discontinuous where

$$4 x-x^3=0$$or $x\left(4-x^2\right)=0$or $x=0, x= \pm 2$

So, $f(x)$ is discontinuous at exactly three points.


8. If $y=\log \left(\frac{1-x^2}{1+x^2}\right)$, then find $\frac{d y}{d x}$

Ans. We have,

$$y=\log \left(\frac{1-x^2}{1+x^2}\right)$$

$$\begin{aligned}& =\log \left(1-x^2\right)-\log \left(1+x^2\right) \\& \therefore \frac{d y}{d x}=\frac{-2 x}{1-x^2}-\frac{2 x}{1+x^2} \\& =-2x\left(\frac{2}{\left(1-x^2\right)\left(1+x^2\right)}\right) \\& =-\frac{4 x}{1-x^4}\end{aligned}$$


9. Let $f(x+y)=f(x)+f(y) \forall x, y \in R$. Suppose that $f^{\prime}(0)=1$, then find $f^{\prime}(6)$.

Ans. $$\begin{aligned}& f^{\prime}(6)=\lim _{h \rightarrow 0} \frac{f(6+h)-f(6)}{h} \\= & \lim _{h \rightarrow 0} \frac{f(6+h)-f(6+0)}{h} \\= & \lim ^h \frac{f(6)+f(h)-\{f(6)+f(0)\}}{h} \\= & \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\= & f^{\prime}(0)=1\end{aligned}$$


10. The value of c in Rolle's theorem for the function f (x) = x3 - 3x in the interval [0, 3] is continuous is ........

Ans. We have f (x) = x3 - 3x, which is a polynomial function.

          So, it is continuous and differentiable for all real numbers.

          Also, f (0) = f (3) = 0

So, there exists at least one $c\;\epsilon\;(0,3)$ such that f '(c) = 0

$\Rightarrow$ 3c2 - 3 = 0

$\Rightarrow$ c2 = 1

$\Rightarrow$ c = 1 $\epsilon$ (0, 3)

$\therefore$ c = 1


11. Given the function $f(x)=\frac{1}{x+2}$. Find the points of discontinuity of the composite function $y=f(f(x))$.

Ans. We have, $f(x)=\frac{1}{x+2}$

Clearly $f(x)$ is discontinuous at

$$x=-2$$

Now,

$$f(f(x))=f\left(\frac{1}{x+2}\right)$$

$$\begin{array}{l}=\frac{1}{\frac{1}{x+2}+2} \\=\frac{x+2}{2 x+5}\end{array}$$

Clearly $f(f(x))$ is discontinuous at

$$\begin{array}{c}2 x+5=0 \\x=-\frac{5}{2}\end{array}$$


12. Find $\dfrac{dy}{dx}$  if $in\;x=\frac{2t}{1+t^{2}}$ and $tan;y=\frac{2t}{1-t^{2}}$ .

Ans. We have $in\;x=\dfrac{2t}{1+t^{2}}$ and $tan;y=\dfrac{2t}{1-t^{2}}$

Let t = tan z

$\therefore sin\;x=\dfrac{2\;tan\;z}{1+tan^{2}\;z}=sin2z$

$\therefore$ x = 2z

Also, 

$tan\;y=\dfrac{2\;tan\;z}{1-tan^{2}\;z}=tan\;2\;z$

$\therefore$ y = 2z

$\Rightarrow$ y = x

$\therefore$ $\dfrac{dy}{dx}=1$

13. Let f(x) = |sin x|. Then show that f is everywhere continuous but not differentiable at $x=n\pi ,n\epsilon Z$ .

Ans.  We have, f(x) = |sin x|

We know that |x| and sin x are continuous for all real x.

So, |sin x| is also continuous for all real x. |x| is non-differentiable at x = 0

So, |sin x| is non-differentiable when sin x = 0 or x = n n𝞹  , n $\epsilon$Z

Hence, f(x) is continuous everywhere but not differentiable at 

           $x=n\pi ,n\epsilon Z$.


PDF Summary - Class 12 Maths Continuity and Differentiability Notes (Chapter 5)

Continuity

1. Definition:

A function $f({\text{x}})$ is said to be continuous at ${\text{x}} = {\text{a}}$; where ${\text{a}} \in $ domain of $f({\text{x}})$, if

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$

i.e., ${\text{LHL}} = {\text{RHL}} = $ value of a function at ${\text{x}} = {\text{a}}$

or $\quad \mathop {\lim }\limits_{x \to a} f(x) = f(a)$


1.1 Reasons of discontinuity

If $f({\text{x}})$ is not continuous at ${\text{x}} = {\text{a}}$, we say that $f({\text{x}})$ is discontinuous at $x = a$

There are following possibilities of discontinuity:

1. $\mathop {\lim }\limits_{x \to {a^ - }} f(x)$ and $\mathop {\lim }\limits_{x \to {a^ + }} f(x)$ exist but they are not equal.

2. $\mathop {\lim }\limits_{x \to {a^ - }} f(x)$ and $\mathop {\lim }\limits_{x \to {a^ + }} f(x)$ exists and are equal but not equal to $f(a)$

3. $f(a)$ is not defined.

4. At least one of the limits does not exist. The graph of the function will show a break at the location of discontinuity from a geometric standpoint.


(“Image will be updated soon”)


The graph as shown is discontinuous at $x = 1,2$ and 3 .


2. Properties of Continuous Functions 

Let $f({\text{x}})$ and $g({\text{x}})$ be continuous functions at ${\text{x}} = {\text{a}}$. Then,

1. $\quad {\text{c}}f({\text{x}})$ is continuous at ${\text{x}} = {\text{a}}$, where ${\text{c}}$ is any constant.

2. $f({\text{x}}) \pm g({\text{x}})$ is continuous at ${\text{x}} = {\text{a}}$.

3. $f(x) \cdot g(x)$ is continuous at $x = a$.

4. $\quad f({\text{x}})/{\text{g}}({\text{x}})$ is continuous at ${\text{x}} = {\text{a}}$, provided $g({\text{a}}) \ne 0$.-

5. Assuming $f({\text{x}})$ be continuous on $[{\text{a}},{\text{b}}]$ in such a way that the function $f({\text{a}})$ and $f(\;{\text{b}})$ will be at opposite signs, then there will exists at least one solution of equation $f(x) = 0$ in the open interval $(a,b)$


3. The Intermediate Value Theorem 

Suppose $f({\text{x}})$ is continuous on an interval I, also a and b are any two points of $I$. Then if ${y_0}$ is a number between $f(a)$ and $f(b)$, their exits a number $c$ between $a$ and $b$ such that $f(c) = {y_0}$

The Function $f$, being continuous on $(a,b)$ takes on every value between $f(a)$ and $f(b)$

Note:

That a function $f$ which is continuous in [a, b] possesses the following properties:

(i) If $f(a)$ and $f(b)$ possess opposite signs, then there exists at least one solution of the equation $f(x) = 0$ in the open interval $(a,b)$

(ii) If $K$ is any real number between $f(a)$ and $f(b)$, then there exists at least one solution of the equation $f$ $({\text{x}}) = {\text{K}}$ in the open interval $({\text{a}},{\text{b}})$


4. Continuity In An Interval 

(a) A function $f$ is said to be continuous in (a, b) if $f$ is continuous at each and every point $ \in ({\text{a}},{\text{b}})$

(b) A function $f$ is said to be continuous in a closed interval $[{\text{a}},{\text{b}}]$ if :

(1) $f$ is continuous in the open interval $(a,b)$ and

(2) $f$ is right continuous at 'a' i.e. ${\operatorname{Limit} _{x \to {a^ + }}}$ $f({\text{x}}) = f({\text{a}}) = {\text{a}}$ finite quantity

(3) $f$ is left continuous at 'b'; i.e. $\mathop {{\text{Limit }}}\limits_{x \to {b^ - }} $ $f(x) = f(b) = a$ finite quantity


5. A List of Continuous Functions 


Function $f (x)$

Interval in which f(x) is continuous

1

Constant (c)

$( - \infty ,\infty )$

2

${{{\text{x}}^{\text{n}}},{\text{n is an integer }}}$

$( - \infty ,\infty )$

3

${{{\text{x}}^{ - {\text{n}}}},{\text{n is a positive integer }}}$

${( - \infty ,\infty ) - \{ 0\} } $

4

${|{\text{x}} - {\text{a}}|}$

$( - \infty ,\infty )$

5

${{\text{P}}({\text{x}}) = {{\text{a}}_0}{{\text{x}}^{\text{n}}} + {{\text{a}}_1}{{\text{x}}^{n - 1}} +  \ldots .. + {{\text{a}}_{\text{n}}}}$

$( - \infty ,\infty )$

6

$\sin x$

$( - \infty ,\infty )$

7

$\cos x$

$( - \infty ,\infty )$

8

$\tan x$

${( - \infty ,\infty ) - \left\{ {(2n + 1)\dfrac{\pi }{2}:n \in I} \right\}} $

9

$\cot x$

${( - \infty ,\infty ) - \{ n\pi :n \in I} $

10

$\sec x$

${( - \infty ,\infty ) - \{ (2n + 1)}$

11

${\operatorname{cosec} x}$

${\pi /2:n \in I}$

12

${{{\text{e}}^x}}$

${( - \infty ,\infty ) - \{ n\pi :n \in I\} }$

13

${{{\log }_c}x}$

$( - \infty ,\infty )$

&$(0,\infty )$


6. Types Of Discontinuities 

Type-1 : (Removable type of discontinuities) 

In this case, $\mathop {{\text{Limit}}}\limits_{x \to c} f({\text{x}})$ exists but it will not equal to $f({\text{c}})$ . As a result, the function is said to have a removable discontinuity or discontinuity of the first kind. In such scenario, we can redefine the function such that $\mathop {\operatorname{Limit} }\limits_{x \to c} f(x) = f(c)$ and make it continuous at ${\text{x}} = {\text{c}}$. It can be further categorised as:

(a) Missing Point Discontinuity:

Where $\mathop {Limit}\limits_{x \to a} f(x)$ exists finitely but $f(a)$ is not defined.

E.g. $f(x) = \dfrac{{(1 - x)\left( {9 - {x^2}} \right)}}{{(1 - x)}}$ will have a missing point discontinuity at $x = 1$, and

$f({\text{x}}) = \dfrac{{\sin {\text{x}}}}{{\text{x}}}$ will have a missing point discontinuity at ${\text{x}} = 0$

(Image will be updated soon)


(b) Isolated Point Discontinuity :

Where $\mathop {\operatorname{Limit} }\limits_{x \to a} f(x)$ exists $f$ (a) also exists but;

${\operatorname{Limit} _{{\text{x}} \to {\text{a}}}} \ne f({\text{a}})$

E.g. $f(x) = \dfrac{{{x^2} - 16}}{{x - 4}},x \ne 4$ and $f(4) = 9$ will have an isolated point discontinuity at $x = 4$

In the same way \[f(x) = [x] + [ - x] = \left[ {\begin{array}{*{20}{c}} 0&{{\text{ if }}x \in I} \\ { - 1}&{{\text{ if }}x \notin I} \end{array}} \right] will have an isolated point discontinuity at all x \in I\].

will have an isolated point discontinuity at all x ∈ I.

(Image will be updated soon)


Type-2 : (Non-Removable type of discontinuities)

In case, $\mathop {\operatorname{Limit} }\limits_{x \to a} f(x)$ does not exist, then it is not possible to make the function continuous by redefining it. Such discontinuities are known as non-removable discontinuity or discontinuity of the 2nd kind. Non-removable type of discontinuity can be further classified as:


(a) Finite Discontinuity:

E.g., $f(x) = x - [x]$ at all integral $x;f(x) = {\tan ^{ - 1}}\dfrac{1}{x}$ at $x = 0$ and $f({\text{x}}) = \dfrac{1}{{1 + {2^{\dfrac{1}{{\text{x}}}}}}}$ at ${\text{x}} = 0$ (note that $\left. {f\left( {{0^ + }} \right) = 0;f\left( {{0^ - }} \right) = 1} \right)$ $1 + {2^ - }$


(b) Infinite Discontinuity:

${\text{ E}}{\text{.g}}{\text{., }}f(x) = \dfrac{1}{{x - 4}}{\text{ or }}g(x) = \dfrac{1}{{{{(x - 4)}^2}}}{\text{ at }}x = 4;f(x) = {2^{\tan x}}$

at $x = \dfrac{\pi }{2}$ and $f(x) = \dfrac{{\cos x}}{x}$ at $x = 0$


(c) Oscillatory Discontinuity:

${\text{ E}}{\text{.g}}{\text{., }}f(x) = \sin \dfrac{1}{x}{\text{ at }}x = 0$

In all these cases the value of $f(a)$ of the function at $x = a$ (point of discontinuity) may or may not exist but $\mathop {{\text{ Limit does }}}\limits_{x \to a} $ not exist.


(“Image will be updated soon”)


From the adjacent graph note that

$  f $ is continuous at $x =  - 1$

$ f $ has isolated discontinuity at ${\text{x}} = 1$

$ f $ has missing point discontinuity at $x = 2$

$ f $ has non-removable (finite type) discontinuity at the origin.


Note:

(a) In case of dis-continuity of the second kind the nonnegative difference between the value of the RHL at ${\text{x}} = {\text{a}}$ and ${\text{LHL}}$ at ${\text{x}} = {\text{a}}$ is called the jump of discontinuity. A function having a finite number of jumps in a given interval I is called a piece wise continuous or sectionally continuous function in this interval.

(b) All Polynomials, Trigonometrical functions, exponential and Logarithmic functions are continuous in their domains.

(c) If $f({\text{x}})$ is continuous and $g({\text{x}})$ is discontinuous at ${\text{x}} = {\text{a}}$ then the product function $\phi (x) = f(x) \cdot g(x)$ is not necessarily be discontinuous at $x = a$. e.g.

\[f(x) = x{\text{ and }}g(x) = \left[ {\begin{array}{*{20}{c}} {\sin \dfrac{\pi }{x}}&{x \ne 0} \\ 0&{x = 0} \end{array}} \right.\]

(d) If $f({\text{x}})$ and $g({\text{x}})$ both are discontinuous at ${\text{x}} = {\text{a}}$ then the product function $\phi (x) = f(x) \cdot g(x)$ is not necessarily be discontinuous at ${\text{x}} = {\text{a}}.$ e.g

\[f(x) =  - g(x) = \left[ {\begin{array}{*{20}{c}} 1&{x \geqslant 0} \\ { - 1}&{x < 0} \end{array}} \right.\]

(e) Point functions are to be treated as discontinuous 

eg.$f(x) = \sqrt {1 - x}  + \sqrt {x - 1} $ is not continuous at $x = 1$

(f) A continuous function whose domain is closed must have a range also in the closed interval.

(g) If $f$ is continuous at $x = a$ and $g$ is continuous at ${\text{x}} = f$ (a) then the composite $g[f({\text{x}})]$ is continous at ${\text{x}} = {\text{a}}$ 

E.g $f({\text{x}}) = \dfrac{{{\text{x}}\sin {\text{x}}}}{{{{\text{x}}^2} + 2}}$ and $g({\text{x}}) = |{\text{x}}|$ are continuous at ${\text{x}}$ $ = 0$, hence the composite$(gof)({\text{x}}) = \left| {\dfrac{{{\text{x}}\sin {\text{x}}}}{{{{\text{x}}^2} + 2}}} \right|$ will also be continuous at ${\text{x}} = 0$.


Differentiability

1. Definition 

Let $f({\text{x}})$ be a real valued function defined on an open interval $(a,b)$ where $c \in (a,b)$. Then $f(x)$ is said to be differentiable or derivable at $x = c$

if, $\mathop {\lim }\limits_{{\text{x}} \to {\text{c}}} \dfrac{{f({\text{x}}) - f({\text{c}})}}{{({\text{x}} - {\text{c}})}}$ exists finitely.

This limit is called the derivative or differentiable coefficient of the function $f(x)$ at $x = c$, and is denoted by ${f^\prime }({\text{c}})$ or $\dfrac{{\text{d}}}{{{\text{dx}}}}{(f({\text{x}}))_{{\text{x}} = {\text{c}}}}$


(“Image will be updated soon”)


- Slope of Right hand secant $ = \dfrac{{f({\text{a}} + {\text{h}}) - f({\text{a}})}}{{\text{h}}}$ as ${\text{h}} \to 0,{\text{P}} \to {\text{A}}$ and secant $({\text{AP}}) \to $ tangent at ${\text{A}}$

$ \Rightarrow \quad {\text{ Right hand derivative }} = {\operatorname{Lim} _{{\text{h}} \to 0}}\left( {\dfrac{{f({\text{a}} + {\text{h}}) - f({\text{a}})}}{{\text{h}}}} \right)$

$ = \quad $ Slope of tangent at ${\text{A}}$ (when approached from right) ${f^\prime }\left( {{{\text{a}}^ + }} \right)$

- Slope of Left hand secant $ = \dfrac{{f({\text{a}} - {\text{h}}) - f({\text{a}})}}{{ - {\text{h}}}}$ as h $ \to 0,{\text{Q}} \to {\text{A}}$ and secant ${\text{AQ}} \to $ tangent at ${\text{A}}$

$ \Rightarrow \quad {\text{ Left hand derivative }} = {\operatorname{Lim} _{h \to 0}}\left( {\dfrac{{f(a - h) - f(a)}}{{ - h}}} \right)$

= Slope of tangent at ${\text{A}}$ (when approached from left) ${f^\prime }\left( {{{\text{a}}^ - }} \right)$

Thus, $f({\text{x}})$ is differentiable at ${\text{x}} = {\text{c}}$.

\[\begin{array}{*{20}{l}} { \Leftrightarrow \quad \mathop {\lim }\limits_{ \to c} \dfrac{{f() - f({\text{c}})}}{{( - {\text{c}})}}{\text{ exists finitely }}} \\ { \Leftrightarrow \quad \mathop {\lim }\limits_{ \to {{\text{c}}^ - }} \dfrac{{f() - f({\text{c}})}}{{( - {\text{c}})}} = \mathop {\lim }\limits_{ \to {{\text{c}}^ + }} \dfrac{{f() - f({\text{c}})}}{{( - {\text{c}})}}} \\ { \Leftrightarrow \quad \mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{f({\text{c}} - {\text{h}}) - f({\text{c}})}}{{ - {\text{h}}}} = \mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{f({\text{c}} + {\text{h}}) - f({\text{c}})}}{{\text{h}}}} \end{array}\]

Hence, $\quad \mathop {\lim }\limits_{x \to {c^ - }} \dfrac{{f({\mathbf{x}}) - f({\mathbf{c}})}}{{({\mathbf{x}} - {\mathbf{c}})}} = \mathop {\lim }\limits_{{\mathbf{h}} \to 0} \dfrac{{f({\mathbf{c}} - {\mathbf{h}}) - {\mathbf{f}}({\mathbf{c}})}}{{ - {\mathbf{h}}}}$ is called the left hand derivative of $f(x)$ at $x = c$ and is denoted by ${f^\prime }\left( {{{\text{c}}^ - }} \right)$or ${\text{L}}{f^\prime }({\text{c}})$ While, $\mathop {\lim }\limits_{{\mathbf{x}} \to {{\text{c}}^ + }} \dfrac{{{\mathbf{f}}({\mathbf{x}}) - {\mathbf{f}}({\mathbf{c}})}}{{{\mathbf{x}} - {\mathbf{c}}}} = \mathop {\lim }\limits_{{\mathbf{h}} \to 0} \dfrac{{{\mathbf{f}}({\mathbf{c}} + {\mathbf{h}}) - {\mathbf{f}}({\mathbf{c}})}}{{\mathbf{h}}}$ is called the right hand derivative of $f(x)$ at $x = c$ and is denoted by ${f^\prime }\left( {{{\text{c}}^ + }} \right)$or ${\text{R}}{f^\prime }({\text{c}})$

If ${f^\prime }\left( {{{\text{c}}^ - }} \right) \ne {f^\prime }\left( {{{\text{c}}^ + }} \right)$, we say that $f({\text{x}})$ is not differentiable at $x = c$.


2. Differentiability in a Set 

1. A function $f(x)$ defined on an open interval $(a,b)$ is said to be differentiable or derivable in open interval $(a,b)$, if it is differentiable at each point of $(a,b)$

2. A function $f(x)$ defined on closed interval [a, b] is said to be differentiable or derivable. "If ${\text{f}}$ is derivable in the open interval (a, b) and also the end points a and b, then $f$ is said to be derivable in the closed interval [a, b]"

i.e., $\mathop {\lim }\limits_{ \to {a^ + }} \dfrac{{f() - f(a)}}{{ - a}}$ and $\mathop {\lim }\limits_{ \to {b^ - }} \dfrac{{f() - f(b)}}{{ - b}}$, both exist.

A function $f$ is said to be a differentiable function if it is differentiable at every point of its domain.


Note:

1. If $f(x)$ and $g(x)$ are derivable at $x = $ a then the functions $f(x) + g(x),f(x) - g(x),f(x) \cdot g(x)$ will also be derivable at $x = a$ and if $g(a) \ne 0$ then the function $f({\text{x}})/{\text{g}}({\text{x}})$ will also be derivable at $x = a$

2. If $f(x)$ is differentiable at $x = a$ and $g(x)$ is not differentiable at ${\text{x}} = {\text{a}}$, then the product function ${\text{F}}({\text{x}}) = f({\text{x}}) \cdot g({\text{x}})$ can still be differentiable at ${\text{x}} = {\text{a}}.$ E.g. $f({\text{x}}) = {\text{x}}$ and $g({\text{x}}) = |{\text{x}}|$

3. If $f(x)$ and $g$ (x) both are not differentiable at ${\text{x}} = {\text{a}}$ then the product function; $F({\text{x}}) = f({\text{x}}) \cdot g({\text{x}})$ can still be differentiable at $x = $ a. E.g. $f({\text{x}}) = |{\text{x}}|$ and ${\text{g}}({\text{x}}) = |{\text{x}}|$

4. If $f(x)$ and $g(x)$ both are not differentiable at ${\text{x}} = {\text{a}}$ then the sum function $F({\text{x}}) = f({\text{x}}) + g({\text{x}})$ may be a differentiable function. E.g., $f(x) = |x|$ and $g({\text{x}}) =  - |{\text{x}}|$

5. If $f(x)$ is derivable at $x = a$

$ \Rightarrow {f^\prime }(x)$ is continuous at $x = a$.

e.g. 

\[f(x) = \left[ {\begin{array}{*{20}{c}} 2&{{\text{ if }} \ne 0} \\ 0&{{\text{ if }} = 0} \end{array}} \right.\]


3. Relation Between Continuity and Differentiability

We learned in the last section that if a function is differentiable at a point, it must also be continuous at that point, and therefore a discontinuous function cannot be differentiable. The following theorem establishes this fact.

Theorem: If a function is differentiable at a given point, it must be continuous at that same point. However, the inverse is not always true.

or $\quad f(x)$ is differentiable at $x = c$

$ \Rightarrow \quad f({\text{x}})$ is continuous at ${\text{x}} = {\text{c}}$

Converse: The reverse of the preceding theorem is not always true, i.e., a function might be continuous but not differentiable at a given point.

E.g., The function $f(x) = |x|$ is continuous at $x = 0$ but it is not differentiable at ${\text{x}} = 0$.


Note:

(a) Let ${f^{\prime  + }}(a) = p;\,{f^{\prime  - }}(a) = q$ where $p{\text{ }}q$ are finite then

$ \Rightarrow f$ is derivable at $x = a$

$ \Rightarrow f$ is continuous at $x = a$

(ii) ${\text{p}} \ne {\text{q}}\quad  \Rightarrow f$ is not derivable at ${\text{x}} = {\text{a}}$.

It is very important to note that f may be still continuous at $x = a$

In short, for a function f:

Differentiable $ \Rightarrow $ Continuous;

Not Differentiable $ \ne $ Not Continuous

(i.e., function may be continuous)

But,

Not Continuous $ \Rightarrow $ Not Differentiable.

(b) If a function ${\text{f}}$ is not differentiable but is continuous at ${\mathbf{x}} = $ a it geometrically implies a sharp corner at ${\mathbf{x}} = {\mathbf{a}}$

Theorem 2: Let $f$ and $g$ be real functions such that fog is defined if $g$ is continuous at $x = a$ and $f$ is continuous at $g$.


Differentiation:

1. Definition 

(a) Let us consider a function ${\text{y}} = f({\text{x}})$ defined in a certain interval. It has a definite value for each value of the independent variable $x$ in this interval.

Now, the ratio of the function's increment to the independent variable's increment,

$\dfrac{{\Delta y}}{{\Delta x}} = \dfrac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$

Now, as $\Delta {\text{x}} \to 0,\Delta {\text{y}} \to 0$ and $\dfrac{{\Delta {\text{y}}}}{{\Delta {\text{x}}}} \to $ finite quantity, then derivative $f(x)$ exists and is denoted by ${y^\prime }$ or ${f^\prime }(x)$ or $\dfrac{{dy}}{{dx}}$ Thus, ${f^\prime }(x) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\Delta y}}{{\Delta x}}} \right) = \mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f(x + \Delta x) - f(x)}}{{\Delta x}}$ (if it exits) for the limit to exist,

$\mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{f({\text{x}} + {\text{h}}) - f({\text{x}})}}{{\text{h}}} = \mathop {\lim }\limits_{{\text{h}} \to 0} \dfrac{{f({\text{x}} - {\text{h}}) - f({\text{x}})}}{{ - {\text{h}}}}$

(Right Hand derivative)    (Left Hand derivative)

(b) The derivative of a given function ${\text{f}}$ at a point ${\text{x}} = {\text{a}}$ of its domain is defined as:

$\mathop {\operatorname{Limit} }\limits_{h \to 0} \dfrac{{f({\text{a}} + {\text{h}}) - f({\text{a}})}}{{\text{h}}}$, provided the limit exists is denoted by ${f^\prime }({\text{a}})$

Note that alternatively, we can define

${f^\prime }({\text{a}}) = {\operatorname{Limit} _{{\text{x}} \to {\text{a}}}}\dfrac{{f({\text{x}}) - f({\text{a}})}}{{{\text{x}} - {\text{a}}}}$, provided the limit exists.

This method is called first principle of finding the derivative of $f(x)$


2. Derivative of Standard Function 

(i) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^{\text{n}}}} \right) = {\text{n}} \cdot {{\text{x}}^{{\text{n}} - 1}};{\text{x}} \in {\text{R}},{\text{n}} \in {\text{R}},{\text{x}} > 0$

(ii) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{e}}^{\text{x}}}} \right) = {{\text{e}}^{\text{x}}}$

(iii) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{a}}^{\text{x}}}} \right) = {{\text{a}}^{\text{x}}} \cdot \ln {\text{a}}({\text{a}} > 0)$

(iv) $\dfrac{{\text{d}}}{{{\text{dx}}}}(\ln |{\text{x}}|) = \dfrac{1}{{\text{x}}}$

(v) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\log }_{\text{a}}}|{\text{x}}|} \right) = \dfrac{1}{{\text{x}}}{\log _{\text{a}}}{\text{e}}$

(vi) $\dfrac{{\text{d}}}{{{\text{dx}}}}(\sin {\text{x}}) = \cos {\text{x}}$

(vii) $\dfrac{{\text{d}}}{{{\text{dx}}}}(\cos {\text{x}}) =  - \sin {\text{x}}$

(viii) $\dfrac{{\text{d}}}{{{\text{dx}}}}(\tan {\text{x}}) = {\sec ^2}{\text{x}}$

(ix) $\dfrac{{\text{d}}}{{{\text{dx}}}}(\sec {\text{x}}) = \sec {\text{x}} \cdot \tan {\text{x}}$

(x) $\dfrac{{\text{d}}}{{{\text{dx}}}}(\operatorname{cosec} x) =  - \operatorname{cosec} x \cdot \cot x$

(xi) $\dfrac{d}{{dx}}(\cot x) =  - {\operatorname{cosec} ^2}x$

(xii) $\dfrac{{\text{d}}}{{{\text{dx}}}}($ constant $) = 0$

(xiii) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\sin }^{ - 1}}{\text{x}}} \right) = \dfrac{1}{{\sqrt {1 - {{\text{x}}^2}} }},\quad  - 1 < {\text{x}} < 1$

(xiv) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\cos }^{ - 1}}{\text{x}}} \right) = \dfrac{{ - 1}}{{\sqrt {1 - {{\text{x}}^2}} }},\quad  - 1 < {\text{x}} < 1$

(xv) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\tan }^{ - 1}}{\text{x}}} \right) = \dfrac{1}{{1 + {{\text{x}}^2}}},\quad {\text{x}} \in {\text{R}}$

(xvi) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\cot }^{ - 1}}{\text{x}}} \right) = \dfrac{{ - 1}}{{1 + {{\text{x}}^2}}},\quad {\text{x}} \in {\text{R}}$

(xvii) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\sec }^{ - 1}}{\text{x}}} \right) = \dfrac{1}{{|{\text{x}}|\sqrt {{{\text{x}}^2} - 1} }},\quad |{\text{x}}| > 1$

(xviii) $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\operatorname{cosec} }^{ - 1}}{\text{x}}} \right) = \dfrac{{ - 1}}{{|{\text{x}}|\sqrt {{{\text{x}}^2} - 1} }},\quad |{\text{x}}| > 1$

(xix) Results:

If the inverse functions $f(g)$ are defined by $y = f(x);x = g(y).$ Then

$g(f(x)) = x$ $ \Rightarrow \quad {g^\prime }(f(x)) \cdot {f^\prime }(x) = 1$

This result can also be written as, if $\dfrac{{dy}}{{dx}}$ exists and $\dfrac{{dy}}{{dx}} \ne 0$, then $\dfrac{{{\text{dx}}}}{{{\text{dy}}}} = 1/\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right)$ or $\dfrac{{{\text{dy}}}}{{{\text{dx}}}} \cdot \dfrac{{{\text{dx}}}}{{{\text{dy}}}} = 1$ or $\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = 1/\left( {\dfrac{{{\text{dx}}}}{{{\text{dy}}}}} \right)\left[ {\dfrac{{{\text{dx}}}}{{{\text{dy}}}} \ne 0} \right]$


3. Theorems On Derivatives 

If $u$ and $v$ are derivable functions of $x$, then,

(i) $\quad $ Term by term differentiation : $\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{u}} \pm {\text{v}}) = \dfrac{{{\text{du}}}}{{{\text{dx}}}} \pm \dfrac{{{\text{dv}}}}{{{\text{dx}}}}$

(ii) Multiplication by a constant $\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{Ku}}) = {\text{K}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}$, where ${\text{K}}$ is any constant

(iii) "Product Rule" $\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{u}}.{\text{v}}) = {\text{u}}\dfrac{{{\text{dv}}}}{{{\text{dx}}}} + {\text{v}}\dfrac{{{\text{du}}}}{{{\text{dx}}}}$ known as In general,

(a) If ${u_1},{u_2},{u_3},{u_4}, \ldots ,{u_n}$ are the functions of $x$, then

\[\begin{array}{*{20}{l}} {\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{u}}_1} \cdot {{\text{u}}_2} \cdot {{\text{u}}_3} \cdot {{\text{u}}_4} \ldots .{{\text{u}}_{\text{n}}}} \right)} \\ { = \left( {\dfrac{{{\text{d}}{{\text{u}}_1}}}{{{\text{dx}}}}} \right)\left( {{{\text{u}}_2}{{\text{u}}_3}{{\text{u}}_4} \ldots {{\text{u}}_{\text{n}}}} \right) + \left( {\dfrac{{{\text{d}}{{\text{u}}_2}}}{{{\text{dx}}}}} \right)\left( {{{\text{u}}_1}{{\text{u}}_3}{{\text{u}}_4} \ldots {{\text{u}}_{\text{n}}}} \right)} \end{array}\] \[\begin{array}{*{20}{l}} { + \left( {\dfrac{{d{u_3}}}{{dx}}} \right)\left( {{u_1}{u_2}{u_4} \ldots {u_n}} \right) + \left( {\dfrac{{d{u_4}}}{{dx}}} \right)\left( {{u_1}{u_2}{u_3}{u_5} \ldots {u_n}} \right)} \\ { +  \ldots  + \left( {\dfrac{{d{u_n}}}{{dx}}} \right)\left( {{u_1}{u_2}{u_3} \ldots {u_{n - 1}}} \right)} \end{array}\]

(iv) Quotient Rule

$\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{\text{u}}}{{\text{v}}}} \right) = \dfrac{{{\text{v}}\left( {\dfrac{{{\text{du}}}}{{{\text{dx}}}}} \right) - {\text{u}}\left( {\dfrac{{{\text{dv}}}}{{{\text{dx}}}}} \right)}}{{{{\text{v}}^2}}}$ where ${\text{v}} \ne 0$ known as

(b) Chain Rule : If $y = f({\text{u}}),{\text{u}} = g({\text{w}}),{\text{w}} = h({\text{x}})$ then $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dw}} \cdot \dfrac{{dw}}{{dx}}$

${\text{ or }}\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = {f^\prime }({\text{u}}) \cdot {g^\prime }(\quad ) \cdot {h^\prime }({\text{x}})$


Note:

In general if ${\text{y}} = f({\text{u}})$ then $\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = {f^\prime }({\text{u}}) \cdot \dfrac{{{\text{du}}}}{{{\text{dx}}}}$


4. Methods of Differentiation 

4.1 Derivative by using Trigonometrical Substitution

The use of trigonometrical transforms before differentiation greatly reduces the amount of labour required. The following are some of the most significant findings:

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

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

(iii) $\tan 2{\text{x}} = \dfrac{{2\tan {\text{x}}}}{{1 - {{\tan }^2}{\text{x}}}},{\tan ^2}{\text{x}} = \dfrac{{1 - \cos 2{\text{x}}}}{{1 + \cos 2{\text{x}}}}$

(iv) $\sin 3x = 3\sin x - 4{\sin ^3}x$

(v) $\quad \cos 3x = 4{\cos ^3}x - 3\cos x$

(vi) $\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}$

(vii) $\tan \left( {\dfrac{\pi }{4} + x} \right) = \dfrac{{1 + \tan x}}{{1 - \tan x}}$

(viii) $\tan \left( {\dfrac{\pi }{4} - {\text{x}}} \right) = \dfrac{{1 - \tan {\text{x}}}}{{1 + \tan {\text{x}}}}$

(ix) $\quad \sqrt {(1 \pm \sin x)}  = \left| {\cos \dfrac{x}{2} \pm \sin \dfrac{x}{2}} \right|$

(x) $\quad {\tan ^{ - 1}}x \pm {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x \pm y}}{{1 \mp xy}}} \right)$

(xi) $\quad {\sin ^{ - 1}}x \pm {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left\{ {x\sqrt {1 - {y^2}}  \pm y\sqrt {1 - {x^2}} } \right\}$

(xii) ${\cos ^{ - 1}}x \pm {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left\{ {xy \mp \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right\}$

(xiii) ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x = {\sec ^{ - 1}}x + {\operatorname{cosec} ^{ - 1}}x = \pi /2$

(xiv) ${\sin ^{ - 1}}x = {\operatorname{cosec} ^{ - 1}}(1/x);{\cos ^{ - 1}}x = {\sec ^{ - 1}}(1/x);{\tan ^{ - 1}}x = {\cot ^{ - 1}}(1/x)$


Note:

Some standard substitutions:

Expressions             Substitutions

(i) $\sqrt {\left( {{{\text{a}}^2} - {{\text{x}}^2}} \right)} \quad {\text{x}} = {\text{a}}\sin \theta {\text{ or a}}\cos \theta $

(ii) $\sqrt {\left( {{{\text{a}}^2} + {{\text{x}}^2}} \right)} {\text{x}} = {\text{a}}\tan \theta $ or ${\text{a}}\cot \theta $

(iii) $\sqrt {\left( {{{\text{x}}^2} - {{\text{a}}^2}} \right)} {\text{x}} = {\text{a}}\sec \theta $ or ${\text{a}}\operatorname{cosec} \theta $

(iv) $\sqrt {\left( {\dfrac{{{\text{a}} + {\text{x}}}}{{{\text{a}} - {\text{x}}}}} \right)} $ or $\sqrt {\left( {\dfrac{{{\text{a}} - {\text{x}}}}{{{\text{a}} + {\text{x}}}}} \right)} {\text{x}} = {\text{a}}\cos \theta $ or ${\text{a}}\cos 2\theta $

(v) $\sqrt {({\text{a}} - {\text{x}})({\text{x}} - {\text{b}})} $ or $\quad {\text{x}} = {\text{a}}{\cos ^2}\theta  + {\text{b}}{\sin ^2}\theta $

(vi) $\sqrt {\left( {\dfrac{{a - x}}{{x - b}}} \right)} $ or $\sqrt {\left( {\dfrac{{x - }}{{a - x}}} \right)} $

(vii)$\sqrt {(x - a)(x - b)} $ or $\quad x = a{\sec ^2}\theta  - b{\tan ^2}\theta $

(viii)$\sqrt {\left( {\dfrac{{x - a}}{{x - b}}} \right)} $ or $\sqrt {\left( {\dfrac{{x - }}{{x - a}}} \right)} $

(ix) $\sqrt {\left( {2ax - {x^2}} \right)} x = a(1 - \cos \theta )$


4.2 Logarithmic Differentiation

To find the derivative of:

If ${\text{y}} = {\left\{ {{f_1}({\text{x}})} \right\}^{{f_2}({\text{x}})}}$ or ${\text{y}} = {f_1}({\text{x}}) \cdot {f_2}({\text{x}}) \cdot {f_3}({\text{x}}) \ldots $

or \[\quad y = \dfrac{{{f_1}(x) \cdot {f_2}(x) \cdot {f_3}(x) \ldots }}{{{g_1}(x) \cdot {g_2}(x) \cdot {g_3}(x) \ldots }}\] then it's easier to take the function's logarithm first and then differentiate. This is referred to as the logarithmic function's derivative.


Important Notes (Alternate methods)

1. If \[{\text{y}} = {\{ f({\text{x}})\} ^{g({\text{x}})}} = {{\text{e}}^{g({\text{x}})\ln f({\text{x}})}}\left( {{{({\text{ variable }})}^{{\text{varable }}}}} \right)\left\{ {\because {\text{x}} = {{\text{e}}^{\ln {\text{x}}}}} \right\}\]

$$\begin{array}{*{20}{l}} {\therefore \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = {{\text{e}}^{g({\text{x}})\ln f({\text{x}})}} \cdot \left\{ {g({\text{x}}) \cdot \dfrac{{\text{d}}}{{{\text{dx}}}}\ln f({\text{x}}) + \ln f({\text{x}}) \cdot \dfrac{{\text{d}}}{{{\text{dx}}}}g({\text{x}})} \right\}} \\ { = {{\{ f({\text{x}})\} }^{g({\text{x}})}} \cdot \left\{ {g({\text{x}}) \cdot \dfrac{{{f^\prime }({\text{x}})}}{{f({\text{x}})}} + \ln f({\text{x}}) \cdot {g^\prime }({\text{x}})} \right\}} \end{array}$$

2. If \[{\text{y}} = {\{ f({\text{x}})\} ^{g({\text{x}})}}\]

\[\therefore \dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \] Derivative of y treating \[f({\text{x}})\] as constant + Derivative of y treating $g({\text{x}})$ as constant 

$ = {\{ f({\text{x}})\} ^{g({\text{x}})}} \cdot \ln f({\text{x}}) \cdot \dfrac{{\text{d}}}{{{\text{dx}}}}g({\text{x}}) + g({\text{x}}){\{ f({\text{x}})\} ^{g({\text{x}}) - 1}} \cdot \dfrac{{\text{d}}}{{{\text{dx}}}}f({\text{x}})$

$ = {\{ f({\text{x}})\} ^{g({\text{x}})}} \cdot \ln f({\text{x}}) \cdot {g^\prime }({\text{x}}) + g({\text{x}}) \cdot {\{ f({\text{x}})\} ^{g({\text{x}}) - 1}} \cdot {f^\prime }({\text{x}})$


4.3 Implicit Differentiation: 

$\phi ({\mathbf{x}},{\mathbf{y}}) = {\mathbf{0}}$

(i) To get ${\text{dy}}/{\text{dx}}$ with the use of implicit function, we differentiate each term w.r.t. ${\text{x}}$, regarding y as a function of ${\text{x}}$ & then collect terms in dy/dx together on one side to finally find ${\text{dy}}/{\text{dx}}$.

(ii) In answers of dy/dx in the case of implicit function, both $x\,\& \,y$are present.

Alternate Method: If $f(x,y) = 0$

then \[\dfrac{{dy}}{{dx}} =  - \dfrac{{\left( {\dfrac{{\partial f}}{{\partial x}}} \right)}}{{\left( {\dfrac{{\partial f}}{{\partial y}}} \right)}} =  - \dfrac{{{\text{ diff of }}f{\text{ w}}{\text{.r}}{\text{.t}}{\text{. }}x{\text{ treating }}y{\text{ as constant }}}}{{{\text{ diff }}.{\text{ of }}f{\text{ w}}{\text{.r}}{\text{.t}}{\text{. }}y{\text{ treating }}x{\text{ as constant }}}}\]


4.4 Parametric Differentiation

If ${\text{y}} = f({\text{t}});{\text{x}} = g({\text{t}})$ where ${\text{t}}$ is a Parameter, then

$\dfrac{{dy}}{{dx}} = \dfrac{{dy/dt}}{{dx/dt}}$


Note:

1. \[\dfrac{{{\text{dy}}}}{{{\text{dx}}}} = \dfrac{{{\text{dy}}}}{{{\text{dt}}}} \cdot \dfrac{{{\text{dt}}}}{{{\text{dx}}}}\]

2. \[\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right) = \dfrac{{\text{d}}}{{{\text{dt}}}}\left( {\dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right) \cdot \dfrac{{{\text{dt}}}}{{{\text{dx}}}}\left( {\because \dfrac{{{\text{dy}}}}{{{\text{dx}}}}} \right.\] in terms of \[{\text{t}}\] )

\[\begin{array}{*{20}{l}} { = \dfrac{{\text{d}}}{{{\text{dt}}}}\left( {\dfrac{{{f^\prime }({\text{t}})}}{{{g^\prime }({\text{t}})}}} \right) \cdot \dfrac{1}{{{f^\prime }({\text{t}})}}\{ {\text{ From }}(1)\} } \\ { = \dfrac{{{f^{\prime \prime }}({\text{t}}){g^\prime }({\text{t}}) - {g^{\prime \prime }}({\text{t}}){f^\prime }({\text{t}})}}{{\left\{ {{f^\prime }({\text{t}})} \right\}}}}  \end{array}\]


4.5 Derivative of a Function w.r.t. another Function

Let \[{\text{y}} = f({\text{x}});{\text{z}} = g({\text{x}})\] then \[\dfrac{{{\text{dy}}}}{{{\text{dz}}}} = \dfrac{{{\text{dy}}/{\text{dx}}}}{{{\text{dz}}/{\text{dx}}}} = \dfrac{{{f^\prime }({\text{x}})}}{{{g^\prime }({\text{x}})}}\]


4.6 Derivative of Infinite Series

When one or more terms are removed from an infinite series, the series stays unaltered. as a result.

(A) If $y = \sqrt {f(x) + \sqrt {f(x) + \sqrt {f(x) +  \ldots  \ldots \infty } } } $

then ${\text{y}} = \sqrt {f({\text{x}}) + {\text{y}}}  \Rightarrow \left( {{{\text{y}}^2} - {\text{y}}} \right) = f({\text{x}})$

Differentiating both sides w.r.t. $x$, we get $(2y - 1)\dfrac{{dy}}{{dx}} = {f^\prime }(x)$

(B) If \[{\text{y}} = {\{ {\text{f}}({\text{x}})\} ^{{{\{ {\text{f}}({\text{x}})\} }^{{{\{ {\text{f}}({\text{x}})\} }^{ - 1}}}}}}\] then [\{\text{y}} = {\{ f({\text{x}})\} ^{\text{y}}} \Rightarrow {\text{y}} = {{\text{e}}^{y\ln f({\text{x}})}}\]

Differentiating both sides w.r.t. $x$, we get

\[\dfrac{{dy}}{{dx}} = \dfrac{{y{{\{ f(x)\} }^{y - 1}} \cdot {f^\prime }(x)}}{{1 - {{\{ f(x)\} }^y} \cdot \ell \operatorname{n} f(x)}} = \dfrac{{{y^2}{f^\prime }(x)}}{{f(x)\{ 1 - y\ell nf(x)\} }}\]


5. Derivative of Order Two & Three

Let us assume a function ${\text{y}} = f({\text{x}})$ be defined on an open interval $(a,b)$. It's derivative, if it exists on $(a,b)$, is a certain function ${f^\prime }({\text{x}})\left[ {} \right.$ or $({\text{dy}}/{\text{dx}})$ or $\left. {{{\text{y}}^\prime }} \right]$ is called the first derivative of y w.r.t. ${\text{x}}$. If it occurs that the first derivative has a derivative on (a, b) then this derivative is called the second derivative of y w.r.t. x is denoted by ${f^{\prime \prime }}(x)$ or $\left( {{d^2}y/d{x^2}} \right)$ or ${y^{\prime \prime }}$.

Similarly, the ${3^{{\text{rd }}}}$ order derivative of $y$ w.r.t. $x$, if it exists, is defined by $\dfrac{{{{\text{d}}^3}{\text{y}}}}{{{\text{dx}}}} = \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}}} \right)$ it is also denoted by ${f^{\prime \prime }}({\text{x}})$ or ${{\text{y}}^{\prime \prime \prime }}$ Some Standard Results :

(i)\[\dfrac{{{{\text{d}}^{\text{n}}}}}{{{\text{d}}{{\text{x}}^{\text{n}}}}}{({\text{ax}} + {\text{b}})^{\text{m}}} = \dfrac{{{\text{m}}!}}{{({\text{m}} - {\text{n}})!}} \cdot {{\text{a}}^{\text{n}}} \cdot {({\text{ax}} + {\text{b}})^{{\text{m}} - {\text{n}}}},{\text{m}} \geqslant {\text{n}}\]

(ii)\[\dfrac{{{{\text{d}}^{\text{n}}}}}{{{\text{d}}{{\text{x}}^{\text{n}}}}}{{\text{x}}^{\text{n}}} = {\text{n}}!\]

(iii)\[\dfrac{{{{\text{d}}^{\text{n}}}}}{{{\text{d}}{{\text{x}}^{\text{n}}}}}\left( {{{\text{e}}^{{\text{mx}}}}} \right) = {{\text{m}}^{\text{n}}} \cdot {{\text{e}}^{{\text{mx}}}},{\text{m}} \in {\text{R}}\]

(iv)\[\dfrac{{{{\text{d}}^{\text{n}}}}}{{{\text{d}}{{\text{x}}^{\text{n}}}}}(\sin ({\text{ax}} + {\text{b}})) = {{\text{a}}^{\text{n}}}\sin \left( {{\text{ax}} + {\text{b}} + \dfrac{{{\text{n}}\pi }}{2}} \right),{\text{n}} \in {\text{N}}\]

(v)\[\dfrac{{{{\text{d}}^{\text{n}}}}}{{{\text{d}}{{\text{x}}^{\text{n}}}}}(\cos ({\text{ax}} + {\text{b}})) = {{\text{a}}^{\text{n}}}\cos \left( {{\text{ax}} + {\text{b}} + \dfrac{{{\text{n}}\pi }}{2}} \right),{\text{n}} \in {\text{N}}\]

(vi)\[\dfrac{{{{\text{d}}^{\text{n}}}}}{{{\text{d}}{{\text{x}}^{\text{n}}}}}\left\{ {{{\text{e}}^{{\text{ax}}}}\sin ({\text{bx}} + {\text{c}})} \right\} = {{\text{r}}^{\text{n}}} \cdot {{\text{e}}^{{\text{ax}}}} \cdot \sin ({\text{bx}} + {\text{c}} + {\text{n}}\phi ),{\text{n}} \in {\text{N}}\]

where $r = \sqrt {\left( {{a^2} + {b^2}} \right)} ,\phi  = {\tan ^{ - 1}}(b/a)$

(vii)\[\dfrac{{{{\text{d}}^{\text{n}}}}}{{{\text{d}}{{\text{x}}^{\text{n}}}}}\left\{ {{{\text{e}}^{{\text{ax}}}} \cdot \cos ({\text{bx}} + {\text{c}})} \right\} = {{\text{r}}^{\text{n}}} \cdot {{\text{e}}^{{\text{ax}}}} \cdot \cos ({\text{bx}} + {\text{c}} + {\text{n}}\phi ),{\text{n}} \in {\text{N}}\]

where $r = \sqrt {\left( {{a^2} + {b^2}} \right)} ,\phi  = {\tan ^{ - 1}}(b/a)$


6. Differentiation Of Determinants 

\[{\text{ If }}F({\text{X}}) = \left| {\begin{array}{*{20}{c}} {f({\text{x}})}&{g({\text{x}})}&{h({\text{x}})} \\ {\ell ({\text{x}})}&{m({\text{x}})}&{n({\text{x}})} \\ {u({\text{x}})}&{v({\text{x}})}&{w({\text{x}})} \end{array}} \right|\]

where $f,g,h,\ell ,m,n,u,v,w$ are differentiable function of $x$ then

\[\begin{array}{*{20}{l}} {{F^\prime }(x) = \left| {\begin{array}{*{20}{c}} {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\  {\ell (x)}&{m(x)}&{n(x)} \\  {u(x)}&{v(x)}&{w(x)} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{\ell ^\prime }(x)}&{{m^\prime }(x)}&{n(x)} \\ {u(x)}&{v(x)}&{w(x)} \end{array}} \right|} \\  { + \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\  {\ell (x)}&{m(x)}&{n(x)} \\   {{u^\prime }(x)}&{{v^\prime }(x)}&{{w^\prime }(x)} \end{array}} \right|} \end{array}\]


7. L' Hospital’s Rule

If $f(x)\,\& g(x)$ are functions of $x$ such that :

(i) $\mathop {\lim }\limits_{x \to a} f(x) = 0 = \mathop {\lim }\limits_{x \to a} g(x)$ or $\mathop {\lim }\limits_{x \to a} f(x) = \infty  = \mathop {\lim }\limits_{x \to a} g(x)$$f(x)$ and

(ii) $\operatorname{Both} f(x)\,\& g(x)$ are continuous at $x = a$ and

(iii) Both $f(x)\,\& g(x)$ are differentiable at $x = a$ and

(iv) $\operatorname{Both} f(x)\,\& g(x)$ are continuous at ${\text{x}} = {\text{a}}$, Then ${\operatorname{Limit} _{x \to a}}\dfrac{{f(x)}}{{g(x)}} = {\operatorname{Limit} _{x \to a}}\dfrac{{{f^\prime }(x)}}{{{g^\prime }(x)}} = {\operatorname{Limit} _{x \to a}}\dfrac{{{f^{\prime \prime }}(x)}}{{{g^{\prime \prime }}(x)}}$ & so on till determinant form vanishes.


Benefits of Revision Notes for Class 12 Maths Chapter 5 - Continuity and Differentiability 

The following are a few among the several benefits that you will enjoy while preparing with Vedantu’s Revision Notes on Class 12 Maths Chapter 5 - Continuity and Differentiability.

  • Vedantu’s Revision Notes for Class 12 Maths Chapter 5 - Continuity and Differentiability have been prepared by subject experts keeping in mind the need for quick and effective revision before the exams.

  • These are concise, precise, and well-structured notes with sufficient use of bulleted points, tables, and others to ensure students can revise the entire chapter and retain all important information at their fingertips.

  • The Class 12 Maths Chapter 5 Revision Notes also contain important mathematical formulae that you need to know by heart in order to be able to solve complex problems and application-based problems.

  • While going through the notes you will be able to identify the important concepts, topics, etc. You will find it easier to grasp areas that you need help with or are facing issues while learning.

  • The various concepts, definitions, and explanations also come with a note at the end of it to help students be aware of common mistakes or areas of confusion. These aid in helping students correctly apply their knowledge in the exams with caution.


What is the Mean Value Theorem?

The mean value theorem states that if a function f is defined on the closed interval [a,b] in such way that it satisfies the two conditions: (i) f is continuous on the closed interval [a,b] and (ii) f is differentiable on the open interval (a,b), then there exists a value x = c such that f'(c) = [f(b) – f(a)]/(b – a). 

 

What is Rolle’s Theorem?

The Rolle’s Theorem states that if a function f is defined in the closed interval [a, b] satisfying the three conditions: (i) f is continuous on the closed interval [a,b], (ii) f is differentiable on the open interval (a,b), and (iii) if f(a) = f(b), then there exists a value of x = c that lies between a and b, i.e., (a < c < b ) such that f‘(c) = 0. 


Conclusion

Vedantu's Continuity and Differentiability Class 12 Revision Notes for CBSE Maths Chapter 5 provide a comprehensive and effective study resource for students. The notes cover essential concepts, formulas, and examples related to continuity and differentiability, ensuring a thorough understanding of the topic. With clear explanations and step-by-step solutions, these notes enable students to grasp the fundamentals and enhance their problem-solving skills. The organised structure and concise presentation make it easier for students to revise and review the material. Overall, Vedantu's Continuity and Differentiability Class 12 Revision Notes serve as a valuable aid in preparing for exams and consolidating knowledge in this important area of mathematics.

FAQs on Continuity and Differentiability Class 12 Notes CBSE Maths Chapter 5 (Free PDF Download)

1. How to differentiate logarithmic and exponential Functions?

The derivatives of exponential and logarithmic functions can be found using the following formulas:


Exponential function:

  • $\dfrac{d}{dx} (e^x) = e^x$


Logarithmic function (natural logarithm):

  • $\dfrac{d}{dx} (ln x) = \dfrac{1}{x}$


Logarithmic function (base b):

  • $\dfrac{d}{dx} (log_b x) = \dfrac{1}{(x ln b)}$


Where e is the natural logarithm constant (approximately 2.718).

To differentiate an exponential function, we can use the power rule, which states that the derivative of $x^n$ is $nx^{(n-1)}$. In the case of the exponential function, n is simply the constant e.

To differentiate a logarithmic function, we can use the following identity:

$\dfrac{d}{dx} (ln x) = \dfrac{1}{x}$

This identity can be derived using the chain rule.

2. How to find the point of discontinuity?

In most of the problems, the point of discontinuity would be the point a,b etc where Function is of form

F(x) =      f1(x)  b>x>a 

                f1(x)  x ≤ a so on

Also, check for the points where the Function is not defined in the Domain.

3. When is fog continuous?

The function fog is continuous at x=a if and only if the following two conditions are satisfied:

The function g is continuous at x=a.

The function f is continuous at g(a).

In other words, for fog to be continuous at x=a, both g(a) and f(g(a)) must be defined, and the two-sided limit lim_{x->a} fog(x) must exist and be equal to f(g(a)).


Examples when fog is continuous:

If f and g are both continuous functions, then fog is continuous.

If f is continuous and g(x) is a constant function, then fog is continuous.

If f(x) = $x^2$ and g(x) = x, then fog(x) = $x^3$ is continuous.


Examples when fog is not continuous:

If f(x) = $\dfrac{1}{x}$ and g(x) = 0, then fog(x) is not continuous at x=0.

If f(x) = |x| and g(x) = $x^2$, then fog(x) is not continuous at x=0.

If f(x) = sin(x) and g(x) = $x^2$, then fog(x) is not continuous at x=$\dfrac{\pi}{2}$.

4. How to differentiate a matrix F(x)= f(x) g(x) h(x) l(x) m(x) n(x) u(x) v(x) w(x)  ?

F’(x)= f'(x) g'(x) h'(x) l(x) m(x) n(x) u(x) v(x) w(x)  +  f(x) g(x) h(x) l'(x) m'(x) n'(x) u(x) v(x) w(x) + f(X) g(x) h(x) l(x) m(x) n(x) u'(x) v'(x) w'(x) 

Where f(x),g(x),h(x),l(x),m(x),n(x),u(x),v(x),w(x) are differentiable functions.

5. What is L’Hospital’s rule?

L’Hospital’s rule :

         If functions f(x) and g(x) are such that

  1.  f(x)= 0  = g(x)  

Or  f(x) = ∞ = g(x) 

  1. Both f(x) and g(x) are continuous at x=a. 

  2. Both f(x) and g(x) are differentiable at x=a

  3. Both f’(x) and g’(x) are continuous at x=a. 

f(x)/g(x) =f'(x)/g'(x) =f''(x)/g''(x)   and so on.

6. Where can I download the latest revision notes for Chapter 5 “Continuity And Differentiability” of Class 12 Maths?

Nowadays, PDFs of revision notes are available on different websites all over the internet. But, for Chapter 5 'Continuity and Differentiability' of Class 12 Maths, the best one out there would be the unparalleled notes that Vedantu provides for the students. They are expert-created, short, simply structured, and excellent for last-minute revisions as well. The Revision Notes for Chapter 5 of Class 12 Maths are available free of cost on the Vedantu website and the Vedantu app.

7. How are revision notes useful for learning Chapter 5 of Class 12 Maths?

Students may grasp the concepts taught in Chapter 5 'Continuity and Differentiability,' by using Vedantu's Revision Notes for Chapter 5 of Class 12 Maths. Because professionals write the answers to most tasks, these notes can greatly help students out if they do not understand the core ideas in the Chapter. The key formulae for solving problems are presented for students to quickly learn and use to prepare properly and achieve good outcomes with their studies.

8. What are the concepts covered in Continuity and Differentiability in Class 12?

'Continuity and Differentiability' is an important Chapter of Class 12 from where questions have been repeatedly asked for the boards. As such, students need to be familiar with the different concepts covered in the Chapter which are listed down below:

  • Algebra of continuous functions

  • Composite, implicit, and inverse trigonometric function derivatives

  • Exponential Functions

  • Logarithmic functions

  • Parametric form derivatives

  • Derivatives of Second orders

  • Mean Value Theorem

9. Is Chapter 5 of Class 12 Maths difficult for students?

Chapter 5 'Continuity and Differentiability' of Class 12 Maths is kind of new to students. The difficulty of this Chapter ranges from moderate to difficult because once students are through with the formulae, problems can be easily solved. But, the difficult part comes in their exams because a lot of questions are asked from this Chapter and the majority of them come in the form of application types. Students should check out revision notes for the important questions and formulae.

10. What are the important questions asked for Class 12 boards from Chapter 5 of Maths?

A large part of the CBSE Board exam Math paper is a contribution from Chapter 5 'Continuity and Differentiability'. So students should know which are the important questions asked from this Chapter and using revision notes can help them out with this.

  • Continuous function problems

  • Composite function problems

  • Differentiable function problems

  • Inverse trigonometric function derivatives

  • Logarithmic differentiation

  • Rolle's Theorem problems

  • MVT problems

  • Parametric form derivatives