CBSE Class 11 Maths Important Questions - Chapter 12 Limits and Derivatives

1 Mark Questions

1. What is the limit’s value $\underset{x\to 3}{lim}\left[{\displaystyle \frac{x^{\mathbf{2}}-9}{x-3}}\right]$.

Ans:

Here, we can see that the limit $\underset{x\to 3}{lim}{\displaystyle \frac{x^2-9}{x-3}}$ is in the form ${\displaystyle \frac{0}{0}}$.

By representing the numerator as the product of two terms, we get,

$$\underset{x\to 3}{lim}{\displaystyle \frac{(x+3)(x-3)}{(x-3)}}$$ 

$$=3+3$$

$$=6$$

2. What is the limit’s value $\underset{x\to \mathbf{0}}{lim}\left[{\displaystyle \frac{\sin 5x}{3x}}\right]$?

Ans:

Multiply and divide the numerator and denominator of the given limit with 5, then we get,

$\underset{x\to 3}{lim}{\displaystyle \frac{\sin 5x}{3x}}$  

$=\underset{x\to 3}{lim}{\displaystyle \frac{\sin 5x}{3x}}\times {\displaystyle \frac{5}{5}}$ 

$=1\times {\displaystyle \frac{5}{3}}$

$={\displaystyle \frac{5}{3}}[\underset{x\to 0}{\because lim}{\displaystyle \frac{\sin ax}{ax}}=1]$

3. What is the result of the derivative of $2^x$ with respect to $x$.

Ans: 

Let us assume the given expression as,

$y=2^x$

Now, differentiating on both sides with respect to $x$ then we get, 

${\displaystyle \frac{dy}{dx}}$ 

$={\displaystyle \frac{d}{dx}}\left(2^x\right)$

$=2^x\log 2$

4. The result when the expression $\sqrt{\sin 2x}$ when it is differentiated with respect to $x$ is.

Ans: 

By using the chain rule of differentiation the derivative of given expression is given as,

${\displaystyle \frac{d}{dx}}\sqrt{\sin 2x}$ 

$={\displaystyle \frac{1}{2\sqrt{\sin 2x}}}{\displaystyle \frac{d}{dx}}\sin 2x$

$={\displaystyle \frac{1}{2\sqrt{\sin 2x}}}\times 2\cos 2x$

$={\displaystyle \frac{\cos 2x}{\sqrt{\cos 2x}}}$

5. What is the limit’s value  $\underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{2}4x}{x^2}}$ .

Ans:

Let us multiply and divide the given expression with $4^2$ then we get,

$\underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{2}4x}{x^2{4}^2}}\times 4^2$ 

$\underset{4x\to 0}{=lim}{\left({\displaystyle \frac{{\sin }^{2}4x}{4x}}\right)}^2\times 16$ 

$=1\times 16$

$=16$

6. What is the value of $\underset{x\to a}{lim}\left({\displaystyle \frac{x^2-a^n}{x-a}}\right)$ ?

Ans:

By using the L’ Hospital rule that is differentiating the numerator and denominator with respect to $x$ we get,

$\underset{x\to a}{lim}{\displaystyle \frac{x^n-a^n}{x-a}}$ 

$=\underset{x\to a}{lim}{\displaystyle \frac{n{x}^{n-1}}{1}}$

$=n{a}^{n-1}$

7. What is the derivative of  ${\displaystyle \frac{2^x}{x}}$ with respect to $x$?

Ans: 

By using the ${\displaystyle \frac{u}{v}}$ formula of differentiating for the given expression, we get,

${\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{2^x}{x}}\right)$ 

$={\displaystyle \frac{x{\displaystyle \frac{d}{dx}}\left(2^x\right)-2^x{\displaystyle \frac{d}{dx}}\left(x\right)}{x^2}}$

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

$=2^x{\displaystyle \frac{[x\ln 2-1]}{x^2}}$

8. If the expression is $y=e^{\sin x}$, then  find the value of ${\displaystyle \frac{dy}{dx}}$ . 

Ans:

We are given the expression as,

$y=e^{\sin x}$ 

Now, by differentiating on both sides with respect to $x$ then we get,

$${\displaystyle \frac{dy}{dx}}$$ 

$$={\displaystyle \frac{d}{dx}}\left(e^{\sin x}\right)$$

$$=e^{\sin x}\times \cos x$$

$$=\cos x{e}^{\sin x}$$

9. What is the limit’s value  $\underset{x\to 1}{lim}{\displaystyle \frac{x^{15}-1}{x^{10}-1}}$

Ans: 

By using the L’ Hospital rule that is differentiating the numerator and denominator with respect to $x$ we get,

$\underset{x\to 1}{lim}{\displaystyle \frac{x^{15}-1}{x^{10}-1}}$ 

$=\underset{x\to 1}{lim}{\displaystyle \frac{15x^{14}}{10x^9}}$ 

$=\underset{x\to 1}{lim}{\displaystyle \frac{3}{2}}{x}^5$

$={\displaystyle \frac{3}{2}}$

10. Differentiate the expression $x\sin x$ with respect to $x$ 

Ans: 

By using the chain rule of differentiation, we get the derivative of the given expression as,

${\displaystyle \frac{d}{dx}}(x\sin x)$ 

$=x\left({\displaystyle \frac{d}{dx}}(\sin x)\right)+\sin x\left({\displaystyle \frac{d}{dx}}\left(x\right)\right)$

$=x(\cos x)+\sin x\left(1\right)$

$=x\cos x+\sin x$ 

11. What is the limit’s value $\underset{x\to 1}{lim}{\displaystyle \frac{x^2+1}{x+100}}$ ?

Ans:

By using the definition of limit, that is by directly substituting $x=1$ in the expression we get,

$\underset{x\to 1}{lim}{\displaystyle \frac{x^2+1}{x+100}}$ 

$={\displaystyle \frac{1+1}{1+100}}$

$={\displaystyle \frac{2}{101}}$

12. What is the limit’s value $\underset{x\to \mathrm{\infty }}{lim}[\text{cosec}x-\cot x]$ ?

Ans:

Rewriting $\cos ecx$ and $\cot x$ in terms of $\sin x$ and $\cos x$ we get,

$\underset{x\to \mathrm{\infty }}{lim}[\text{cosec}x-\cot x]$ 

$=\underset{x\to \mathrm{\infty }}{lim}[{\displaystyle \frac{1}{\sin x}}-{\displaystyle \frac{\cos x}{\sin x}}]$ 

Now, by using the half angle formula of $\sin x$ and $\cos x$ we get,

$=\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2{\sin }^2{\displaystyle \frac{x}{2}}}{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}}$

$=\underset{x\to \mathrm{\infty }}{lim}\tan {\displaystyle \frac{x}{2}}$

$=0$

13. Find the value of $f^{\prime }(x)$ at $x=100$ if $f(x)=99x$ 

Ans:

By differentiating the given function with respect to $x$ we get,

$f^{\prime }(x)=99$ 

Now, by substituting $x=100$ in the above derivative we get,

$f^{\prime }\left(100\right)=99$

14. What is the limit’s value $\underset{x\to -2}{lim}{\displaystyle \frac{\tan \pi x}{x+2}}$?

Ans: 

By using the L’ Hospital rule that is differentiating the numerator and denominator with respect to $x$ we get,

$\underset{x\to -2}{lim}{\displaystyle \frac{\tan \pi x}{x+2}}$

$=\underset{x\to -2}{lim}{\displaystyle \frac{\pi {\sec }^2\left(\pi x\right)}{1}}$

$=\pi {\sec }^2(-2\pi )$

$=\pi$

15. Find the derivative of expression ${\sin }^{n}x$ with respect to $x$. 

Ans:

Now, by using the power rule and chain rule of differentiation, we get the derivative of the given expression as,

${\displaystyle \frac{d}{dx}}\left({\sin }^{n}x\right)$ 

$=\left(n{\sin }^{n-1}x\right){\displaystyle \frac{d}{dx}}(\sin x)$ 

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

16. Find derivative of the expression $1+x+x^2+x^3+\dots +x^{50}$ at $x=1$ .

Ans: 

By using the power rule of differentiation, we get the derivative of given function as,

$f^1(x)=1+2x+3x^2+\dots +50x^{49}$

Now, by substituting $x=100$ in the above derivative, we get,

$f^{\prime }(1)$

$=1+2+3+\dots +50$

$={\displaystyle \frac{50(50+1)}{2}}$

$=25\times 51$

$=1275$

17. What is the limit’s value $\underset{h\to 0}{lim}{\displaystyle \frac{e^{2h}-1}{h}}$ ?

Ans:

By using the L’ Hospital rule that is differentiating the numerator and denominator with respect to $h$ we get,

$\underset{h\to 0}{lim}{\displaystyle \frac{e^{2h}-1}{h}}$ 

$=\underset{h\to 0}{lim}{\displaystyle \frac{2e^{2h}}{1}}$

$=2$ 

18. What is the limit’s value $\underset{x\to 0}{lim}{\displaystyle \frac{{(1+x)}^6-1}{{(1+x)}^2-1}}$ ?

Ans: 

By using the L’ Hospital rule that is differentiating the numerator and denominator with respect to $x$ we get,

$\underset{x\to 0}{lim}\left[{\displaystyle \frac{{(1+x)}^6-1}{{(1+x)}^2-1}}\right]$ 

$=\underset{x\to 0}{lim}\left[{\displaystyle \frac{6{(1+x)}^5}{2(1+x)}}\right]$

$={\displaystyle \frac{3{(1+0)}^5}{(1+0)}}$

$=3$ 

19. Find the value of the variable $a$ if the limit value is $\underset{x\to a}{lim}{\displaystyle \frac{x^7+a^7}{x+a}}=7$.

Ans:

By using the definition of limit, that is by directly substituting $x=a$ in the expression we get,

$\underset{x\to a}{lim}{\displaystyle \frac{x^7+a^7}{x+a}}=7$ 

$$\Rightarrow {\displaystyle \frac{a^7+a^7}{a+a}}=7$$ 

$$\Rightarrow a^6=7$$

$$\Rightarrow a=\sqrt[6]{7}$$

20. Find the derivative of expression $x^{-3}(5+3x)$ with respect to $x$.

Ans:

First multiply $x^{-3}$ to each term in brackets and then differentiate the expression we get,

${\displaystyle \frac{d}{dx}}{x}^{-3}(5+3x)$ 

$={\displaystyle \frac{d}{dx}}[5x^{-3}+3x^{-2}]$ 

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

$=-{\displaystyle \frac{15}{x^4}}-{\displaystyle \frac{6}{x^3}}$

4 Marks Questions

1. Prove that the value of the limit is, $\underset{x\to 0}{lim}\left({\displaystyle \frac{e^x-1}{x}}\right)=1$ .

Ans: 

First let us take the left hand side (LHS) of the given equation that is,

$\underset{x\to 0}{lim}{\displaystyle \frac{e^x-1}{x}}$ .

First by using the definition of limit substitute $x=0$ in the expression then we get,

$\Rightarrow \underset{x\to 0}{lim}{\displaystyle \frac{e^x-1}{x}}={\displaystyle \frac{e^0-1}{0}}$

$\Rightarrow \underset{x\to 0}{lim}{\displaystyle \frac{e^x-1}{x}}={\displaystyle \frac{0}{0}}$

Here, we can see that the limit is in undetermined form.

So, by using the L’ Hospital rule, that is differentiating both numerator and denominator with respect to $x$ then we get,

$\Rightarrow \underset{x\to 0}{lim}{\displaystyle \frac{e^x-1}{x}}$

$=\underset{x\to 0}{lim}{\displaystyle \frac{e^x}{1}}$

$=e^0$

$=1$

Hence we can say that the required result has been proved.

2. What is the limit’s value  $\underset{x\to 1}{lim}{\displaystyle \frac{(2x-3)(\sqrt{x}-1)}{(2x^2+x-3)}}$ ?

Ans:

First let us represent the denominator as the product of two factors then we get,

$\underset{x\to 1}{lim}{\displaystyle \frac{(2x-3)(\sqrt{x}-1)}{(2x^2+x-3)}}$

$=\underset{x\to 1}{lim}{\displaystyle \frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(x-1)}}$

Now, Let us use the formula of difference of squares of two numbers that is,

$a^2-b^2=(a+b)(a-b)$

Using this formula for the expression $x-1$ in the denominator we get,

$\underset{x\to 1}{lim}{\displaystyle \frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(x-1)}}$

$=\underset{x\to 1}{lim}{\displaystyle \frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)}}$

$=\underset{x\to 1}{lim}{\displaystyle \frac{(2x-3)}{(2x+3)(\sqrt{x}+1)}}$

Now by using the definition of limit substitute $x=1$ in the expression then we get,

$=\underset{x\to 1}{lim}{\displaystyle \frac{(2x-3)}{(2x+3)(\sqrt{x}+1)}}$

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

$={\displaystyle \frac{-1}{10}}$

3. What is the limit’s value  $\underset{x\to 0}{lim}{\displaystyle \frac{x\tan 4x}{1-\cos 4x}}$ ?

Ans: 

We know the half angle formula of trigonometric ratio $\cos 2x$ is given as,

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

$\Rightarrow 1-\cos 2x=2{\sin }^{2}x$

We also know the half-angle formula of $\sin 2x$ as.

$\sin 2x=2\sin x\cos x$

Now, by using these formulas for the angle $4x$ for the numerator and the denominator of given expression we get,

$\underset{x\to 0}{lim}{\displaystyle \frac{x\tan 4x}{1-\cos 4x}}$

$=\underset{x\to 0}{lim}{\displaystyle \frac{x\sin 4x}{\cos 4x[2{\sin }^{2}2x]}}$ 

$=\underset{x\to 0}{lim}{\displaystyle \frac{2x\sin 2x\cos 2x}{\cos 4x[2{\sin }^{2}2x]}}$

$=\underset{x\to 0}{lim}({\displaystyle \frac{\cos 2x}{\cos 4x}}\times {\displaystyle \frac{2x}{\sin 2x}}\times {\displaystyle \frac{1}{2}})$

Now, by separating the limits to each expression that is present then we get,

$={\displaystyle \frac{1}{2}}{\displaystyle \frac{\underset{x\to 0}{lim}\cos 2x}{\underset{x\to 0}{lim}\cos 4x}}\times \underset{x\to 0}{lim}\left({\displaystyle \frac{2x}{\sin 2x}}\right)$

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

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

4. If the expression is given as $y={\displaystyle \frac{(1-\tan x)}{(1+\tan x)}}$ , then show that the derivative is ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-2}{(1+\sin 2x)}}$ .

Ans:

First let us differentiate the given expression with respect to $x$ on both sides, then we get,

${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{(1-\tan x)}{(1+\tan x)}}\right)$

Now, by using the formula of derivative of the form ${\displaystyle \frac{u}{v}}$ for the above expression, then we get,

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

$${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{(1+\tan x)(-{\sec }^{2}x)-(1-\tan x)({\sec }^{2}x)}{{(1+\tan x)}^2}}$$

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

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

Now, let us represent $\sec x$ and $\tan x$ in terms of $\cos x$ and $\sin x$ then we get,

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

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

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

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

Therefore, the required result has been proved that is, if $y={\displaystyle \frac{(1-\tan x)}{(1+\tan x)}}$ then the derivative is,

$${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-2}{1+{\sin }^{2}x}}$$

5. Find the derivative of the expression  $e^{\sqrt{\cot x}}$ with respect to $x$. 

Ans:

First by using the chain rule of differentiation for the given expression the derivative will be,

${\displaystyle \frac{d}{dx}}\left(e^{\sqrt{\cot x}}\right)$ 

$$=e^{\sqrt{\cot x}}{\displaystyle \frac{d}{dx}}\left(\sqrt{\cot x}\right)$$

Now, again by using the chain rule for the remained expression then we get,

$=e^{\sqrt{\cot x}}\times \left({\displaystyle \frac{1}{2\sqrt{\cot x}}}\right)\cdot {\displaystyle \frac{d}{dx}}(\cot x)$

$={\displaystyle \frac{e^{\sqrt{\cot x}}}{2\sqrt{\cot x}}}(-\text{cose}{\text{c}}^{2}x)$

$=-{\displaystyle \frac{\left(\text{cose}{\text{c}}^{2}x\right)e^{\sqrt{\cot x}}}{2\sqrt{\cot x}}}$

6. Let $f(x)=\{\begin{array}{rl}& a+bx,x<1\\ & 4,x=1\\ & b-ax,x>1\end{array}$ and if $\underset{x\to 1}{lim}f(x)=f(1)$. What are the possible value of $a$ and $b$ ?

Ans:

We know that the limit of a function exists when the left hand limit and right hand limit both equal to value of function at that point that is,

$\underset{x\to a^{+}}{lim}f\left(x\right)=\underset{x\to a^{-}}{lim}f\left(x\right)$

Now, by using the above equation for the given function at $x=1$ then we get,

$\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)=4$…….. $\left(1\right)$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=f\left(1\right)=4$…….. $\left(2\right)$

For $$x>1$$ we have the given function as,

$$f(x)=a+bx$$

Now, by using the equation $\left(1\right)$ we get,

$$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}(a+bx)$$

$\Rightarrow 4=a+b$…….. $\left(3\right)$

Now, for $x<1$ we have the given function as,

$f\left(x\right)=b-ax$

Now, by using the equation $\left(2\right)$ we get,

$$\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}(b-ax)$$

$\Rightarrow 4=b-a$………. $\left(4\right)$

Now, by adding both equation $\left(3\right)$ and equation $\left(4\right)$ then we get,

$4+4=(a+b)+(b-a)$

$\Rightarrow 8=2b$

$\Rightarrow b=4$

Similarly we get the value of other variable as,

$\Rightarrow a=0$

7. If the function is given as $y={\displaystyle \frac{1}{\sqrt{a^2-x^2}}}$ , then find the value of ${\displaystyle \frac{dy}{dx}}$ .

Ans:

First let us take the denominator of given function to numerator by assuming the negative power then we get,

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

Now, by using the power rule and chain rule of differentiation then we get the derivative of above function as, 

${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-1}{2}}{(a^2-x^2)}^{{\displaystyle \frac{-1}{2}}-1}{\displaystyle \frac{d}{dx}}(a^2-x^2)$

$=\left({\displaystyle \frac{-1}{{(a^2-x^2)}^{{\displaystyle \frac{3}{2}}}}}\right){\displaystyle \frac{d}{dx}}(a^2-x^2)$

Now, again by using the power rule for the above equation then we get,

$=\left({\displaystyle \frac{-1}{{(a^2-x^2)}^{{\displaystyle \frac{3}{2}}}}}\right)(-2x)$

$=\left({\displaystyle \frac{2x}{{(a^2-x^2)}^{{\displaystyle \frac{3}{2}}}}}\right)$

Therefore, we can conclude the value of ${\displaystyle \frac{dy}{dx}}$ for the given function $y={\displaystyle \frac{1}{\sqrt{a^2-x^2}}}$ is given as,

${\displaystyle \frac{dy}{dx}}=\left({\displaystyle \frac{2x}{{(a^2-x^2)}^{{\displaystyle \frac{3}{2}}}}}\right)$

8. Differentiate $\sqrt{{\displaystyle \frac{1-\tan x}{1+\tan x}}}$ 

Ans:

First, consider the given expression,

$y=\sqrt{{\displaystyle \frac{1-\tan x}{1+\tan x}}}$ 

Now, let us assume the expression inside square root as,

${\displaystyle \frac{1-\tan x}{1+\tan x}}=t$ 

Differentiating on both sides with respect to $x$, using quotient rule, we get,

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

$$={\displaystyle \frac{(1+\tan x)(0-{\sec }^{2}x)-(1-\tan x)(0+{\sec }^{2}x)}{{(1+\tan x)}^2}}$$

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

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

Next, we have the function of $y$ in terms of $t$ as,
$y=\sqrt{t}$ 

Differentiating on both sides with respect to $t$, we get,

$${\displaystyle \frac{dy}{dt}}={\displaystyle \frac{d}{dt}}{t}^{{\displaystyle \frac{1}{2}}}$$

$$={\displaystyle \frac{1}{2}}{t}^{{\displaystyle \frac{1}{2}}-1}$$

$$={\displaystyle \frac{1}{2\sqrt{t}}}$$

Substituting ${\displaystyle \frac{1-\tan x}{1+\tan x}}=t$, we get,

$={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{1+\tan x}{1-\tan x}}}$

Using the standard definition of chain rule for differentiation, we get,
$${\displaystyle \frac{dy}{dt}}={\displaystyle \frac{dy}{dt}}\times {\displaystyle \frac{dt}{dx}}$$

$$={\displaystyle \frac{-2{\sec }^{2}x}{{(1+\tan x)}^2}}\times {\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{1+\tan x}{1-\tan x}}}$$

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

9. Differentiate 

i) $\left({\displaystyle \frac{\sin x+\cos x}{\sin x-\cos x}}\right)$

Ans:  

We are given the expression,

${\displaystyle \frac{\sin x+\cos x}{\sin x-\cos x}}$,

which we have to differentiate. 

Using quotient rule, we can derivate the given expression as,${\displaystyle \frac{dy}{dx}}\left({\displaystyle \frac{\sin x+\cos x}{\sin x-\cos x}}\right)={\displaystyle \frac{(\sin x-\cos x)\cdot {\displaystyle \frac{d}{dx}}(\sin x+\cos x)-(\sin x+\cos x)\cdot {\displaystyle \frac{d}{dx}}(\sin x-\cos x)}{{(\sin x-\cos x)}^2}}$

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

On simplifying, we get

${\displaystyle \frac{dy}{dx}}\left({\displaystyle \frac{\sin x+\cos x}{\sin x-\cos x}}\right)={\displaystyle \frac{-[{(\sin x-\cos x)}^2+{(\sin x+\cos x)}^2]}{{(\sin x-\cos x)}^2}}$ 

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

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

ii) $\left({\displaystyle \frac{\sin x-1}{\sec x+1}}\right)$

Ans:

We are given the expression, 

${\displaystyle \frac{\sin x-1}{\sec x+1}}$

which we have to differentiate. 

Using quotient rule, we can differentiate the given function as,

${\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{\sin x-1}{\sec x+1}}\right)={\displaystyle \frac{(\sec x+1){\displaystyle \frac{d}{dx}}(\sin x-1)-(\sin x-1){\displaystyle \frac{d}{dx}}(\sec x+1)}{{(\sec x+1)}^2}}$ 

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

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

10. Evaluate $\underset{x\to {\displaystyle \frac{\pi }{4}}}{lim}{\displaystyle \frac{\sin x-\cos x}{(x-{\displaystyle \frac{\pi }{4}})}}$

Ans:

We have to determine the limit for the given expression. We will see the substitution method to do so.
Put $(x-{\displaystyle \frac{\pi }{4}})=y$ , so that when $x\to {\displaystyle \frac{\pi }{4}}$ then $y\to 0$.

Now, replace the variable $x$, with the new definition for $y$.

$\underset{x\to {\displaystyle \frac{\pi }{4}}}{lim}{\displaystyle \frac{\sin x-\cos x}{(x-{\displaystyle \frac{\pi }{4}})}}=\underset{y\to 0}{lim}{\displaystyle \frac{[\sin ({\displaystyle \frac{\pi }{4}}+y)-\cos ({\displaystyle \frac{\pi }{4}}+y)]}{y}}$

Then, using the addition formula for trigonometric functions, we get,

$\underset{x\to {\displaystyle \frac{\pi }{4}}}{lim}{\displaystyle \frac{\sin x-\cos x}{(x-{\displaystyle \frac{\pi }{4}})}}=\underset{y\to 0}{lim}{\displaystyle \frac{(\sin {\displaystyle \frac{\pi }{4}}\cos y+\cos {\displaystyle \frac{\pi }{4}}\sin y-\cos {\displaystyle \frac{\pi }{4}}\cos y+\sin {\displaystyle \frac{\pi }{4}}\sin y)}{y}}$ 

$=\underset{y\to 0}{lim}{\displaystyle \frac{((\sin {\displaystyle \frac{\pi }{4}}-\cos {\displaystyle \frac{\pi }{4}})\cos y+(\cos {\displaystyle \frac{\pi }{4}}+\sin {\displaystyle \frac{\pi }{4}})\sin y)}{y}}$

Evaluating further, 

$\underset{x\to {\displaystyle \frac{\pi }{4}}}{lim}{\displaystyle \frac{\sin x-\cos x}{(x-{\displaystyle \frac{\pi }{4}})}}={\displaystyle \frac{2}{\sqrt{2}}}\times \underset{y\to 0}{lim}\left({\displaystyle \frac{\sin y}{y}}\right)$

$=\sqrt{2}\times 1$

$=\sqrt{2}$

11.Evaluate $$\underset{x\to 0}{lim}{\displaystyle \frac{{(1+x)}^6-1}{{(1+x)}^5-1}}$$.

Ans:

We have to determine the limit for the given expression. We will see the substitution method to do so. 

Put $(1+x)=y$ , so that when $x\to 0$ then $y\to 1$ .

Now, consider the given expression, and substitute the new variable, 

$\underset{x\to 0}{lim}{\displaystyle \frac{{(1+x)}^6-1}{{(1+x)}^5-1}}=\underset{y\to 1}{lim}\left[{\displaystyle \frac{y^6-1}{y^5-1}}\right]$

Evaluating this limit, using L ’Hospital’s rule,

$$\underset{x\to 0}{lim}{\displaystyle \frac{{(1+x)}^6-1}{{(1+x)}^5-1}}=\underset{y\to 1}{lim}{\displaystyle \frac{{\displaystyle \frac{d}{dy}}(y^6-1)}{{\displaystyle \frac{d}{dy}}(y^5-1)}}$$ 

$$=\underset{y\to 1}{lim}{\displaystyle \frac{\left(6y^5\right)}{\left(5y^4\right)}}$$

$$={\displaystyle \frac{6}{5}}$$

12. Evaluate $\underset{x\to a}{lim}{\displaystyle \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}}$.

Ans:

The given expression is 

$\underset{x\to a}{lim}{\displaystyle \frac{(\sqrt{a+2x}-\sqrt{3x})}{(\sqrt{3a+x}-2\sqrt{x})}}$ 

We will evaluate its limit, using the rationalization method,

$$\underset{x\to a}{lim}{\displaystyle \frac{(\sqrt{a+2x}-\sqrt{3x})}{(\sqrt{3a+x}-2\sqrt{x})}}=\underset{x\to a}{lim}{\displaystyle \frac{(\sqrt{a+2x}-\sqrt{3x})}{(\sqrt{3a+x}-2\sqrt{x})}}\times {\displaystyle \frac{(\sqrt{3a+x}+2\sqrt{x})}{(\sqrt{3a+x}+2\sqrt{x})}}\times {\displaystyle \frac{(\sqrt{a+2x}+\sqrt{3x})}{(\sqrt{a+2x}+\sqrt{3x})}}$$

$$=\underset{x\to a}{lim}{\displaystyle \frac{[(a+2x)-3x]\times (\sqrt{3a+x}+2\sqrt{x})}{[(3a+x)-4x]\times (\sqrt{a+2x}+\sqrt{3x})}}$$

$$=\underset{x\to a}{lim}{\displaystyle \frac{(a-x)\times (\sqrt{3a+x}+2\sqrt{x})}{3(a-x)\times (\sqrt{a+2x}+\sqrt{3x})}}$$

$$=\underset{x\to a}{lim}{\displaystyle \frac{(\sqrt{3a+x}+2\sqrt{x})}{3(\sqrt{a+2x}+\sqrt{3x})}}$$

On simplifying, we get, 

$\underset{x\to a}{lim}{\displaystyle \frac{(\sqrt{a+2x}-\sqrt{3x})}{(\sqrt{3a+x}-2\sqrt{x})}}={\displaystyle \frac{(\sqrt{4a}+2\sqrt{a})}{3(\sqrt{3a}+\sqrt{3a})}}$ 

$={\displaystyle \frac{4\sqrt{a}}{6\sqrt{3}\sqrt{a}}}$

$={\displaystyle \frac{2}{3\sqrt{3}}}$

13. Find the derivative of $f(x)=1+x+x^2+x^3+\dots +x^{50}$  at $x=1$ .

Ans:

We are required to find the derivative of $f(x)=1+x+x^2+x^3+\dots +x^{50}$