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

1 Mark Questions

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

Ans:

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

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

\[\underset{x\to 3}{\mathop{\lim }}\,\dfrac{(x+3){(x-3)}}{{(x-3)}}\] 

\[=3+3\]

\[=6\]

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

Ans:

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

$\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sin 5x}{3x}$  

$=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sin 5x}{3x}\times \dfrac{5}{5}$ 

$=1\times \dfrac{5}{3}$

$=\dfrac{5}{3}\left[ \underset{x\to 0}{\mathop{\because \lim }}\,\dfrac{\sin ax}{ax}=1 \right]$

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, 

$\dfrac{dy}{dx}$ 

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

$={{2}^{x}}\log 2$

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,

$\dfrac{d}{dx}\sqrt{\sin 2x}$ 

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

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

$=\dfrac{\cos 2x}{\sqrt{\cos 2x}}$

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

Ans:

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

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}4x}{{{x}^{2}}{{4}^{2}}}\times {{4}^{2}}$ 

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

$=1\times 16$

$=16$

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

Ans:

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

$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}$ 

$=\underset{x\to a}{\mathop{\lim }}\,\dfrac{n{{x}^{n-1}}}{1}$

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

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

Ans: 

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

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

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

$=\dfrac{\left( x\times {{2}^{x}}\ln 2 \right)-\left( {{2}^{x}}\times 1 \right)}{{{x}^{2}}}$ 

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

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

Ans:

We are given the expression as,

$y={{e}^{\sin x}}$ 

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

\[\dfrac{dy}{dx}\] 

\[=\dfrac{d}{dx}\left( {{e}^{\sin x}} \right)\]

\[={{e}^{\sin x}}\times \cos x\]

\[=\cos x{{e}^{\sin x}}\]

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

Ans: 

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

$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{{{x}^{15}}-1}{{{x}^{10}}-1}$ 

$=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{15{{x}^{14}}}{10{{x}^{9}}}$ 

$=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{3}{2}{{x}^{5}}$

$=\dfrac{3}{2}$

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,

$\dfrac{d}{dx}\left( x\sin x \right)$ 

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

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

$=x\cos x+\sin x$ 

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

Ans:

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

$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{{{x}^{2}}+1}{x+100}$ 

$=\dfrac{1+1}{1+100}$

$=\dfrac{2}{101}$

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

Ans:

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

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

$=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x} \right]$ 

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

$=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}$

$=\underset{x\to \infty }{\mathop{\lim }}\,\tan \dfrac{x}{2}$

$=0$

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

Ans:

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

${f}'(x)=99$ 

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

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

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

Ans: 

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

$\underset{x\to -2}{\mathop{\lim }}\,\dfrac{\tan \pi x}{x+2}$

$=\underset{x\to -2}{\mathop{\lim }}\,\dfrac{\pi {{\sec }^{2}}\left( \pi x \right)}{1}$

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

$=\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,

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

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

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

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

Ans: 

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

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

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

${f}'(1)$

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

$=\dfrac{50(50+1)}{2}$

$=25\times 51$

$=1275$

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

Ans:

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

$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{2h}}-1}{h}$ 

$=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2{{e}^{2h}}}{1}$

$=2$ 

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

Ans: 

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

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

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

$=\dfrac{3{{\left( 1+0 \right)}^{5}}}{\left( 1+0 \right)}$

$=3$ 

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

Ans:

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

$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{7}}+{{a}^{7}}}{x+a}=7$ 

\[\Rightarrow \dfrac{{{a}^{7}}+{{a}^{7}}}{a+a}=7\] 

\[\Rightarrow {{a}^{6}}=7\]

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

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,

$\dfrac{d}{dx}{{x}^{-3}}(5+3x)$ 

$=\dfrac{d}{dx}\left[ 5{{x}^{-3}}+3{{x}^{-2}} \right]$ 

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

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

4 Marks Questions

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

Ans: 

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

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-1}{x}$ .

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

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-1}{x}=\dfrac{{{e}^{0}}-1}{0}$

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-1}{x}=\dfrac{0}{0}$

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,

$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-1}{x}$

$=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}}{1}$

$={{e}^{0}}$

$=1$

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

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

Ans:

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

$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{(2x-3)(\sqrt{x}-1)}{(2{{x}^{2}}+x-3)}$

$=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{(2x-3)(\sqrt{x}-1)}{\left( 2x+3 \right)\left( x-1 \right)}$

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

${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

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

$\underset{x\to 1}{\mathop{\lim }}\,\dfrac{(2x-3)(\sqrt{x}-1)}{\left( 2x+3 \right)\left( x-1 \right)}$

$=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{(2x-3){\left( \sqrt{x}-1 \right)}}{\left( 2x+3 \right){\left( \sqrt{x}-1 \right)}\left( \sqrt{x}+1 \right)}$

$=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{(2x-3)}{\left( 2x+3 \right)\left( \sqrt{x}+1 \right)}$

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

$=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{(2x-3)}{\left( 2x+3 \right)\left( \sqrt{x}+1 \right)}$

$=\dfrac{\left( 2-3 \right)}{\left( 2+3 \right)\left( 1+1 \right)}$

$=\dfrac{-1}{10}$

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

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,

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\tan 4x}{1-\cos 4x}$

$=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\sin 4x}{\cos 4x[2{{\sin }^{2}}2x]}$ 

$=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2x{\sin 2x}\cos 2x}{\cos 4x[2{{\sin }^{{{2}}}}2x]}$

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

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

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

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

$=\dfrac{1}{2}$

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

Ans:

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

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

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

\[\dfrac{dy}{dx}=\dfrac{(1+\tan x)\dfrac{d}{dx}(1-\tan x)-(1-\tan x)\dfrac{d}{dx}(1+\tan x)}{{{(1+\tan x)}^{2}}}\] 

\[\dfrac{dy}{dx}=\dfrac{(1+\tan x)(-{{\sec }^{2}}x)-(1-\tan x)({{\sec }^{2}}x)}{{{(1+\tan x)}^{2}}}\]

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

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

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

\[=\dfrac{-2}{{{\cos }^{2}}x{{\left[ 1+\dfrac{\sin x}{\cos x} \right]}^{2}}}\]

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

\[=\dfrac{-2}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}\]

\[=\dfrac{-2}{1+{{\sin }^{2}}x}\]

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

\[\dfrac{dy}{dx}=\dfrac{-2}{1+{{\sin }^{2}}x}\]

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,

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

\[={{e}^{\sqrt{\cot x}}}\dfrac{d}{dx}\left( \sqrt{\cot x} \right)\]

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

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

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

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

6. Let $f(x)=\left\{ \begin{align} & a+bx\text{  },x<1 \\  & 4\text{         },x=1 \\ & b-ax\text{  },x>1 \\  \end{align} \right.$ and if $\underset{x\to 1}{\mathop{\lim }}\,f(x)=f(1)$. 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,

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

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

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

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

For \[\text{ }x>1\] we have the given function as,

\[f(x)=a+bx\]

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

\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,(a+bx)\]

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

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,

\[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,(b-ax)\]

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

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

$4+4=\left( a+b \right)+\left( b-a \right)$

$\Rightarrow 8=2b$

$\Rightarrow b=4$

Similarly we get the value of other variable as,

$\Rightarrow a=0$

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

Ans:

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

$y={{\left( {{a}^{2}}-{{x}^{2}} \right)}^{\dfrac{-1}{2}}}$

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

$\dfrac{dy}{dx}=\dfrac{-1}{2}{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{\dfrac{-1}{2}-1}}\dfrac{d}{dx}\left( {{a}^{2}}-{{x}^{2}} \right)$

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

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

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

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

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

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

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

Ans:

First, consider the given expression,

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

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

$\dfrac{1-\tan x}{1+\tan x}=t$ 

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

\[\dfrac{dt}{dx}=\dfrac{(1+\tan x)\dfrac{d}{dx}(1-\tan x)-(1-\tan x)\dfrac{d}{dx}(1+\tan x)}{{{(1+\tan x)}^{2}}}\]

\[=\dfrac{(1+\tan x)(0-{{\sec }^{2}}x)-(1-\tan x)(0+{{\sec }^{2}}x)}{{{(1+\tan x)}^{2}}}\]

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

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

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,

\[\dfrac{dy}{dt}=\dfrac{d}{dt}{{t}^{\dfrac{1}{2}}}\]

\[=\dfrac{1}{2}{{t}^{\dfrac{1}{2}-1}}\]

\[=\dfrac{1}{2\sqrt{t}}\]

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

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

Using the standard definition of chain rule for differentiation, we get,
\[\dfrac{dy}{dt}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}\]

\[=\dfrac{-{2}{{\sec }^{2}}x}{{{(1+\tan x)}^{2}}}\times \dfrac{1}{{{2}}}\sqrt{\dfrac{1+\tan x}{1-\tan x}}\]

\[=\dfrac{-{{\sec }^{2}}x}{{{(1+\tan x)}^{\dfrac{3}{2}}}{{(1-\tan x)}^{\dfrac{1}{2}}}}\]

9. Differentiate 

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

Ans:  

We are given the expression,

$\dfrac{\sin x+\cos x}{\sin x-\cos x}$,

which we have to differentiate. 

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

$=\dfrac{(\sin x-\cos x)\left( \cos x-\sin x \right)-(\sin x+\cos x)\left( \cos x+\sin x \right)}{{{(\sin x-\cos x)}^{2}}}$

On simplifying, we get

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

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

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

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

Ans:

We are given the expression, 

$\dfrac{\sin x-1}{\sec x+1}$

which we have to differentiate. 

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

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

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

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

10. Evaluate $\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\sin x-\cos x}{\left( x-\dfrac{\pi }{4} \right)}$

Ans:

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

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

$\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\sin x-\cos x}{\left( x-\dfrac{\pi }{4} \right)}=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{\left[ \sin \left( \dfrac{\pi }{4}+y \right)-\cos \left( \dfrac{\pi }{4}+y \right) \right]}{y}$

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

$\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\dfrac{\sin x-\cos x}{\left( x-\dfrac{\pi }{4} \right)}=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{\left( \sin \dfrac{\pi }{4}\cos y+\cos \dfrac{\pi }{4}\sin y-\cos \dfrac{\pi }{4}\cos y+\sin \dfrac{\pi }{4}\sin y \right)}{y}$ 

$=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{\left( \left( \sin \dfrac{\pi }{4}-\cos \dfrac{\pi }{4} \right)\cos y+\left( \cos \dfrac{\pi }{4}+\sin \dfrac{\pi }{4} \right)\sin y \right)}{y}$

Evaluating further, 

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

$=\sqrt{2}\times 1$

$=\sqrt{2}$

11.Evaluate \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{(1+x)}^{6}}-1}{{{(1+x)}^{5}}-1}\].

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, 

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

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

\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{(1+x)}^{6}}-1}{{{(1+x)}^{5}}-1}=\underset{y\to 1}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dy}\left( {{y}^{6}}-1 \right)}{\dfrac{d}{dy}\left( {{y}^{5}}-1 \right)}\] 

\[=\underset{y\to 1}{\mathop{\lim }}\,\dfrac{\left( 6{{y}^{5}} \right)}{\left( 5{{y}^{4}} \right)}\]

\[=\dfrac{6}{5}\]

12. Evaluate $\underset{x\to a}{\mathop{\lim }}\,\dfrac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$.

Ans:

The given expression is 

$\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left( \sqrt{a+2x}-\sqrt{3x} \right)}{\left( \sqrt{3a+x}-2\sqrt{x} \right)}$ 

We will evaluate its limit, using the rationalization method,

\[\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left( \sqrt{a+2x}-\sqrt{3x} \right)}{\left( \sqrt{3a+x}-2\sqrt{x} \right)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left( \sqrt{a+2x}-\sqrt{3x} \right)}{\left( \sqrt{3a+x}-2\sqrt{x} \right)}\times \dfrac{\left( \sqrt{3a+x}+2\sqrt{x} \right)}{\left( \sqrt{3a+x}+2\sqrt{x} \right)}\times \dfrac{\left( \sqrt{a+2x}+\sqrt{3x} \right)}{\left( \sqrt{a+2x}+\sqrt{3x} \right)}\]

\[=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left[ (a+2x)-3x \right]\times \left( \sqrt{3a+x}+2\sqrt{x} \right)}{\left[ (3a+x)-4x \right]\times \left( \sqrt{a+2x}+\sqrt{3x} \right)}\]

\[=\underset{x\to a}{\mathop{\lim }}\,\dfrac{{(a-x)}\times \left( \sqrt{3a+x}+2\sqrt{x} \right)}{3{(a-x)}\times \left( \sqrt{a+2x}+\sqrt{3x} \right)}\]

\[=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left( \sqrt{3a+x}+2\sqrt{x} \right)}{3\left( \sqrt{a+2x}+\sqrt{3x} \right)}\]

On simplifying, we get, 

$\underset{x\to a}{\mathop{\lim }}\,\dfrac{\left( \sqrt{a+2x}-\sqrt{3x} \right)}{\left( \sqrt{3a+x}-2\sqrt{x} \right)}=\dfrac{\left( \sqrt{4a}+2\sqrt{a} \right)}{3\left( \sqrt{3a}+\sqrt{3a} \right)}$ 

$=\dfrac{4\sqrt{a}}{6\sqrt{3}\sqrt{a}}$

$=\dfrac{2}{3\sqrt{3}}$

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

Ans:

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

at the point $x=1$.

Differentiating the function, we get, 

$f'(x)=\dfrac{d}{dx}\left( 1+x+{{x}^{2}}+{{x}^{3}}+\ldots +{{x}^{50}} \right)$

$=0+1+2x+3{{x}^{2}}+\ldots +50{{x}^{49}}$

$=1+2x+3{{x}^{2}}+\ldots +50{{x}^{49}}$

Now, evaluating the derivative at $x=1$, and using the formula for sum of first $n$ natural numbers, $\left[ 1+2+3+\cdots +n=\dfrac{n(n+1)}{2} \right]$
$f'(1)=1+2+3+\cdots +50$ 

$=~\dfrac{{{{50}}^{25}}(50+1)}{{{2}}}$

$=25\times 51$

$=1305$

14. Find the derivative of ${{\sin }^{2}}x$ with respect to $x$ using product rule.

Ans: 

Let the given function be $y={{\sin }^{2}}x$. 

We can write it as product of two functions. 

$\Rightarrow y=\sin x\times \sin x$

Differentiating, using product rule,

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sin x\times \sin x \right)$

$\text{=}\sin x\dfrac{d}{dx}\sin x+\sin x\dfrac{d}{dx}\sin x$

$\text{      =}\sin x\cos x+\sin x\cos x$

$\text{=}\sin 2x$

15. Find the derivative of $\dfrac{{{x}^{5}}-\cos x}{\sin x}$ with respect to$x$ .

Ans:

Let the given expression be  $y=\dfrac{{{x}^{5}}-\cos x}{\sin x}$.
Differentiating using quotient rule,

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{{{x}^{5}}-\cos x}{\sin x} \right)$ 

$\text{    }=\dfrac{\sin x\dfrac{d}{dx}\left( {{x}^{5}}-\cos x \right)-\left( {{x}^{5}}-\cos x \right)\dfrac{d}{dx}\sin x}{{{\sin }^{2}}x}$

$\text{    }=\dfrac{\sin x\left( 5{{x}^{4}}+\sin x \right)-\left( {{x}^{5}}-\cos x \right)\cos x}{{{\sin }^{2}}x}$

On simplifying, we get, 

$\text{    }=\dfrac{5{{x}^{4}}\sin x+{{\sin }^{2}}x-{{x}^{5}}\cos x+{{\cos }^{2}}x}{{{\sin }^{2}}x}$

$\text{    }=\dfrac{5{{x}^{4}}\sin x-{{x}^{5}}\cos x+1}{{{\sin }^{2}}x}$

16. Find $\underset{x\to 0}{\mathop{\lim }}\,f(x)$ when $f(x)=\left\{ \begin{align} & \dfrac{\left| x \right|}{x},x\ne 0 \\  & 0,x=0 \\ \end{align} \right.$.

Ans:

We are given the function as 

$f(x)=\left\{ \begin{align} & \dfrac{\left| x \right|}{x},x\ne 0 \\  & 0,x=0 \\ \end{align} \right.$ 

We know that, by definition,

$\left| x \right|=\left\{ \begin{align} & x,x\ge 0 \\  & -x,x<0 \\ \end{align} \right.$ 

$\Rightarrow f(x)=\left\{ \begin{align} & \dfrac{x}{x}=1,x>0 \\  & \dfrac{-x}{x}=-1,x<0 \\  & 0,x=0 \\ \end{align} \right.$

Now, we evaluate the limits separately:

L.H.L as  $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,\,\,-1=-1$

and

R.H.L as $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,1=1$ 

As, we can see that L.H.L$\ne $ R.H.L,

$\therefore \underset{x\to 0}{\mathop{\lim }}\,f(x)$ does not exist.

17. Find the derivative of the function $f(x)=2{{x}^{2}}+3x-5$ at $x=-1$ . 

Also show that $f'(0)+3f'(-1)=0$

Ans:

We are required to find the derivative of the function$f(x)=2{{x}^{2}}+3x-5$

at the point  $x=-1$. 

Differentiating, we get, 

${{f}^{1}}(x)=\dfrac{d}{dx}\left( 2{{x}^{2}}+3x-5 \right)$ 

$\text{        }=4x+3$

At $x=-1$ , the derivative is obtained to be

${{f}^{1}}(-1)=4\times (-1)+3$

$\text{         }=-4+3$

$\text{         }=-1$

Similarly, we evaluate at $x=0$,

${{f}^{1}}(0)=4\times 0+3$

$\text{       }=3$

This gives us,

${{f}^{1}}(0)+3{{f}^{1}}(-1)=3+3\times -1$

$\text{                       }=3-3$

$\text{                       }=0$

Hence proved.

18. Evaluate$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{ax+x\cos x}{b\sin x}$ .

Ans:

We have to evaluate the expression

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{ax+x\cos x}{b\sin x}$ 

First, we will simplify the expression.

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{ax+x\cos x}{b\sin x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x[a+\cos x]}{xb\dfrac{\sin x}{x}}$

$=\dfrac{\underset{x\to 0}{\mathop{\lim }}\,\left( a+\cos x \right)}{\underset{x\to 0}{\mathop{\lim }}\,b\dfrac{\sin x}{x}}$

$=\dfrac{a+1}{b\times 1}$

$=\dfrac{a+1}{b}$

Using the formula,  $\underset{x\to 0}{\mathop{\lim }}\,\cos x=1$ and $\underset{x\to 0}{\mathop{\text{  }\lim }}\,\dfrac{\sin x}{x}=1$, we get
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{ax+x\cos x}{b\sin x}=\dfrac{a+1}{b\times 1}$

$=\dfrac{a+1}{b}$

20. Find the derivative of $\tan x$ by the first principle.

Ans:

Let the given function be $f(x)=\tan x$. From this, we get

$f(x+h)=\tan (x+h)$.

Using the limit formula for derivative, we get, 

${{f}^{1}}(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$

$\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan (x+h)-\tan (x)}{h}$

$\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}}{h}$

Simplifying, we get,  

$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (x+h)\cos x-\cos (x+h)\sin x}{h\cos (x+h)\cos x}$

$\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (x+h-x)}{h\cos (x+h)\cos x}$

$\text{        }=\dfrac{\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \,h}{h}}{\underset{h\to 0}{\mathop{\lim }}\,\cos (x+h)\cos xh}$

Evaluating the  limits, 

$f'(x)=\dfrac{1}{\cos (x+0)\cos x}$

$\text{        }=\dfrac{1}{{{\cos }^{2}}x}$

$\text{        }={{\sec }^{2}}x$

21. Evaluate  $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+{{x}^{3}}+\cdots +{{x}^{n}}-n}{(x-1)}$ .

Ans:

We are required to evaluate the limit of the expression,  $\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+{{x}^{3}}+\cdots +{{x}^{n}}-n}{(x-1)}$ 

Separating the $n$ into $1$’s,

\[\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+{{x}^{3}}+\cdots +{{x}^{n}}-n}{(x-1)}\]

\[=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{(x-1)+({{x}^{2}}-1)+\left( ({{x}^{3}}-1+\cdots +({{x}^{n}}-1) \right)}{(x-1)}\]

\[=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{{(x-1)}\left[ 1+(x+1)+({{x}^{2}}+x+1)+\cdots +{{x}^{n-1}}+{{x}^{n-2}}+\cdots +1 \right]}{{(x-1)}}\]

\[=1+2+3+\cdots +n\]

\[=\dfrac{n(n+1)}{2}\]

22. Evaluate$\underset{x\to 4}{\mathop{\lim }}\,\dfrac{\left| 4-x \right|}{x-4}$ (if it exist).

Ans:

We are required to evaluate the limit of the expression:

$\underset{x\to 4}{\mathop{\lim }}\,\dfrac{\left| 4-x \right|}{x-4}$

We will evaluate the limits  separately:

L.H.L as $\underset{x\to {{4}^{-}}}{\mathop{\lim}}\,\dfrac{-(4-x)}{x-4}=\underset{x\to 4}{\mathop{\lim }}\,\dfrac{-({4}-x)}{-({4}-x)}=1$

and

R.H.L as $\underset{x\to {{4}^{+}}}{\mathop{\lim }}\,\dfrac{4-x}{x-4}=\underset{x\to 4}{\mathop{\lim }}\,\dfrac{-(x-4)}{(x-4)}=-1$

Clearly, we can see that L.H.L $\ne $ R.H.L.

$\therefore \underset{x\to 4}{\mathop{\lim }}\,\dfrac{\left| 4-x \right|}{x-4}$ does not exist.

23. For what integers $m$and $n$ does both $\underset{x\to 4}{\mathop{\lim }}\,f(x)$ and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$ exist if $f(x)=\left\{ \begin{align} & m{{x}^{2}}+n,x<0 \\  & nx+m,0\le x\le 1 \\  & n{{x}^{3}}+m,x>1 \\  \end{align} \right.$ .

Ans:

Consider the point $x=0$. 

The left-hand limit at this point is 

$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,m{{x}^{2}}+n$

$\text{             }=m$

Similarly, the right-hand limit at the point is 

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,nx+m$

$\text{             }=m$

As, \[L.H.L.=R.H.L.\]

We get that,

\[\underset{x\to 0}{\mathop{\lim }}\,f(x)\] exist, and is obtained to be,

\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\]

\[\text{           }n=m\]

Therefore, for all real numbers $m=n,\,$the limit $\,\underset{x\to 0}{\mathop{\lim }}\,f(x)$ exists.

Consider the point $x=1$ 

The left hand limit at this point is
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,nx+m$

$\text{             }=n+m$

Similarly, the right hand limit at the point is 

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,n{{x}^{3}}+m$

$=n+m$

As, \[L.H.L.=R.H.L.\]

We get that,

\[\underset{x\to 1}{\mathop{\lim }}\,f(x)\] exist, and is obtained to be,

\[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)\]

\[n+m=n+m\]

At all integral values of $m+n,$ the $\underset{x\to 1}{\mathop{\lim }}\,f(x)$ exists.

24. If $y=\sqrt{x}+\dfrac{1}{\sqrt{x}}$ , prove that $2x\dfrac{dy}{dx+y}+y=2\sqrt{x}$ .

Ans:

We are given the function:

$y=\sqrt{x}+\dfrac{1}{\sqrt{x}}={{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{-1}{2}}}$

Differentiating with respect to $x$, we get,

$\dfrac{dy}{dx}=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}+\left( \dfrac{-1}{2} \right){{x}^{\dfrac{-3}{2}}}$  

$\text{    }=\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2{{x}^{\dfrac{3}{2}}}}$

Simplifying, 

$2x\dfrac{dy}{dx}=\sqrt{x}-\dfrac{1}{\sqrt{x}}$

$2x\dfrac{dy}{dx}+y=\left( \sqrt{x}-\dfrac{1}{\sqrt{x}} \right)+\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)$

$2x\dfrac{dy}{dx}+y=2\sqrt{x}$

Hence proved.

25. Evaluate  $\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{1+\cos 2x}{{{\left( \pi -2x \right)}^{2}}}$ .

Ans:

We have to find the value of the limit,

$\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{1+\cos 2x}{{{\left( \pi -2x \right)}^{2}}}$. 

We will use the substitution, $\pi -2x=y$

$\Rightarrow 2x=\pi -y$ 

and the limits become,

$x\to \dfrac{\pi }{2}$ , $y\to 0$ 

Substituting these values, in the given function, 

\[\underset{y\to 0}{\mathop{\lim }}\,\dfrac{1+\cos (\pi -y)}{{{y}^{2}}}=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{1}{2}\dfrac{1-\cos y}{{{y}^{2}}}\]

\[\text{                           }=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{2{{\sin }^{2}}\dfrac{{{y}^{2}}}{2}}{4\times \dfrac{{{y}^{2}}}{4}}\]

Simplifying,

\[\underset{x\to \dfrac{\pi }{2}}{\mathop{\lim }}\,\dfrac{1+\cos 2x}{{{\left( \pi -2x \right)}^{2}}}=\underset{y\to 0}{\mathop{\lim }}\,\dfrac{1}{2}\times \dfrac{{{\sin }^{2}}\dfrac{{{y}^{2}}}{2}}{{{\left( \dfrac{y}{4} \right)}^{2}}}\] 

\[\text{                     }=\dfrac{1}{2}\underset{y\to 0}{\mathop{\lim }}\,{{\left[ \dfrac{\sin \dfrac{y}{2}}{\dfrac{y}{2}} \right]}^{2}}\]

\[\text{                     }=\dfrac{1}{2}\times 1\]

\[\text{                     }=\dfrac{1}{2}\]

26. Differentiate the function $y=\dfrac{(x+2)(3x-1)}{(2x+5)}$ with respect to$x$.

Ans: 

We are given the function, 

$y=\dfrac{(x+2)(3x-1)}{(2x+5)}$. 

Differentiating using the quotient rule, 

$\dfrac{dy}{dx}=\dfrac{d}{dx}\dfrac{(x+2)(3x-1)}{(2x+5)}$

$\text{    }=\dfrac{(2x+5)\dfrac{d}{dx}(x+2)(3x-1)-(x+2)(3x-1)\dfrac{d}{dx}(2x+5)}{{{(2x+5)}^{2}}}$

$\text{    }=\dfrac{(2x+5)\left[ (x+2)\dfrac{d}{dx}(3x-1)-(3x-1)\dfrac{d}{dx}(x+2) \right]-(x+2)(3x-1)\left[ 2+0 \right]}{{{(2x+5)}^{2}}}$

Simplifying,

$\dfrac{dy}{dx}=\dfrac{(2x+5)\left[ (x+2)\times 3+(3x-1)\times 1 \right]-2\left[ 3{{x}^{2}}+6x-x-2 \right]}{{{(2x+5)}^{2}}}$

$\text{    }=\dfrac{(2x+5)\left[ 3x+6+3x-1 \right]-6{{x}^{2}}-12x+2x+4}{{{(2x+5)}^{2}}}$

$\text{    }=\dfrac{12{{x}^{2}}+30x+{10x}+25-6{{x}^{2}}-{10x}+4}{{{(2x+5)}^{2}}}$

$\text{    }=\dfrac{6{{x}^{2}}+30x+29}{{{(2x+5)}^{2}}}$

27. Find $\underset{x\to 5}{\mathop{\lim }}\,\left| x \right|=5$ .

Ans:

First, we will evaluate the left hand limit, i.e., $\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)$.

Let $x=5-h$. Then, we get $x\to 5,$ as $h\to 0$.

Evaluating the limits, 

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(5-h)$

$=\underset{h\to 0}{\mathop{\lim }}\,\left| 5-h \right|-5$

$=0$

Next, we will evaluate the right hand limit, i.e., $\underset{h\to {{5}^{+}}}{\mathop{\lim }}\,f(x)$.

Let $x=5+h$. Then,  as $h\to 0$,$x\to 5,$

Evaluating the limits, 

$\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(5+h)$

$=\underset{h\to 0}{\mathop{\lim }}\,\left| 5+h \right|-5$

$=0$

As, L.H.L.= R.H.L., 

$\underset{x\to 5}{\mathop{\lim }}\,f(x)$ exists and $\underset{x\to 5}{\mathop{\lim }}\,f(x)=0$ 

28. Find$\underset{x\to 0}{\mathop{\lim }}\,f(x)$ and $\underset{x\to 1}{\mathop{\lim }}\,f(x)$ where $f(x)=\left\{ \begin{align} & 2x+3,x\le 0 \\  & 3(x+1),x>0 \\  \end{align} \right.$ .

Ans:

Given function is $f(x)=\left\{ \begin{align} & 2x+3,x\le 0 \\  & 3(x+1),x>0 \\ \end{align} \right.$

We can calculate the limits separately as:

At the point $x=0$ ,

L.H.L. as $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,2x+3=3$

and 

R.H.L. as $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,3(x+1)=3$

As, L.H.L. = R.H.L., we have $\underset{x\to 0}{\mathop{\lim }}\,f(x)\text{ }$exists and $\underset{x\to 0}{\mathop{\lim }}\,f(x)=3$.

At the point $x=1$ ,

L.H.L. as $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,3(x+1)=3(1+1)=6$

and 

R.H.L. as $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 1}{\mathop{\lim }}\,3(x+1)=3(1+1)=6$

As, L.H.L. = R.H.L., we have $\underset{x\to 1}{\mathop{\lim }}\,f(x)$exists and 
$\underset{x\to 1}{\mathop{\lim }}\,f(x)=6$.

29. Find the derivative of $\sec x$ by the first principle. Ans:

Let the given function be  $f(x)=\sec x$ . 

From this, we can write,

\[f(x+h)=\sec (x+h)\]

By definition, we have, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]

Substituting the values of the functions, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec (x+h)-\sec x}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\cos (s+h)}-\dfrac{1}{\cos x}}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos x-\cos (x+h)}{\cos (x+h)\cos xh}\]

Simplifying, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-2\sin \left[ \dfrac{2x+h}{2} \right]\sin \left( \dfrac{-h}{2} \right)}{\cos (x+h)\cos xh}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sin \left[ \dfrac{2x+h}{2} \right]\sin \dfrac{h}{2}}{\cos (x+h)\cos xh}\]

Using the formula \[\sin (-\theta )=-\sin \theta \] , we get,

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sin \dfrac{h}{2}}{2\dfrac{h}{2}}\times \dfrac{\underset{h\to 0}{\mathop{\lim }}\,\sin \dfrac{(2x+h)}{2}}{\underset{h\to 0}{\mathop{\lim }}\,\cos (x+h)\cos x}\]

\[\text{        }=1\times \dfrac{\sin \left( \dfrac{2x+0}{2} \right)}{\cos (x+0)\cos x}\]

\[\text{        }=\dfrac{\sin x}{\cos x\cos x}\]

\[\text{        }=\tan x\sec x\]

30. Find derivative of $f(x)=\dfrac{4x+5\sin x}{3x+7\cos x}$ . Ans: 

We have to find the derivative of the function 

$f(x)=\dfrac{4x+5\sin x}{3x+7\cos x}$ .

Differentiating using the quotient rule, 

${{f}^{1}}(x)=\dfrac{(3x+7\cos x)\dfrac{d}{dx}(4x+5\sin x)-(4x+5\sin x)\dfrac{d}{dx}(3x+7\cos x)}{{{(3x+7\cos x)}^{2}}}$

$\text{         }=\dfrac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{{{(3x+7\cos x)}^{2}}}$

Simplifying,

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

$\text{         }=\dfrac{15x\cos x+35[{{\sin }^{2}}x+{{\cos }^{2}}x]+28\cos x+43\sin x}{{{(3x+7\cos x)}^{2}}}$

$\text{         }=\dfrac{15x\cos x+35+28\cos x+43\sin x}{{{(3x+7\cos x)}^{2}}}$

31. Find derivative of $\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}$ . Ans:

We have to find the derivative of $\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}$.

Differentiating using the quotient rule, 

$\dfrac{d}{dx}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=\dfrac{(x-a)\dfrac{d}{dx}({{x}^{n}}-{{a}^{n}})-({{x}^{n}}-{{a}^{n}})\dfrac{d}{dx}(x-a)}{{{(x-a)}^{2}}}$ 

$=\dfrac{(x-a)\left[ n{{x}^{n-1}}-0 \right]-({{x}^{n}}-{{a}^{n}})[1-0]}{{{(x-a)}^{2}}}$

Simplifying,

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

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

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

6 Marks Questions

1. Differentiate $\tan x$ from first principle.

Ans:

Let the given function be $f(x)=\tan x$. From this we get,
\[f(x+h)=\tan (x+h)\]

By definition, we have, 

$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$

Substituting the values of the functions, we get,

\[{{f}^{'}}(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\tan (x+h)-\tan x}{h}\] 

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (x+h)\cos x-\cos (x+h)\sin x}{h\cos (x+h)\cos x}\]

Simplifying, we get, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (x+h-x)}{h\cos (x+h)\cos x}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin h}{h\cos (x+h)\cos x}\]

\[\text{        }=\dfrac{\underset{h\to 0}{\mathop{\lim }}\,\sin h}{\underset{h\to 0}{\mathop{\lim }}\,\cos (x+h)\cos x}\]

Applying the formula \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin h}{h}=1\], we get, 

\[f'(x)=\dfrac{1}{\cos x\cos x}\]

\[\text{       }=\dfrac{1}{{{\cos }^{2}}x}\]

\[\text{       }={{\sec }^{2}}x\]

2. Differentiate ${{(x+4)}^{5}}$ from first principle.

Ans: 

Let the given function be$f(x)={{(x+4)}^{5}}$ . From this we get,

$f(x+h)={{(x+h+4)}^{5}}$

By definition, we have, 

$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$

Substituting the values of the functions, and applying the formula 

$\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$, we get, 

$\text{  }f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(x+h+4)}^{5}}-{{(x+4)}^{5}}}{h}$ 

$\text{         }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(x+h+4)}^{5}}-{{(x+4)}^{5}}}{(x+h+4)-(x+4)}$

$\text{         }=6{{(x+4)}^{(6-1)}}$

$\text{         }=6{{(x+4)}^{5}}$

3. Differentiate $\text{cosec}x$ from first principle.

Ans: 

Let the given function be $f(x)\,\text{=}\,\text{cosec}x$.

By definition, 

$f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$ .

Therefore, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\text{cosec}(x+h)-\text{cosec}(x)}{h}\]

\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\sin (x+h)}-\dfrac{1}{\sin x}}{h}\]

\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin x-\sin (x+h)}{h\sin (x+h)\sin x}\]

Simplifying, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \dfrac{x+x+h}{2}\sin \dfrac{x-x+h}{2}}{h\sin (x+h)\sin x}\]

\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \left( x+\dfrac{h}{2} \right)\sin \left( \dfrac{-h}{2} \right)}{h\sin (x+h)\sin x}\]

\[=\dfrac{\underset{h\to 0}{\mathop{\lim }}\,\cos \left( x+\dfrac{h}{2} \right)}{\cos x\underset{h\to 0}{\mathop{\lim }}\,\sin (x+h)}\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \dfrac{h}{2}}{\dfrac{h}{2}}\]

Evaluating the limits, 

\[f'(x)=-\dfrac{\cos x}{\sin x\sin x}\cdot 1\] 

\[=-\text{cosec }x\cot x\]

4. Find the derivatives of the following functions.

i) ${{\left( x-\dfrac{1}{x} \right)}^{3}}$

Ans:

Let the given function be $f(x)={{\left( x-\dfrac{1}{x} \right)}^{3}}$ 

Expanding the function, 

$f(x)={{x}^{3}}-\dfrac{1}{{{x}^{3}}}-3\left( x.\dfrac{1}{x} \right)\left( x-\dfrac{1}{x} \right)$

$={{x}^{3}}-{{x}^{-3}}-3x+3{{x}^{-1}}$

Differentiating with respect to $x$ , we get

$f'(x)=3{{x}^{2}}-(-3){{x}^{-4}}-3+3(-1){{x}^{-2}}$

$\text{        }=3{{x}^{2}}+\dfrac{3}{{{x}^{4}}}-3-\dfrac{3}{{{x}^{2}}}$

ii) $\dfrac{(3x+1)(2\sqrt{x-1})}{\sqrt{x}}$

Ans:

Let the given function be $f(x)=\dfrac{(3x+1)(2\sqrt{x-1})}{\sqrt{x}}$.

On simplifying, 

$f(x)\text{ =}\dfrac{6x\sqrt{x-1}+2\sqrt{x-1}}{\sqrt{x}}$ 

$f(x)\text{ =}6\sqrt{{{x}^{2}}-x}+2\sqrt{1-\dfrac{1}{x}}$

Differentiating with respect to $x$, we get

\[f'(x)\text{ = }6\dfrac{d}{dx}\left( \sqrt{{{x}^{2}}-x} \right)+2\dfrac{d}{dx}\left( \sqrt{1-\dfrac{1}{x}} \right)\]

\[=6\dfrac{1}{2}{{\left( {{x}^{2}}-x \right)}^{-\dfrac{1}{2}}}\dfrac{d}{dx}\left( {{x}^{2}}-x \right)+2\dfrac{1}{2}{{\left( 1-\dfrac{1}{x} \right)}^{-\dfrac{1}{2}}}\dfrac{d}{dx}\left( 1-\dfrac{1}{x} \right)\]

\[=\dfrac{3\left( 2x-1 \right)}{\sqrt{\left( {{x}^{2}}-x \right)}}+\left( -\dfrac{1}{{{x}^{2}}} \right)\sqrt{\dfrac{x}{x-1}}\]

5. If $f(x)=\left\{ \begin{align} & \left| x \right|+a,x<0 \\ & 0,x=0 \\  & \left| x \right|-a,x>0 \\  \end{align} \right.$ , for what values of $a$ does $\underset{x\to 0}{\mathop{\lim }}\,f(x)$ exist?

Ans: 

Given that$f(x)=\left\{ \begin{align} & \left| x \right|+a,x<0 \\  & 0,x=0 \\  & \left| x \right|-a,x>0 \\  \end{align} \right.$ .

At \[a=0\], the limits calculated separately are, 

L.H.L as 

\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left| x \right|+a\]

\[\text{                        }=\underset{x\to 0}{\mathop{\lim }}\,-x+a=a\]

and

R.H.L. as 

\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\left| x \right|-a\]

\[\text{                        }=\underset{x\to 0}{\mathop{\lim }}\,-x-a=-a\]

It is given that the limits exists, that means 

\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\]

\[\Rightarrow a=-a\]

\[\Rightarrow 2a=0\]

\[\Rightarrow a=0\]

At $a=0$ , \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\] exists.

6. Find the derivative of  $\sin (x+1)$ , with respect to $x$, from first principle.

Ans:

Let the given function be $f(x)=\sin (x+1)$. From this we get,

\[f(x+h)=\sin (x+h+1)\]. 

Obtaining the derivative, using the limits definition, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (x+h+1)-\sin (x+1)}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \left[ \dfrac{x+h+1+x+1}{2} \right]\sin \left[ \dfrac{x+h+1-x-1}{2} \right]}{h}\]

On simplifying, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \left[ x+1+\dfrac{h}{2} \right]\sin \left[ \dfrac{h}{2} \right]}{h}\] 

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,{2}\cos \left( x+1+\dfrac{h}{2} \right)\times \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \dfrac{h}{2}}{{2}\dfrac{h}{2}}\]

\[\text{       }=\cos (x+1)\times 1\]

\[\text{       }=\cos (x+1)\]

7. Differentiate $\sin x+\cos x$ from first principle.

Ans: 

Let he given function be$f(x)=\sin x+\cos x$.  

This gives us, 

\[f(x+h)=\sin (x+h)+\cos (x+h)\] .

Evaluating the derivative using limits definition,

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left[ \sin (x+h)+\cos (x+h) \right]-\left[ \sin x+\cos x \right]}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left[ \sin (x+h)-\sin x \right]-\left[ \cos (x+h)+\cos x \right]}{h}\]

On simplifying, 

\[f'(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \left( \dfrac{x+h+x}{2} \right)\sin \dfrac{\left( x+h-x \right)}{2}-2\sin \dfrac{\left( x+h-x \right)}{2}\times \sin \left( \dfrac{x+h+x}{2} \right)}{h}\]

\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\cos \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{h}+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-2\sin \left( x+\dfrac{h}{2} \right)\sin \dfrac{h}{2}}{h}\]\[\text{        }=\underset{h\to 0}{\mathop{\lim }}\,{2}\cos \left( x+\dfrac{h}{2} \right)\dfrac{\sin \dfrac{h}{2}}{{2}\dfrac{h}{2}}+\underset{h\to 0}{\mathop{\lim }}\,-{2}\sin \left( x+\dfrac{h}{2} \right)\dfrac{\sin \dfrac{h}{2}}{{2}\dfrac{h}{2}}\]\[\text{        }=\cos (x+0)\times 1-\sin (0+x)\times 1\]

\[=\cos x-\sin x\]

8. Find derivative of 

i) $\dfrac{x\sin x}{1+\cos x}$

Ans:

We have to find the derivative of  

$\dfrac{x\sin x}{1+\cos x}$.

Using the quotient rule, we get,                     

$\dfrac{d}{dx}\dfrac{x\sin x}{1+\cos x}$

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

$=\dfrac{(1+\cos x)\left[ x\dfrac{d}{dx}(\sin x)+\sin x\dfrac{d}{dx}(x) \right]-x\sin x[0-\sin x]}{{{(1+\cos x)}^{2}}}$

$=\dfrac{(1+\cos x)\left[ x\cos x+\sin x\times 1 \right]+x{{\sin }^{2}}x}{{{(1+\cos x)}^{2}}}$

$\dfrac{d}{dx}\dfrac{x\sin x}{1+\cos x}$

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

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

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

ii) $(ax+b){{(cx+d)}^{2}}$

Ans:

We need to find the derivative of the expression:

$(ax+b){{(cx+d)}^{2}}$

Differentiating using product rule, we get, 

$\dfrac{d}{dx}(ax+b){{(cx+d)}^{2}}$

$=(ax+b)\dfrac{d}{dx}{{(cx+d)}^{2}}+{{(cx+d)}^{2}}\dfrac{d}{dx}(ax+b)$

$=2(ax+b)(cx+d)\dfrac{d}{dx}(cx+d)+{{(cx+d)}^{2}}\times a$

$=2(ax+b)(cx+d)\times c+a{{(cx+d)}^{2}}$

\[\dfrac{d}{dx}(ax+b){{(cx+d)}^{2}}\]

\[=(cx+d)\left[ 2c(ax+b)+a(cx+d) \right]\]

\[=(cx+d)\left[ 2acx+2bc+acx+ad \right]\]

\[=(cx+d)\left[ 3acx+2bc+ad \right]\]

9. Evaluate  $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}$ .

Ans:  

We have to evaluate the limit of the expression, 

$\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}$

Using the formula for the derivative, obtained from limits, we get,

\[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}\]

\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{({{a}^{2}}+2ah+{{h}^{2}})\sin (a+h)-{{a}^{2}}\sin a}{h}\]

\[=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{2}}\sin (a+h)+2ah\sin (a+h)+{{h}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}\]

On simplifying,

\[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{a}^{2}}2\cos \left( \dfrac{2a+h}{2} \right)\sin \dfrac{h}{2}}{2\dfrac{h}{2}}+\underset{h\to 0}{\mathop{\lim }}\,2ah\sin (a+h)+\underset{h\to 0}{\mathop{\lim }}\,h\sin (a+h)\] 

\[={{a}^{2}}2\cos \left( \dfrac{2a+0}{2} \right)\times 1+2a\sin (a+0)+0\times \sin a\]

\[={{a}^{2}}2\cos a+2a\sin a\]

10. Differentiate

i) $\left( \dfrac{a}{{{x}^{4}}} \right)-\dfrac{b}{{{x}^{2}}}+\cos x$ 

Ans:

Differentiating the given expression with respect to $x$, 

$\dfrac{d}{dx}\left[ \left( \dfrac{a}{{{x}^{4}}} \right)-\dfrac{b}{{{x}^{2}}}+\cos x \right]=\dfrac{d}{dx}a{{x}^{-4}}-\dfrac{d}{dx}b{{x}^{-2}}+\dfrac{d}{dx}\cos x$

$=a(-4{{x}^{-5}})-b(-2{{x}^{-3}})-\sin x$

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

ii) $(x+\cos x)(x-\tan x)$ 

Ans: 

Differentiating the given expression with respect to $x$, using product rule,

\[\dfrac{d}{dx}(x+\cos x)(x-\tan x)\]

\[=(x+\cos x)\dfrac{d}{dx}(x-\tan x)+(x-\tan x)\dfrac{d}{dx}(x+\cos x)\]

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

On simplifying,

\[\dfrac{d}{dx}(x+\cos x)(x-\tan x)\]

\[=x-x{{\sec }^{2}}x+\cos x-\cos x{{\sec }^{2}}x+x-x\sin x-\tan x+\tan x\sin x\]

$=2x-x{{\sec }^{2}}x+{cosx}-{cosx}-x\sin x-\tan x+\tan x\sin x$

$=2x-x{{\sec }^{2}}x-x\sin x-\tan x+\tan x\sin x$

Important Ponits for Class 11 Maths Chapter 12 Limits and Derivatives

Meaning of Limit and Derivative

Calculus mainly deals with the study of change in the value of a function as the points in the domain change. The limit is used when we have to find the value of a function near to some value. If right and left-hand limits coincide, we call that common value as the limit of f(x) at x = a & denote it by \[\lim_{x\rightarrow a} f(x)\]. The derivative is used to measure the instantaneous rate of change of the function, as distinct from its average rate of change. The derivative can also be defined as the limit of the average rate of change in the function as the length of the interval on which the average is computed tends to zero.

Limits and Derivatives Introduction

The basics of differentiation and calculus are the foundation for advanced mathematics, modern physics and various other branches of modern sciences and engineering. Limits and derivatives Class 11 is the entry point to calculus for CBSE students. Hence the concept of Limits and Derivative is very important.

Limits of a Function

Limit of a function f(x) is defined as a value, where the function reaches as the limit reaches some value. Limits are used to define other topics like integration, integral calculus and continuity of the function.

Limit Formula

Consider f(y) is a function, then the limit of the function can be represented as;

\[\lim_{y \rightarrow b}\]

Properties of Limits

Let p and q be two functions and a be a value such that \[\lim_{x \rightarrow a} p(x)\] and \[\lim_{x \rightarrow a} q(x)\] exists:

1. \[\lim_{x \rightarrow a} [p(x) + q(x)] = \lim_{x \rightarrow a} p(x) + \lim_{x \rightarrow a} q(x)\]

2. \[\lim_{x \rightarrow a} [p(x) - q(x)] = \lim_{x \rightarrow a} p(x) - \lim_{x \rightarrow a} q(x)\]

For every real number k

3. \[\lim_{x \rightarrow a} [kp(x)] = k \lim_{x \rightarrow a} p(x)\]

4. \[\lim_{x \rightarrow a} [p(x) q(x)] = \lim_{x \rightarrow a} p(x) \times \lim_{x \rightarrow a} q(x)\]

5. \[\lim_{x \rightarrow a} \frac{p(x)}{q(x)} = \frac{\lim_{x \rightarrow a} p(x)}{\lim_{x \rightarrow a} q(x)}\]

Derivatives of a Function

Instantaneous rate of change of a quantity with respect to the other is known as derivative. The derivative of a function is represented by the below formula.

Derivative Formula

\[\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}\]

Properties of Derivatives

Algebra of the derivative of the function is given below:

Consider f and g be two functions such that their derivatives are defined in a common domain.

(i) The derivative of the sum of two functions is the sum of the derivatives of the functions.

\[\frac{d}{d(x)} [p(x) + q(x)] = \frac{d}{d(x)} (p(x)) + \frac{d}{d(x)} (q(x))\]

(ii) The derivative of the difference of two functions is the difference of the derivatives of the functions

\[\frac{d}{d(x)} [p(x) - q(x)] = \frac{d}{d(x)} (p(x)) - \frac{d}{d(x)} (q(x))\]

(iii) The derivative of product of two functions is given by the following product rule

\[\frac{d}{d(x)} [p(x) \times q(x)] = \frac{d}{d(x)} [p(x)]q(x) + p(x) \frac{d}{d(x)}[q(x)]\]

(iv) The derivative of the quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero).

\[\frac{d}{d(x)} [\frac{p(x)}{q(x)}] = \frac{\frac{d}{d(x)} [p(x)]q(x) - p(x)\frac{d}{d(x)} [q(x)]}{(g(x))^{2}}\]

Steps to Find the Derivative:

1. Change x by the smallest possible value and let that be ‘h’ and so the function becomes f(x+h).

2. Get the change in value of function that is : f(x + h) – f(x)

3. The rate of change in function f(x) on changing from ‘x’ to ‘x+h’ will be

\[\frac{dy}{dx} = \frac{f(x+h)-f(x)}{h}\]

We can ignore d(x) because it is considered to be too small.

Types of Derivative

Derivatives can be classified into different types based on their order such as first and second-order derivatives. These can be defined as given below.

First-Order Derivative

The first-order derivatives are used to find the direction of the function whether the function is increasing or decreasing. The first-order derivative can be interpreted as an instantaneous rate of change. The slope of the tangent line is used to predict the first-order derivative.

Second-Order Derivative

The second-order derivatives are used to get an idea of the shape of the graph of the given function. If the value of second-order derivatives is positive, then the graph of a function is upwardly concave. If the value of the second-order derivative is negative, then the graph of a function is downwardly open.

What are the Benefits of Important Questions from Vedantu for Class 11 Maths Chapter 12 - Limits and Derivatives

• Focus on key topics for efficient studying.

• Prepares students for exams and reduces anxiety.

• Reinforces understanding of fundamental concepts.

• Teaches effective time management.

• Enables self-assessment and progress tracking.

• Strategic approach for higher scores.

• Covers a wide range of topics for comprehensive understanding.

• Supports exam preparation and boosts confidence.

Conclusion:

After solving above Class 11 Maths Limits and Derivatives Important Questions students can get an idea about the type of questions asked in the examination. Here, the questions are based on the NCERT textbook and as per the latest syllabus of the CBSE board. The questions provided include all types of questions such as 1 mark, 2 marks, 4 marks, 6 marks. Practice Important Questions for Class 11 Maths Chapter 12 provided here to achieve a good score in the final examination.

Important Study Materials for Class 11 Maths Chapter 12 Limits and Derivatives

CBSE Class 11 Maths Chapter-wise Important Questions

CBSE Class 11 Maths Chapter-wise Important Questions and Answers cover topics from all 14 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

Important Related Links for CBSE Class 11 Maths