RD Sharma Class 12 Exercise 10.1 Differentiation Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 10.1 Differentiation Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 20, 2022 05:28 PM IST

RD Sharma Class 12th Exercise 10.1 is the most preferred reference material by many CBSE schools as well as the students. The high standard of the solved sums present in this book makes the students understand every concept in-depth. For many years, the questions for the board examinations have been taken from the RD Sharma books. When the students use this book for their practice, they would be exam-ready effortlessly. RD Sharma Solutions
Chapter 10 for class 12 mathematics, Differentiation has 8 exercises in total. The first exercise, ex 10.1 has basic differentiation functions learned in the previous academic year. Using the RD Sharma Class 12th Exercise 10.1 book, the students will be able to solve all the problems easily. The simple tricks and tips to solve the differentiation sums are present in the Class 12th RD Sharma Chapter 10 Exercise 10.1 Solutions material. Concepts like logarithmic differentiation, differentiation of inverse trigonometric functions, and various other differentiating methods are included in this exercise.

## RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

Differentiation Excercise: 10.1

Differentiation exercise 10 .1 question 1

$\left ( -e^{-x} \right )$
Hint:
Use the first principle to find the differentiation of
$\left ( -e^{-x} \right )$
Given:
$\left ( -e^{-x} \right )$

Solution:
Let
$f\left ( x \right )= e^{-x}$
$f \left ( x+h \right )= e^{-\left ( x+h \right )}$
So, we will use formula of differentiation by first principle,
$\therefore \frac{d}{dx}\left ( f\left ( x \right ) \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
$\frac{d}{dx}= \lim_{h\rightarrow o}\frac{e^{-\left ( x+h \right )}-e^{-x}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{-x}\times e^{-h}-e^{-x}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{-x}\left ( e^{-h}-1 \right )}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{-x}\left ( e^{-h}-1 \right )}{-h}\times \left ( -1 \right )$
$= \lim_{h\rightarrow 0}e^{-x}\times \left ( \frac{e^{-h}-1}{-h} \right )\times \left ( -1 \right )$

$= \lim_{h\rightarrow 0}e^{-x}\left ( 1 \right )\times \left ( -1 \right )$ $\left [ \therefore \lim_{h\rightarrow 0}\frac{e^{x}-1}{x} \right ]= 1$
$=\left ( -e^{-x} \right )$
Hence, the differentiation of $e^{-x}$ is $\left ( -e^{-x} \right )$
$\left ( 3e^{3x} \right )$
Hint:
Use first principle to find the differentiation of $\left ( e^{3x} \right )$
Given:
$\left ( e^{3x} \right )$
Solution:
Let
$f\left ( x \right )= e^{3x}$
$f\left ( x+h \right )= e^{3\left ( x+h \right )}$
So, now we will use formula of differentiation by first principle
$\frac{d}{dx}f\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
$\frac{d}{dx}\left ( e^{3x} \right )=\lim_{h\rightarrow 0}\frac{e^{3\left ( x+h \right )}-e^{3x}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{3x}\times e^{3h}-e^{3x}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{3x}\left ( e^{3h}-1 \right )}{h}$
Multiply and divide by 3
$=\lim_{h\rightarrow 0}e^{3x}\frac{\left ( e^{3h}-1 \right )}{3h}\times 3$
$=\lim_{h\rightarrow 0}e^{3x}\times 1\times 3$ $\left [ \because \lim_{h\rightarrow 0} \left ( \frac{e^{x}-1}{x} \right )=1\right ]$
$=3e^{3x}$
Hence, the differentiation of $e^{3x}$ is $3e^{3x}$.
$ae^{\left ( ax+b \right )}$
Hint:
Use first principle to find the differentiation
Given:
$\left ( e^{ax+b} \right )$
Solution:
Let

$f\left ( x \right )=e^{ax+b}$
$f\left ( x+h \right )=e^{a\left ( x+h \right )+b}$
Now, we will use formula of differentiation by first principle
$\frac{d}{dx}\left ( f\left ( x \right ) \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
$\frac{d}{dx}\left ( e^{ax+b} \right )=\lim_{h\rightarrow 0}\frac{e^{a\left ( x+h \right )+b}-e^{ax+b}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{ax+ah+b}-e^{ax+b}}{h}$
$=\lim_{h\rightarrow 0}\frac{\left ( e^{ax+b} \right )\left ( e^{ah} \right )-e^{ax+b}}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{ax+b}\left ( e^{ah}-1 \right )}{h}$
Multiply and divide by a
$=\lim_{h\rightarrow 0}e^{ax+b}\frac{\left ( e^{ah}-1 \right )}{ah}\times a$
$=\lim_{h\rightarrow 0}\left ( e^{ax+b} \right )\times a$ $\left [ \because \lim_{h\rightarrow 0}\frac{e^{x}-1}{x}=1\right ]$
$=ae^{ax+b}$
Hence, the differentiation of $e^{ax+b}$ is $ae^{ax+b}$.
$-\left ( \sin x \right )\cdot e^{\cos x}$
Hint:
Use first principle to find the differentiation
Given:
$\left ( e^{\cos x} \right )$
Solution:
Let
$f\left ( x \right )= e^{\cos x}$
$f\left ( x+h \right )= e^{\cos \left ( x+h \right )}$
Now, we will use formula of differentiation by first principle
$\frac{d}{dx}\left ( f\left ( x \right ) \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
$\frac{d}{dx}\left ( e^{\cos x}\right )=\lim_{h\rightarrow 0}\frac{e^{\cos \left ( x+h \right )}-e^{\cos x}}{h}$
Take $e^{\cos x}$ common from the numerator
$=\lim_{h\rightarrow 0}\left ( e^{\cos x} \right )\frac{\left [ \frac{e^{cos\left ( x+h \right )}}{e^{\cos x}}-1 \right ]}{h}$
$=\lim_{h\rightarrow 0}\frac{e^{\cos x}\left [ e^{\cos x\left ( x+h \right )-\cos x} -1\right ]}{h}$ $\left [ \because \frac{a^{m}}{a^{n}}=a^{m-n} \right ]$

Multiply and divide $\left [ \cos \left ( x+h \right )-\cos x \right ]$
$=\lim_{h\rightarrow 0}\frac{e^{\cos x}\left [ e^{\cos \left ( x+h \right )-\cos x} -1\right ]}{\left ( \cos \left ( x+h \right )-\cos x \right )\times h}\times\left [ \cos \left ( x+h \right )-\cos x \right ]$

$=\lim_{h\rightarrow 0}e^{\cos x}\times \left ( \frac{\cos \left ( x+h \right )-\cos x}{h} \right )$ $\left [ \because \lim_{h\rightarrow 0}\frac{e^{x}-1}{x}=1 \right ]$

$=\lim_{h\rightarrow 0}e^{\cos x}\times \left ( \frac{\cos \left ( x+h \right )-\cos x}{h} \right )$ $\left [ \because \cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2} \right ]$
$=\lim_{h\rightarrow 0}e^{\cos x}\times \left [ \frac{-2\sin \frac{x+h+x}{2}\times \sin \frac{x+h-x}{2}}{h} \right ]$
$e^{\cos x}$ Is a function of $f\left ( x \right )$ and limit is applied on variable ‘h’
So,$e^{\cos x}$ can be taken outside
$= e^{\cos x}\times \lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{2x+h}{2}\ \right )\times \sin \left ( \frac{h}{2} \right )}{h}$
Multiply the denominator by 2 and divide the denominator by 2
$= e^{\cos x}\times \lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{2x+h}{2} \right )}{2}\times \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}$
$= e^{\cos x}\times \lim_{h\rightarrow 0}\left [ -\sin \left ( \frac{2x+h}{2} \right ) \right ]\times 1$ $\left [ \because \lim_{h\rightarrow 0}\frac{\sin x}{x}=1 \right ]$
$=e^{\cos x}\times\left ( -\sin x \right )=-\sin xe^{\cos x}$
Hence, the differentiation of $e^{\cos x}$ is $\left [ -\sin xe^{\cos x} \right ]$

Differentiability exercise 10.1 question 5

$\frac{e\sqrt{2x}}{\sqrt{2x}}$
Hint:
Use first principle formula to find the differentiation
Given:
$e\sqrt{2x}$
Solution:
Let,
$f\left ( x \right )=e^{\sqrt{2x}}$
$f\left ( x+h \right )=e^{\sqrt{2\left ( x+h \right )}}$
Now, we will use the formula of first principle to find the differentiation
$\frac{d}{dx}\left ( f\left ( x \right ) \right )=\lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
$\frac{d}{dx}\left ( e^{\sqrt{2x}} \right )=\lim_{h\rightarrow 0}\frac{e^{\sqrt{2\left ( x+h \right )}}-e^{\sqrt{2x}}}{h}$
Take 2$e^{\sqrt{2x}}$ common from numerator
$= \lim_{h\rightarrow 0}\frac{e^{\sqrt{2x}}\left [ \left ( \frac{e^{\sqrt{2\left ( x+h \right )}}}{e^{\sqrt{2x}}} \right )-1 \right ]}{h}$
$=\lim_{h\rightarrow 0}\cdot e^{\sqrt{2x}}\frac{\left [ \left ( \left [ e^{\sqrt{2\left ( x+h \right )}} \right ] \right ) -1\right ]}{h}$ $\left [ \because \frac{a^{m}}{a^{n}} =a^{m-n}\right ]$

Multiply and divide by $\left ( \sqrt{2\left ( x+h \right )}-\sqrt{2x} \right )$
$= \lim_{h\rightarrow 0}e^{\sqrt{2x}}\times \frac{\left [ \left ( e^{^{\sqrt{2\sqrt{\left ( x+h \right )}-\sqrt{2x}}}} \right ) -1\right ]}{\left ( \sqrt{2\left ( x+h \right )}-\sqrt{2x} \right )}\times \left [ \frac{\sqrt{2\left ( x+h \right )-\sqrt{2x}}}{h} \right ]$

$= \lim_{h\rightarrow 0}e^{\sqrt{2x}}\times 1\times \left [ \frac{\sqrt{2\left ( x+h\right )}-\sqrt{2x}}{h} \right ]$ $\left [ \because \lim_{h\rightarrow 0}\frac{e^{x}-1}{x} =1\right ]$

$=\lim_{h\rightarrow 0}e^{\sqrt{2x}}\times \left [ \frac{\sqrt{2\left ( x+h \right )-\sqrt{2x}}}{h} \right ]$
Now, $e^{\sqrt{2x}}$ is a function of x and unit applied on variable ‘h’. So,$e^{\sqrt{2x}}$ can be taken outside became w.r.t variable ‘h’, x will remain constant
$= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\left [ \frac{\sqrt{2\left ( x+h \right )}-\sqrt{2x}}{h} \right ]$

Rationalising the numerator
$= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\left [ \frac{\sqrt{2\left ( x+h \right )}-\sqrt{2x}}{h}\times \frac{\sqrt{2\left ( x+h \right )}+\sqrt{2x}}{\sqrt{2\left ( x+h \right )}+\sqrt{2x}} \right ]$
$= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\left [ \frac{\left ( \sqrt{2\left ( x+h \right )} \right )^{2}-\left ( \sqrt{2x} \right )^{2}}{h\times \left ( \sqrt{2\left ( x+h \right )} \right )} \right ]$ $\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2\left ( x+h \right )-2x}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}$
$= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2x+2h-2x}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}$
$= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2h}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}$
$= e^{\sqrt{2x}}\times \lim_{h\rightarrow 0}\frac{2}{h\left ( \sqrt{2\left ( x+h \right )}+\sqrt{2x} \right )}$
$= e^{\sqrt{2x}}\times \frac{2}{2\sqrt{2x}}$
$=\frac{ e^{\sqrt{2x}}}{\sqrt{2x}}$
Hence, the differentiation of $e^{\sqrt{2x}}$ is$\frac{ e^{\sqrt{2x}}}{\sqrt{2x}}$

$\left ( -\tan x \right )$
Hint:
Use first principle formula to find the differentiation
Given:
$\log \left ( \cos x \right )$
Solution:
Let,
$f x= \log \left ( \cos x \right )$
$f \left ( x+h \right )= \log \left ( \cos \left ( x+h \right ) \right )$
Now, we will use formula of first principle to find the differentiation
$\frac{d}{dx}\left ( f\left ( x \right ) \right )= \lim_{h\rightarrow 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$
$\frac{d}{dx}\left (\log \left ( \cos x \right ) \right )= \lim_{h\rightarrow 0}\frac{\log \left ( \cos \left ( x+h \right ) \right )-\log \left ( \cos x \right )}{h}$
$= \lim_{h\rightarrow 0}\frac{\log \left ( \frac{\cos \left ( x+h \right )}{\cos x} \right )}{h}$ $\left [ \because \log \mathit{A}\log \mathit{B= \log \frac{\mathit{A}}{\mathit{B}}} \right ]$

$= \lim_{h\rightarrow 0}\frac{\log \left [1+\frac{\cos \left ( x+h \right )}{\cos x}-1 \right ]}{h}$
$= \lim_{h\rightarrow 0}\frac{\log \left [1+\frac{\cos \left ( x+h \right )-\cos x}{\cos x}\right ]}{h}$
Multiply and divide by $= \frac{\cos \left ( x+h \right )-\cos x}{h}$
$=\lim_{h\rightarrow 0}\frac{\log \left [ 1+ \frac{\cos \left ( x+h \right )-\cos x}{\cos x} \right ]}{h\times \left [ \frac{\cos \left ( x+h \right )-\cos x}{\cos x} \right ]}\times \left [ \frac{\cos \left ( x+h \right )-\cos x}{\cos x} \right ]$
$=\lim_{h\rightarrow 0}\left ( \frac{1}{h} \right )\times \frac{\log \left ( 1+ \frac{\cos \left ( x+h \right )-\cos x}{\cos x} \right )}{ \frac{\cos \left ( x+h \right )-\cos x}{\cos x}}\times \frac{\cos \left ( x+h \right )-\cos x}{\cos x}$
$=\lim_{h\rightarrow 0}\left ( \frac{1}{h} \right )\times1\times \frac{\cos \left ( x+h \right )-\cos x}{\cos x}$ $\left [ \because \lim_{h\rightarrow 0}\frac{\log \left ( 1+x \right )}{x}=1 \right ]$
$=\lim_{h\rightarrow 0} \frac{\cos \left ( x+h \right )-\cos x}{h\times \cos x}$

$=\lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{\left ( x+h \right )+x}{2} \right )\sin\left ( \frac{\left ( x+h \right )-x}{2} \right )}{h\times \cos x}$ $\left [ \because \cos \mathit{A}-\cos \mathit{B}=2\sin \left ( \frac{\mathit{A+B}}{2} \right )\sin \left ( \frac{\mathit{A-B}}{2} \right ) \right ]$
$=\lim_{h\rightarrow 0}\frac{-2\sin \left ( \frac{2 x+h }{2} \right )\sin\left ( \frac{h}{2} \right )}{h\cdot \cos x}$
Multiply the denominator and divide the denominator by 2
$=\lim_{h\rightarrow 0}\frac{-2\left ( \frac{2x+h}{2} \right )}{\cos x}\cdot \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )\times 2}$ $\left [ \because \lim_{h\rightarrow 0}\frac{\sin x}{x}=1 \right ]$
$=\lim_{h\rightarrow 0}\frac{-\sin \left ( \frac{2x+h}{2} \right )}{\cos x}\times 1$
$=\frac{-\sin \left ( \frac{2x}{2} \right )}{\cos x}$
$=\frac{-\sin x }{\cos x}$
$=-\tan x$
Hence, the differentiation of $\log \left ( \cos x \right )= \left ( -\tan x \right )$

$\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$
Hint:
Use first principle formula to find the differentiation
Given:
$e^{\sqrt{\cot x}}$
Solution:
Let,
\begin{aligned} &f(x)=e^{\sqrt{\cot x}} \\ &f(x+h)=e^{\sqrt{\cot (x+h)}} \end{aligned}
Now, we will use the formula of first principle to find the differentiation
\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}\left(e^{\sqrt{\cot x}}\right)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot (x+h)}}-e^{\sqrt{\cot x}}}{h} \end{aligned}
Take $e^{\sqrt{\cot x}}$ common from numerator
$=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(\frac{e^{\sqrt{\cot (x+h)}}}{e^{\sqrt{\cot x}}}-1\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right)}{h}$ $\left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right]$
$\because e^{\sqrt{\cot x}}$ Is a function of variable x and limit is applied on variable ‘h’
So,
$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{e^{(\sqrt{\cot (x+h)}-\sqrt{\cot x})}-1}{h}$
Multiply and divide by $(\sqrt{\cot (x+h)}-\sqrt{\cot x})$
$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\left[e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right]}{(\sqrt{\cot (x+h)}-\sqrt{\cot x})} \times\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}\right]$
$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0}(1) \times\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h}\right]$ $\left[\because \lim _{h \rightarrow 0} \frac{e^{x}-1}{x}=1\right]$
Rationalising the numerator
$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0}\left[\frac{\sqrt{\cot (x+h)}-\sqrt{\cot x}}{h} \times \frac{\sqrt{\cot (x+h)}+\sqrt{\cot x}}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}\right]$
$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$ $\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$$\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\frac{\cot (x+h) \times \cot x+1}{\cot (x+h-x)}}{h(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$$\left[\cot (A+B)=\frac{\cot A \cot B+1}{\cot A-\cot B}\right]$

$=\left(e^{\sqrt{\cot x}}\right) \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{h(\operatorname{coth})(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{\left(\frac{h}{\tanh }\right)(\sqrt{\cot (x+h)}+\sqrt{\cot x})}$$\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$

$=e^{\sqrt{\cot x}} \times \lim _{h \rightarrow 0} \frac{\tanh }{h} \times\left[\frac{\cot (x+h) \cot x+1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}\right]$
$=e^{\sqrt{\cot x}} \times 1 \times \lim _{h \rightarrow 0} \frac{\cot (x+h) \cot x+1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}$ $\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]$
$=e^{\sqrt{\cot x}} \times \frac{\cot ^{2} x+1}{2 \sqrt{\cot x}}$
$=e^{\sqrt{\cot x}} \times \frac{\cos e c^{2} x}{2 \sqrt{\cot x}}$ $\left[\because 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\right]$
$\therefore \frac{d}{d x}\left(e^{\sqrt{\cot x}}\right)=\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$
Hence, the differentiation of $e^{\sqrt{\cot x}}$ is $\frac{e^{\sqrt{\cot x}} \times \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$

$e^{x^{2}}\left ( x^{2}+2x \right )$
Hint:
Use first principle formula to find the differentiation
Given:
${x^{2}}e^{x}$
Solution:
Let,\begin{aligned} &f(x)=x^{2} e^{x} \\ &f(x+h)=(x+h)^{2} \cdot e^{(x+h)} \end{aligned}

Now, we will use the formula of first principle

\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}\left(x^{2} e^{x}\right)=\lim _{h \rightarrow 0} \frac{(x+h)^{2} \cdot e^{(x+h)}-x^{2} e^{x}}{h} \end{aligned}

$=\lim _{h \rightarrow 0} \frac{\left(x^{2}+2 h x+h^{2}\right) e^{x+h}-x^{2} e^{x}}{h}$ $\left [ \because \left ( a+b \right )^{2}= a^{2}+2ab+b^{2} \right ]$

\begin{aligned} &=\lim _{h \rightarrow 0}\left[\frac{\left(x^{2} e^{x+h}-x^{2} e^{x}\right)}{h}+\frac{2 h x e^{x+h}}{h}+\frac{h^{2} e^{x+h}}{h}\right] \\ &=\lim _{h \rightarrow 0}\left[\frac{x^{2} e^{x} \cdot e^{h}-x^{2} e^{x}}{h}+2 x e^{x+h}+h e^{x+h}\right] \\ &=\lim _{h \rightarrow 0}\left[\frac{x^{2} e^{x}\left(e^{h}-1\right)}{h}+2 x e^{x+h}+h e^{x+h}\right] \\ &=\lim _{h \rightarrow 0}\left[x^{2} e^{x} \cdot\left(\frac{e^{h}-1}{h}\right)+2 x e^{x+h}+h e^{x+h}\right] \end{aligned}

$=\lim _{h \rightarrow 0}\left[\left(x^{2} e^{x}\right) \times 1+2 x e^{x+h}+h e^{x+h}\right]$ $\left [ \because \lim _{h \rightarrow 0}\frac{e^{x}-1}{x}=1 \right ]$

\begin{aligned} &=x^{2} e^{x}+2 x e^{x}+\left(0 \times e^{x}\right) \\ &=x^{2} e^{x}+2 x e^{x} \\ &=e^{x}\left(x^{2}+2 x\right) \\ &\therefore \frac{d}{d x}\left(x^{2} e^{x}\right)=e^{x}\left(x^{2}+2 x\right) \end{aligned}

Hence, the differentiation of $x^{2} e^{x}$ is $e^{x}\left ( x^{2} +2x \right )$

$-\cot x$

Hint:

Use first principle formula to find the differentiation

Given:

$\log \left ( \cos ecx \right )$

Solution:

Let,

\begin{aligned} &f(x)=\log (\cos e c x) \\ &f(x+h)=\log (\cos e c(x+h)) \end{aligned}

Now, we will use the formula of first principle

\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}(\log (\operatorname{cosec} x))=\lim _{h \rightarrow 0} \frac{\log (\operatorname{cosec}(x+h)-\log (\cos e c x))}{h} \end{aligned}

$=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\cos e c(x+h)}{\cos e c x}\right)}{h}$$\left [ \because \log a-\log b=\log \left ( \frac{a}{b} \right ) \right ]$

$=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\frac{1}{\sin (x+h)}}{\frac{1}{\sin x}}\right)}{h}$$\left [ \because \cos ecx=\frac{1}{\sin x} \right ]$

$=\lim _{h \rightarrow 0} \frac{\log \left(\frac{\sin x}{\sin (x+h)}\right)}{h}$

Add 1 and subtract 1 from argument of the numerator

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\sin x}{\sin (x+h)}-1\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right)}{h} \end{aligned}

Multiply and divide by $\frac{\sin x-\sin (x+h)}{\sin (x+h)}$

$=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right)}{\left[\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right]} \times \frac{\left[\frac{\sin x-\sin (x+h)}{\sin (x+h)}\right]}{h}$

$=\lim_{h\rightarrow 0}\times \frac{\sin x-\sin (x+h)}{h\sin (x+h)}$$\left [ \because \lim_{h\rightarrow 0}\frac{\log \left ( 1+x \right )}{x}=1 \right ]$

$=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \times \sin \left(\frac{-h}{2}\right)}{h \cdot \sin (x+h)}$ $\left [ \because \sin \mathit{A}-\sin \mathit{B}=2\cos \left ( \frac{\mathit{A+B}}{2} \right )\sin \left ( \frac{\mathit{A-B}}{2} \right ) \right ]$

$=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \times \sin \left(\frac{-h}{2}\right)}{h \cdot \sin (x+h)}$

Divide and multiply the denominator by (-2)

$=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \cdot \sin \left(\frac{-h}{2}\right)}{(-2) \times \sin (x+h)\left(\frac{-h}{2}\right)}$

$=\lim _{h \rightarrow 0} \frac{-\cos \left(\frac{2 x+h}{2}\right)}{\sin (x+h)} \times\left(\frac{\sin \left(\frac{-h}{2}\right)}{\left(\frac{-h}{2}\right)}\right)$ $\left [ \lim _{h \rightarrow 0}\frac{\sin x}{x}=1 \right ]$

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\cos \left(\frac{2 x+h}{2}\right)}{\sin (x+h)} \\ &=-\cot x \\ &\therefore \frac{d}{d x}(\log \cos e c x)=-\cot x \end{aligned}

Hence, the differentiation of $-\cot x$

$\frac{2}{\sqrt{1-\left ( 2x+3 \right )^{2}}}$
Hint:
Use first principle formula to find the differentiation
Given:
$\sin ^{-1}\left ( 2x+3 \right )$
Solution:
Let,
\begin{aligned} &f(x)=\sin ^{-1}(2 x+3) \\ &f(x+h)=\sin ^{-1}(2(x+h)+3)=\sin ^{-1}(2 x+2 h+3) \end{aligned}
Now, we will use formula of first principle to find the differentiation
\begin{aligned} &\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\frac{d}{d x}\left(\sin ^{-1}(2 x+3)\right)=\lim _{h \rightarrow 0} \frac{\sin ^{-1}(2 x+2 h+3)-\sin ^{-1}(2 x+3)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\sin ^{-1}\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}{h} \end{aligned}
$\left [ \because \sin ^{-1}x-\sin ^{-1}y=\sin ^{-1}\left [ x\sqrt{1-y^{2}}-y\sqrt{1-x^{2}} \right ] \right ]$
Let,

\begin{aligned} &{\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]=z} \\ &=\lim _{h \rightarrow 0} \frac{\sin ^{-1} z}{h} \end{aligned}
Multiply and divide by z
$=\lim _{h \rightarrow 0} \frac{\sin ^{-1} z}{h}\times \frac{z}{h}$
$=\lim _{h \rightarrow 0} \left ( 1 \right )\times \frac{z}{h}$ $\left [ \because \lim_{h\rightarrow 0}\frac{\sin ^{-1}x}{x}=1 \right ]$
$=\lim _{h \rightarrow 0} \frac{z}{h}$
Put the value of z now
$=\lim _{h \rightarrow 0} \frac{(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}}{h}$
Multiply and divide by $(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}$
\begin{aligned} &\quad\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}-(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right] \\ &=\lim _{h \rightarrow 0} \frac{\times\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \\ &=\lim _{h \rightarrow 0} \frac{\left((2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}\right)^{2}-\left((2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right)^{2}}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \end{aligned}
$\because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2}$
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[(2 x+2 h+3)^{2}\left(1-(2 x+3)^{2}\right)\right]-\left[(2 x+3)^{2}\left(1-(2 x+2 h+3)^{2}\right)\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \\ &=\lim _{h \rightarrow 0} \frac{[(2 x+3)+2 h]^{2}\left[1-(2 x+3)^{2}\right]-\left[(2 x+3)^{2}\right]\left[1-((2 x+3)+2 h)^{2}\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]} \end{aligned}
$=\lim _{h \rightarrow 0} \frac{\left\{\left[(2 x+3)^{2}+4 h^{2}+4 h(2 x+3)\right]\left[1-(2 x+3)^{2}\right]\right\}\left\{(2 x+3)^{2}\left(1-(2 x+3)^{2}-4 h^{2}-4 h(2 x+3)\right)\right\}}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}$
$\left ( \left ( a^{2}+b^{2}\right )=a^{2}+b^{2}+2ab \right )$
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[\begin{array}{l} (2 x+3)^{2}+4 h^{2}+4 h(2 x+3)-(2 x+3)^{4}-4 h^{2}(2 x+3)^{2}-4 h(2 x+3)^{3}-(2 x+3)^{2} \\ +(2 x+3)^{4}+4 h^{2}(2 x+3)^{2}+4 h(2 x+3)^{3} \end{array}\right]}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}\\ \end{aligned}
$=\lim _{h \rightarrow 0} \frac{4 h^{2}+4 h(2 x+3)}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}\\ =\lim _{h \rightarrow 0} \frac{4 h(h+(2 x+3))}{h\left[(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}\right]}\\ =\lim _{h \rightarrow 0} \frac{4(h+(2 x+3))}{(2 x+2 h+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+2 h+3)^{2}}}\\$
$=\frac{4(2 x+3)}{(2 x+3) \sqrt{1-(2 x+3)^{2}}+(2 x+3) \sqrt{1-(2 x+3)^{2}}}\\ =\frac{4(2 x+3)}{2 \times(2 x+3) \sqrt{1-(2 x+3)^{2}}}\\ =\frac{2}{\sqrt{1-(2 x+3)^{2}}}\\ \frac{d}{d x}\left(\sin ^{-1}(2 x+3)\right)=\frac{2}{\sqrt{1-(2 x+3)^{2}}}$
Hence, the differentiation of $\sin ^{-1}\left ( 2x+3 \right )$ Is $\frac{2}{\sqrt{1-(2 x+3)^{2}}}$.

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