RD Sharma Class 12 Exercise 10.5 Differentiation Solutions Maths-Download PDF Free Online

RD Sharma Class 12 Exercise 10.5 Differentiation Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 20, 2022 05:35 PM IST

RD Sharma books are well known for their accuracy, subject knowledge and exam-oriented questions. There is always a high possibility that the questions in CBSE exams might appear from this book as many schools refer to them for setting up question papers.
RD Sharma Class 12th Exercise 10.5 Is designed specifically for students to prepare for exams. It is created by a team of subject experts that have years of experience with exam-oriented materials. These solutions will help students get a better understanding of the subject and score good marks in exams.

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RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
Differentiation Excercise: 10.5
Differentiation exercise 10.5 question 30
Differentiation exercise 10.5 question 33
Differentiation exercise 10.5 question 55
Differentiation exercise 10.5 question 58
RD Sharma Chapter-wise Solutions

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

Differentiation Excercise: 10.5

Differentiation exercise 10.5 question 1

Answer: $\chi^{\frac{1}{x}} \frac{(1-\ln x)}{x^{2}}$
Hint: Differentiate by function $x^{n}$
Given: $x^{\frac{1}{x}}$
Solution: Let, $y=x^{\frac{1}{x}}$
Taking log on both sides,
$\begin{aligned} &\log \mathrm{y}=\log x^{\frac{1}{x}} \\\\ &\Rightarrow \log y=\frac{1}{x} \log x \end{aligned}$ $\left[\because \log a^{b}=b \log a\right]$
Differentiate both sides,
$\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \frac{d}{d x}(\log x)+\log x \frac{d}{d x}\left(x^{-1}\right)$ [using product rule]
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \times \frac{1}{x}+(\log x) \times\left(-\frac{1}{x^{2}}\right) \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}}-\frac{\log x}{x^{2}} \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{(1-\log x)}{x^{2}} \\\\ &\frac{d y}{d x}=y \frac{(1-\log x)}{x^{2}} \end{aligned}$
put value of $y=x^{\frac{1}{x}}$
$\frac{d y}{d x}=x^{\frac{1}{x}} \cdot\left[\frac{(1-\log x)}{x^{2}}\right]$

Differentiation exercise 10.5 question 2

Answer: $x^{\sin x}\left(\cos x \ln x+\left(\frac{\sin x}{x}\right)\right)$
Hint: Differentiate by $x^{n}$
Given: $x^{\operatorname{Sin} x}$
Solution: Let $\mathrm{y}=x^{\operatorname{Sin} x}$
Take natural log to both sides
$\begin{aligned} &\ln y=\ln x^{\sin x} \\\\ &\ln y=\sin x \ln x \end{aligned}$ $\left[\because \log a^{b}=b \log a\right]$

Diff both side w.r.t $x$

$\frac{d}{d x}(\ln y)=\frac{d}{d x}(\sin x \ln x)$
Using implicit diff on LHS, product rule on RHS
$\begin{aligned} &\frac{1}{y}\left(\frac{d y}{d x}\right)=\cos x \ln x+\frac{\sin x}{x} \\\\ &\frac{d y}{d x}=y\left(\cos x \ln x+\frac{\sin x}{x}\right) \end{aligned}$
Substituting back for y
$\frac{d y}{d x}=x^{\sin x}\left(\cos x \ln x+\frac{\sin x}{x}\right)$

Differentiation exercise 10.5 question 3

Answer: $(1+\cos x)^{x}\left[\left(-\frac{x \sin x}{1+\cos x}\right)+\log (1+\cos x)\right.$
Hint: Differentiate by $\cos ^{n}x$
Given: $(1+\cos x)^{x}$
Solution:Let $y=(1+\cos x)^{x}$ ...........(i)
Taking log on both the sides,
$\begin{aligned} &\log y=\log (1+\cos x)^{x} \\\\ &\log y=x \log (1+\cos x) \end{aligned}$
Differentiating with respect to $x$ ,
$\frac{1}{y} \frac{d y}{d x}=x \frac{d}{d x} \log (1+\cos x)+\log (1+\cos x) \frac{d}{d x}(x)$ [Using product rule]
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=x \frac{1}{(1+\cos x)} \frac{d}{d x}(1+\cos x)+\log (1+\cos x)(1) \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{x}{(1+\cos x)}(0-\sin x)+\log (1+\cos x) \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\log (1+\cos x)-\frac{x \operatorname{som} x}{(1+\cos x)} \\\\ &\frac{d y}{d x}=y\left[\log (1+\cos x)-\frac{x \sin x}{1+\cos x}\right] \end{aligned}$
$\frac{d y}{d x}=(1+\cos x)^{x}\left[\log (1+\cos x)-\frac{x \sin x}{(1+\cos x)}\right]$ [Using equation (i)]

Differentiation exercise 10.5 question 4

Answer: $x^{\cos ^{-1} x}\left(\frac{\cos ^{-1} x}{x}-\frac{\log x}{\sqrt{1-x^{2}}}\right)$
Hint: Differentiate by function $x^{n}x$
Given: $x^{\cos ^{-1} x}$
Solution: Let $y=x^{\cos ^{-1} x}$
$y=e^{\cos ^{-1} x \log x}$ $\left[\because a^{b}=e^{b \log x}\right]$
Differentiate w.r.t $x$
$e^{\cos ^{-1} x \log x}\left(\frac{d}{d x}\right) \cos ^{-1} \log x$
$x^{\cos ^{-1} x}\left(\cos ^{-1} x\left(\frac{d}{d x}\right) \log x+\log x \frac{d}{d x} \cos ^{-1} x\right)$
$x^{\cos ^{-1} x}\left[\cos ^{-1} x \cdot \frac{1}{x}+\log x \cdot\left(-\frac{1}{\sqrt{1-x^{2}}}\right)\right]$
$x^{\cos ^{-1} x}\left[\frac{\cos ^{-1} x}{x}-\frac{\log x}{\sqrt{1-x^{2}}}\right]$

Differentiation exercise 10.5 question 5

Answer: $\log x^{x-1}(1+\log x \log \log x)$
Hint: Differentiate by $\log ^{x}\left ( x \right )$
Given: $(\log x)^{x}$
Solution: Let $y=(\log x)^{x}$
Taking log both sides
$\log y=x \log (\log x)$
Differentiate w.r.t $x$ ,
$\frac{1}{y} \frac{d y}{d x}=x \cdot \frac{d}{d x}\{\log (\log x)\}+\log (\log x)\left(\frac{d}{d x}\right)(x)$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=x \frac{1}{\log x} \frac{d}{d x}(\log x)+\log (\log x)(1)$
$\begin{aligned} &\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{x}{\log x}\left(\frac{1}{x}\right)+\log (\log x) \\\\ &\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{\log x}+\log (\log x) \\\\ &\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{\log x}+\log (\log x)\right] \end{aligned}$
Now, put value of $y=(\log x)^{x}$
$\Rightarrow \frac{d y}{d x}=(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]$

Differentiation exercise 10.5 question 6

Answer: $(\log x)^{\cos x}(1+\log x \log \log x)$
Hint: Diff. by applying $x \log x$
Given: $(\log x)^{\cos x}$
Solution: Let $y=(\log x)^{\cos x}$
Taking log both sides
$\log y=\cos x \cdot \log (\log x)$ $\left[a s \log a^{b}=b \log a\right]$
Differentiate w.r.t $x$
$\frac{1}{y} \frac{d y}{d x}=d \frac{(\cos x \cdot \log (\log x)}{d x}$
Using product of rule $\cos x \cdot \log (\log x)$
$\begin{aligned} &(u v)^{\prime}=u^{\prime} v+v^{\prime} u \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{d(\cos x)}{d x} \log (\log x)+d\left(\frac{\log (\log x)}{d x}\right) \cos x \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=-\sin x \cdot \log (\log x) \cos x \\\\ &\frac{1}{y} \frac{d y}{d x}=\sin x \cdot \log (\log x)+\frac{1}{\log x} \cdot \frac{1}{x} \cos x \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\sin x \cdot \log (\log x)+\frac{\cos x}{x \log x} \\\\ &\frac{d y}{d x}=y\left(-\sin x \cdot \log (\log x)+\frac{\cos x}{x \log x}\right) \end{aligned}$
$\frac{d y}{d x}=(\log x)^{\cos x}\left ( -\sin x\cdot \log \left ( \log x \right )+\frac{\cos x}{x\log x} \right )$
$\frac{d y}{d x}=(\log x)^{\cos x}\left ( \frac{\cos x}{x\log\: x} -sin \: x.\log\left ( \log x \right ) \right )$

Differentiation exercise 10.5 question 7

Answer: $(\sin )^{\cos x}(\cos x \cot x-\sin \log (\sin x)$

Given: $(\sin x)^{\cos x}$
Solution: Let $y=(\sin x)^{\cos x}$
Taking log both sides
$\log y=\log (\sin x)^{\cos x}$ $\left[\because \log m^{n}=n \log m\right]$
$\log y=\cos x \cdot \log (\sin x)$
Differentiate w.r.t x,
$\frac{1}{y} \frac{d y}{d x}=\cos x \cdot \frac{d}{d x}\left(\log (\sin x)+\log (\sin x) \frac{d}{d x}(\cos x)\right)$
$\frac{1}{y} \frac{d y}{d x}=\cos x \cdot \frac{1}{\sin x} \cdot \cos x+\log (\sin x)(-\sin x)$
$\begin{aligned} &\frac{d y}{d x}=y(\cos x \cdot \cot x-\sin (\log (\sin x)) \\\\ &(\sin x)^{\cos x}(\cos x \cdot \cot x-\sin \log (\sin x)) \end{aligned}$

Differentiation exercise 10.5 question 8

Answer: $e^{x \log x}(1+\log x)$
Hint: Differentiate by applying $e^{x}$
Given: $e^{x \log x}$
Solution: Let $y=e^{x \log x}$
$y=x^{x}$ $\left[\because a^{b}=e^{b \log a}\right]$
Taking log on both sides
$\begin{aligned} &\log y=\log e^{x \log x} \\\\ &\log y=x \log x \end{aligned}$
Differentiate w.r.t x,
$\frac{1}{y}=x \frac{d}{d x}(\log x)+\log x \frac{d}{d x} x$ [ use multiplication rule]
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=x \cdot \frac{1}{x}+\log x .1 \\\\ &\frac{d y}{d x}=y+y \log x \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=y(1+\log x) \\\\ &=e^{x \log x}(1+\log x) \end{aligned}$ $\begin{aligned} &\frac{d y}{d x}=y(1+\log x) \end{aligned}$
$=e^{x \log x}(1+\log x)$

Differentiation exercise 10.5 question 9

Answer: $(\sin x)^{\log x} \frac{\log (\sin x)}{x}+\log x+\sin x$
Hint: Diff by $\sin x^{x}$
Given: $(\sin x)^{\log x}$
Solution:
Let $y=(\sin x)^{\log x}$ ........(i)
taking log on both sides,
$\begin{aligned} &\log y=\log (\sin x)^{\log x} \\\\ &\log y=\log x \log (\sin x) \end{aligned}$ $\left[\text { Using } \log a^{b}=b \log a\right]$
Differentiate w.r.t x, using product rule and chain rule,
$\frac{1}{y} \frac{d y}{d x}=\log x \frac{d}{d x}(\log \sin x)+\log \sin x \frac{d}{d x}(\log x)$
$=\log _{x}\left(\frac{1}{\sin x}\right) \frac{d}{d x}(\sin x)+\log \sin x\left(\frac{1}{x}\right)$
$=\frac{\log x}{\sin x} \times \cos x+\frac{\log \sin x}{x}$
$\begin{aligned} \frac{1}{y} \frac{d y}{d x} &=\log x \cot x+\frac{\log \sin x}{x} \\\\ \frac{d y}{d x} &=y\left[\log x \cot x+\frac{\log \sin x}{x}\right] \\\\ \frac{d y}{d x} &=(\sin x)^{\log x}\left[\log x \cot x+\frac{\log \sin x}{x}\right] \end{aligned}$ [Using equaton (i)]

Differentiation exercise 10.5 question 10

Answer: $10^{\log \sin x} \times \log 10 \times \cot x$
Hint: Diff by $10^{x}$
Given: $10^{\log \sin x}$
Solution: Let $y=10^{\log \sin x}$ .........(1)
Taking log both sides
$\begin{aligned} &\log y=\log 10^{\log \sin x} \\\\ &\log y=\log \sin x \log 10 \end{aligned}$
Differentiate w.r.t x,
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\log 10 \frac{d}{d x} \log \sin x \\\\ &\frac{1}{y} \frac{d y}{d x}=\log 10 \frac{1}{\sin x} \frac{d}{d x} \sin x \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\log 10\left(\frac{1}{\sin x}\right) \cos x \\\\ &\frac{d y}{d x}=y(\log 10 \cdot \cot x) \\\\ &\frac{d y}{d x}=10^{\log \sin x} \times \log 10 \times \cot x \end{aligned}$ ......[Using (1)]

Differentiation exercise 10.5 question 11

Answer: $(\log x)^{\log x}\left(\frac{1}{x}+\frac{\log (\log x)}{x}\right)$

Hint: Diff by $(\log x)^{\log x}$
Given: $(\log x)^{\log x}$
Solution: Let $y=(\log x)^{\log x}$
Taking log on both sides
$\begin{aligned} &\log y=\log (\log x)^{\log x} \\\\ &\log y=\log x \cdot \log (\log x) \end{aligned}$ $\left[\because \log \left(a^{b}\right)=b \log a\right]$
Differentiate w.r.t x,
$\frac{1}{y} \frac{d y}{d x}=d \frac{(\log x \cdot \log (\log x)}{d x}$
Use product rule
$\begin{aligned} &(u v)^{\prime}=u^{\prime} v+v^{\prime} u \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{d(\log x)}{d x} \cdot \log (\log x)+d \frac{(\log (\log x)}{d x} \cdot \log x \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \log (\log x)+\frac{1}{\log x}+\frac{d(\log x)}{d x} \log x \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \log (\log x)+\frac{d(\log x)}{d x} \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \log (\log x)+\frac{1}{x} \\\\ &\frac{d y}{d x}=y\left(\frac{1}{x}+\frac{\log (\log x)}{x}\right) \\\\ &\frac{d y}{d x}=(\log x)^{\log x}\left(\frac{1}{x}+\frac{\log (\log x)}{x}\right) \end{aligned}$

Differentiation exercise 10.5 question 12

Answer: $10^{10^{x}} \cdot 10^{x} \cdot(\log 10)^{2}$
Hint: Diff by $\left ( 10 \right )^{10x}$
Given: $\left ( 10 \right )^{10x}$
Solution: Let $y=\left ( 10 \right )^{10x}$
Taking log on both sides
$\log y=\log 10^{\left(10^{x}\right)}=10^{x} \log 10$
Differentiate w.r.t x,
$\begin{aligned} \frac{1}{y} \frac{d y}{d x}=& \log 10 \frac{d}{d x} 10^{x} \\\\ &=\log 10 \times 10^{x} \cdot \log 10 \end{aligned}$
$\begin{aligned} \frac{d y}{d x}=y & \times 10^{x} \times(\log 10)^{2} \\\\ &=10^{10^{x}} \cdot 10^{x}(\log 10)^{2} \end{aligned}$

Differentiation exercise 10.5 question 13

Answer: $\sin x^{x}(\log \sin x+x \cot x)$
Hint: Diff equation by $\sin ^{x}$
Given: $\sin x^{x}$
Solution: Let $y=(\sin x)^{x}$
Taking log on both sides
$\begin{aligned} &y=\sin x^{x} \\\\ &\log y=\log \sin x^{x} \\\\ &\log y=x \log \sin x \\\\ &\frac{1}{y} \frac{d y}{d x}=\log \sin x+x \cot x \\\\ &\frac{d y}{d x}=\sin x^{x}(\log \sin x+x \cot x) \end{aligned}$

Differentiation exercise 10.5 question 14

Answer: $\left(x \cdot \frac{1}{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}}\right)+(\sin x)^{x} \log (\sin x)$
Hint: Diff by using $\left(\sin ^{-1} x\right)^{x}$
Given: $\left(\sin ^{-1} x\right)^{x}$
Solution: Let $y=\left(\sin ^{-1} x\right)^{x}$
Taking log on both sides
$\log y=x \cdot \log \left(\sin ^{-1} x\right)$
Differentiate w.r.t x,
$\begin{aligned} &\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log \left(\sin ^{-1} x\right)+x \cdot \frac{1}{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}} \\\\ &\Rightarrow \frac{d y}{d x}=y\left[\log \left(\sin ^{-1} x\right)+x \cdot \frac{1}{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}}\right] \end{aligned}$
Now , put the value of , $y=\left(\sin ^{-1} x\right)^{x}$
$\Rightarrow \frac{d y}{d x}=\left(\sin ^{-1} x\right)^{x}\left[\log \left(\sin ^{-1} x\right)+x \cdot \frac{1}{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}}\right]$

Differentiation exercise 10.5 question 15

Answer: $x^{\sin ^{-1} x}\left[\frac{\sin ^{-1} x}{x}+\frac{\log x}{\sqrt{1-x^{2}}}\right]$
Hint: Diff by $x^{\sin ^{-1} x}$
Given: $x^{\sin ^{-1} x}$
Solution: Let $y=x^{\sin ^{-1} x}$
Taking log on both sides
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\sin ^{-1} x \cdot \log x \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{\log x}{\sqrt{1-x^{2}}}+\frac{\sin ^{-1} x}{x} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=y\left[\frac{\log x}{\sqrt{1-x^{2}}}+\frac{\sin ^{-1} x}{x}\right] \\\\ &\frac{d y}{d x}=x^{\sin ^{-1} x}\left[\frac{\sin ^{-1} x}{x}+\frac{\log x}{\sqrt{1-x^{2}}}\right] \end{aligned}$

Differentiation exercise 10.5 question 16

Answer: $(\tan x)^{\frac{1}{x}} \cdot(x \operatorname{cosec} x \sec x)-\ln \left ( \tan \left ( x \right ).\frac{1}{x^{2}} \right )$
Hint: Diff by $(\tan x)^{\frac{1}{x}}$
Given: $(\tan x)^{\frac{1}{x}}$
Solution: Let $y=(\tan x)^{\frac{1}{x}}$
Taking log on both sides
$\begin{aligned} &\log y=\log (\tan x)^{\frac{1}{x}} \\\\ &\frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \log \tan x \end{aligned}$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{-1}{x^{2}} \ln (\tan x)+\frac{1}{x}\left(\frac{1}{\tan x} \sec ^{2} x\right) \\\\ &\left(\frac{1}{\tan x}=\cot x\right) \end{aligned}$
So,
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=y\left(\frac{-1}{x^{2}} \ln \left(\tan x+\frac{1}{x} \operatorname{cosec} x \cdot \sec x\right)\right. \\\\ &\frac{d y}{d x}=(\tan x)^{\frac{1}{x}}\left(\frac{-1}{x^{2}} \ln \tan x\right)+\frac{1}{x} \cos e c x \sec x \\\\ &\frac{d y}{d x}=(\tan x)^{\frac{1}{x}}\left[(x \operatorname{cosec} x \cdot \sec x)-\ln (\tan x) \cdot \frac{1}{x^{2}}\right] \end{aligned}$
Differentiate w.r.t x,

Differentiation exercise 10.5 question 17

Answer: $\frac{d y}{d x}=\left[\frac{\tan ^{-1} x}{x}+\frac{\log x}{1+x^{2}}\right] \cdot \tan ^{-1} x$
Hint: Diff by $x^{\tan ^{-1} x}$
Given: $x^{\tan ^{-1} x}$
Solution: Let $y=x^{\tan ^{-1} x}$
Taking log on both sides
$\begin{aligned} &\log y=\log x^{\tan ^{-1} x} \\\\ &\frac{1}{y} \frac{d y}{d x}=\tan ^{-1} x \cdot \frac{1}{x}+\log x \cdot \frac{1}{1+x^{2}} \end{aligned}$
$\frac{1}{y} \frac{d y}{d x}=\frac{\tan ^{-1} x}{x}+\frac{\log x}{1+x^{2}}$
$\begin{aligned} &\frac{d y}{d x}=\left(\frac{\tan ^{-1} x}{x}+\frac{\log x}{1+x^{2}}\right) y \\\\ &\frac{d y}{d x}=x^{\tan ^{-1} x}\left[\frac{\tan ^{-1} x}{x}+\frac{\log x}{1+x^{2}}\right] \end{aligned}$

Differentiation exercise 10.5 question 18 (i)

Answer: $x^{x} \sqrt{x}\left(\log x+1+\frac{1}{2 x}\right)$
Hint: $\text { Diff by } x^{x} \sqrt{x}$
Given: $x^{x} \sqrt{x}$
Solution:
Let, $y=x^{x} \sqrt{x}$
Taking log both side
$\log y=\log x^{x} \sqrt{x} \; \; \; \; \; \; \; \; \; \; \; \quad[\log m n=\log m+\log n]$
$\log y=\log x^{x}+\log \sqrt{x}$
$\log y=x \log x+\log (x)^{\frac{1}{2}}$
$\log y=\left(x+\frac{1}{2}\right) \cdot \log x$
$\frac{d}{d x} \log y=\frac{d}{d x}\left(x+\frac{1}{2}\right) \cdot \log x$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}\left(x+\frac{1}{2}\right) \cdot \log x+\left(x+\frac{1}{2}\right) \cdot \frac{d}{d x} \log x \\\\ &\frac{d y}{d x}=x^{x} \sqrt{x}\left(\log x+1+\frac{1}{2 x}\right) \end{aligned}$

Differentiation exercise 10.5 question 18 (ii)

Answer: $x^{\sin x-\cos x}\left[\frac{\sin x-\cos x}{x}+\log x(\sin x+\cos x)\right]+\frac{4 x}{\left(x^{2}+1\right)^{2}}$
Hint: $\text { Diff by } x^{n}$
Given: $y=x^{\sin x-\cos x}+\frac{x^{2}-1}{x^{2}+1}$
Solution:
$y=u+v$
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$ .........(1)
$u=x^{\sin x-\cos x}$
Take log
$\begin{aligned} &\log u=\log x^{\sin x-\cos x} \\\\ &\log u=(\sin x-\cos x) \cdot \log x \\\\ &\frac{1 d u}{u} \frac{d u}{d x}=(\sin x-\cos x) \cdot \frac{1}{x}+\log x(\cos x+\sin x) \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=\left[\frac{\sin x-\cos x}{x}+\log x(\sin x+\cos x) x^{\sin x-\cos x}\right. \\\\ &v=\frac{x^{2}-1}{x^{2}+1} \end{aligned}$
$\begin{aligned} &\log v=\log \left(\frac{x^{2}-1}{x^{2}+1}\right) \\\\ &=\log \left(x^{2}-1\right)-\log \left(x^{2}+1\right) \end{aligned}$
$\begin{aligned} &\frac{1}{v} \frac{d v}{d x}=\frac{1}{x^{2}-1}(2 x) \cdot \frac{-1}{x^{2}+1}(2 x) \\\\ &\frac{d v}{d x}=\left(\frac{2 x}{x^{2}+1}\right)\left(\frac{-2 x}{x^{2}+1}\right)\left(\frac{x^{2}-1}{x^{2}+1}\right) \end{aligned}$
$\begin{aligned} &=2 \times\left(\frac{\left(x^{2}+1\right)-\left(x^{2}-1\right)}{\left(x^{2}-1\right)\left(x^{2}+1\right)}\right) \\\\ &=\frac{2(2 x)}{\left(x^{2}+1\right)^{2}} \end{aligned}$
$=\frac{4 x}{\left(x^{2}+1\right)^{2}}$
Using (1) we get
$\frac{d y}{d x}=x^{\sin x-\cos x}\left[\frac{\sin x-\cos x}{x}+\log x(\sin x+\cos x)\right]+\frac{4 x}{\left(x^{2}+1\right)^{2}}$

Differentiation exercise 10.5 question 18 (iii)

Answer: $x^{x \cos x}\left(\cos x(1+\log x)-x \sin x \log x-\frac{4 x}{\left(x^{2}-1\right)^{2}}\right.$
Hint: $\text { Diff by } x^{\cos x}$
Given: $y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
Solution: $\text { Let } y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}$
$\begin{aligned} &\text { Let } u=x^{x \cos x} \text { and } v=\frac{x^{2}+1}{x^{2}-1} \\ &\qquad y=u+v \end{aligned}$
Diff by w.r.t.x
$\begin{aligned} &\frac{d y}{d x}=\frac{d(u+v)}{d x} \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \end{aligned}$ $\begin{aligned} &\frac{d y}{d x}=\frac{d(u+v)}{d x} \\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \end{aligned}$
Calculate $\frac{d u}{d x}$
$u=x^{x \cos x} \quad\left[\because \log \left(a^{b}\right)=b \log a\right]$
Take log on both sides
$\begin{aligned} &\log u=\log x^{x \cos x} \\\\ &\log u=x \cos x \log x \end{aligned}$
Diff w.r.t x
$\begin{aligned} &\frac{d(\log u)}{d x}=\frac{d(x \cos x \log x)}{d x} \\\\ &\frac{1}{u} \frac{d u}{d x}=\frac{d(x \cos x \log x)}{d x} \end{aligned}$
Use product rule
$(u v)^{\prime}=u^{\prime} v+v^{\prime} u$
Where $u=x \cos x, v=\log x$
$\frac{1}{u} \frac{d u}{d x}=\frac{d(x)}{d x} \cos x+x \frac{d(\cos x)}{d x} \log x+\cos x$
$\frac{1}{u} \frac{d u}{d x}=\cos x \cdot \log x+x(-\sin x) \log x+x \cos x \frac{1}{x}$
$\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=\cos x \cdot \log x+\cos x-x \sin x \log x \\\\ &\frac{1}{u} \frac{d u}{d x}=\cos x(\log x+1)-x \sin x \log x \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=u[\cos x(\log x+1)-x \sin x \log x] \\\\ &\frac{d u}{d x}=x^{x \cos x}(\cos x(\log x+1)-x \sin x \log x) \end{aligned}$

Calculate $\frac{d v}{d x}$ $v=\frac{x^{2}+1}{x^{2}-1}$
Diff w.r.t x
$\begin{aligned} &\frac{d v}{d x}=\frac{d\left(\frac{x^{2}+1}{x^{2}-1}\right)}{d x} \\\\ &\frac{d v}{d x} \cdot=\frac{d\left(\frac{x^{2}+1}{x^{2}-1}\right)}{d x} \\\\ &\frac{d v}{d x}=\frac{d\left(\frac{x^{2}+1}{x^{2}-1}\right)}{d x} \end{aligned}$
Use quotient rule $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-v^{\prime} u}{v^{2}}$
$\begin{aligned} &\frac{d v}{d x}=\frac{\frac{d\left(x^{2}+1\right)}{d x} \cdot\left(x^{2}-1\right)-\frac{d}{d x}\left(x^{2}-1\right)\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}} \\\\ &\frac{d v}{d x}=\frac{(2 x+0)\left(x^{2}-1\right)-(2 x-0)\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}} \\\\ &\frac{d v}{d x}=\frac{2 x\left(x^{2}-1\right)-2 x\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}} \end{aligned}$
$\frac{d v}{d x}=\frac{2 x\left(x^{2}-1-x^{2}-1\right)}{\left(x^{2}-1\right)^{2}}$
$\frac{d v}{d x}=\frac{2 x(-2)}{\left(x^{2}-1\right)^{2}}$
$=\frac{-4 x}{\left(x^{2}-1\right)^{2}}$
Now $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
Put value
$\frac{d y}{d x}=x^{x \cos x}(\cos x(1+\log x)-x \sin x \log x)-\frac{4 x}{\left(x^{2}-1\right)^{2}}$

Differentiation exercise 10.5 question 18 (iv)

Answer: $(x \cos x)^{x}\left\{(1-x \tan x)+\log (x \cos x)+(x \sin x)^{\frac{1}{x}}\left\{\frac{x \cos x}{x^{2} \sin x}-\log \left(\frac{x \sin x}{x^{2}}\right)\right\}\right.$
Hint: Diff by applying $x^{x}$
Given: $(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$
Solution: $\text { Let } y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}$
$u=(x \cos x)^{x}$
Take log both sides
$\log u=\log (x \cos x)^{x}$
$\frac{1}{u} \times \frac{d u}{d x}=(x \cos x)^{x}\left[(\log x+1)+\left\{\log \cos x+\frac{x}{\cos x}(-\sin x)\right\}\right]$
$\frac{1}{u} \frac{d u}{d x}=(x \cos x)^{x}\{(1-x \tan x)+\log (x \cos x)\}$ .....................(1)
$v=(x \sin x)^{\frac{1}{x}}$
Take log both side
$\log v=\frac{1}{x} \log (x \sin x)$
$\frac{1}{v} \frac{d v}{d x}=\frac{1}{x}\left\{\frac{1}{x \sin x}(x \cos x+\sin x \times 1\}+\log (x \sin x) \times \frac{-1}{x^{2}}\right.$
$\frac{1}{x}\left(\frac{x \cos x+\sin x \times 1}{x \sin x}\right)+\log (x \sin x) \times \frac{-1}{x^{2}}$
$\frac{d u}{d x}=(x \sin x)^{\frac{1}{x}}\left\{\frac{x \cos x+\sin x}{x^{2} \sin x}-\log \frac{(x \sin x)}{x^{2}}\right\}$ ............(2)
From (1) and (2)
$\frac{d y}{d x}=(x \cos x)^{x}\left\{(1-x \tan x)+\log (x \cos x)+(x \sin x)^{\frac{1}{x}}\left\{\frac{x \cos x}{x^{2} \sin x}-\log \left(\frac{x \sin x}{x^{2}}\right)\right\}\right.$

Differentiation exercise 10.5 question 18 (v)

Answer: $\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)$
Hint: $\text { Diff by }\left(x+\frac{1}{x}\right)^{x}$
Given: $\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$
Solution: $\text { Let } y=\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}$
$y=u+v$
Diff w.r.t x
$\frac{d y}{d x}=\frac{d(u+v)}{d x}$
$=\frac{d u}{d x}+\frac{d v}{d x}$
Calculate $\frac{d u}{d x}, u=\left(x+\frac{1}{x}\right)^{x}$
Take log on both side
$\begin{aligned} &\log u=\log \left(x+\frac{1}{x}\right)^{x} \\\\ &\log u=x \log \left(x+\frac{1}{x}\right) \end{aligned}$
$\frac{1}{u} \frac{d u}{d x}=\frac{d\left(x \log \left(1+\frac{1}{x}\right)\right.}{d x}$
Use product rule $(u v)^{\prime}=u^{\prime} v+v^{\prime} u$
$\frac{1}{u} \frac{d u}{d x}=\log \left(x+\frac{1}{x}\right)+\frac{1}{\left(x+\frac{1}{x}\right)} \frac{d}{d x}\left(x+\frac{1}{x}\right) x$
$\frac{1}{u} \frac{d u}{d x}=\log \left(x+\frac{1}{x}\right)+\left(\frac{x}{x^{2}+1}\left(\frac{x^{2}-1}{x^{2}}\right) x\right)$
$\frac{1}{u} \frac{d u}{d x}=\log \left(x+\frac{1}{x}\right)+\left(\frac{x^{2}-1}{x^{2}+1}\right)$
$\frac{d u}{d x}=u\left(\log \left(x+\frac{1}{x}\right)\right)+\left(\frac{x^{2}-1}{x^{2}+1}\right)$
$\frac{d u}{d x}=\left(x+\frac{1}{x}\right)^{x} \cdot\left(\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right)$
Calculate $\frac{d v}{d x}$
$v=x^{\left(1+\frac{1}{x}\right)}$
Taking $\log v=\left[1+\frac{1}{x}\right] \log x$
Diff w.r.t. x
Use product rule
$\frac{1}{v} \frac{d v}{d x}=\frac{d\left(1+\frac{1}{x}\right)}{d x} \cdot \log x+\frac{d(\log x)}{d x}\left(1+\frac{1}{x}\right)$
$\frac{1}{v} \frac{d v}{d x}=\frac{d(1)}{d x}+\frac{d\left(\frac{1}{x}\right)}{d x} \cdot \log x+\frac{1}{x}\left(1+\frac{1}{x}\right)$
$\frac{1}{v}\left(\frac{d v}{d x}\right)=\frac{-1}{x^{2}} \cdot \log x+\frac{1}{x}\left(1+\frac{1}{x}\right)$
$\begin{aligned} &\frac{1}{v}\left(\frac{d v}{d x}\right)=\left(\frac{-\log x+x+1}{x^{2}}\right) \\\\ &\frac{d v}{d x}=v\left(\frac{-\log x+x+1}{x^{2}}\right) \end{aligned}$
$\frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)$
Now $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
Put value of $\frac{d u}{d x} \text { and } \frac{d v}{d x}$
$\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)$

Differentiation exercise 10.5 question 18 (vi)

Answer: $e^{\sin x} \cos x+(\tan x)^{x} \cdot\left[\log (\tan x)+\frac{x \sec ^{2} x}{\tan x}\right]$
Hint: Diff by $e^{\sin x}$
Given: $e^{\sin x}+(\tan x)^{x}$
Solution: $y=e^{\sin x}+(\tan x)^{x}$
Let $z=(\tan x)^{x}$
Take log on both sides
$\log z=x \log \tan x$
Diff w.r.t x
$\frac{1}{z} \frac{d z}{d x}=\log \tan x+\frac{x \sec ^{2} x}{\tan x}$
Thus
$\frac{d y}{d x}=e^{\sin x} \cos x+(\tan x)^{x} \cdot\left[\log (\tan x)+\frac{x \sec ^{2} x}{\tan x}\right]$

Differentiation exercise 10.5 question 18 (viii)

Answer: $\left(\frac{x^{2}-3}{x}+2 x \log x\right) x^{x^{2}-3}+\left(\frac{x^{2}}{x-3}+2 x \log (x-3)\right)(x-3)^{x^{2}}$
Hint: Diff by $x^{n-3}$
Given: $y=x^{x^{2}-3}+(x-3)^{x^{2}}$
Solution: $y=u+v$
$u=x^{x^{2}-3}$
$\begin{aligned} &\log u=\log x^{x^{2}-3} \\\\ &\log u=\left(x^{2}-3\right) \log x \end{aligned}$
$\frac{1}{u} \frac{d u}{d x}=\left(x^{2}-3\right) \cdot \frac{1}{x}+\log x(2 x)$
$\frac{d u}{d x}=\left(\frac{x^{2}-3}{x}+2 x \log x\right) \cdot x^{x^{2}-3}$
Now $v=(x-3)^{x^{2}}$
Take log on both sides
$\begin{aligned} &\log v=\log (x-3)^{x^{2}} \\\\ &\frac{1}{v} \frac{d v}{d x}=x^{2} \log (x-3) \end{aligned}$
$\frac{d v}{d x}=\left[\frac{x^{2}}{x-3}+2 x \log (x-3)\right](x-3)^{x^{2}}$
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\left(\frac{x^{2}-3}{x}+2 x \log x\right) x^{x^{2}-3}+\left(\frac{x^{2}}{x-3}+2 x \log (x-3)\right)(x-3)^{x^{2}}$

Differentiation exercise 10.5 question 19

Answer: $\frac{d y}{d x}=e^{x}+10^{x} \log 10+x^{x}(\log n+1)$
Hint: Differentiate the statement taking u and v
Given: $y=e^{x}+10^{x}+x^{x}$
Solution:
Let $x^{x}=u$
$\begin{aligned} &y=e^{x}+10^{x}+u \\\\ &\frac{d y}{d x}=\frac{d\left(e^{x}\right)}{d x}+\frac{d\left(10^{x}\right)}{d x}+\frac{d u}{d x} \end{aligned}$ ..........(1)
For $\mathrm{u}, u=x^{x}$
$\begin{aligned} &\log u=\log x^{x} \\\\ &\frac{d}{d x}(\log u)=x \log x \end{aligned}$
$\frac{1}{u} \frac{d u}{d x}=\log x+x \cdot \frac{d}{d x}(\log x)$
~~$\begin{aligned} &\frac{d u}{d x}=u\left[\log x+\not x \cdot \frac{1}{\not{x}}\right] \\\\ &=x^{x}(\log x+1) \end{aligned}$~~
Put $\frac{d u}{d x}$ in eq (1)
$\begin{aligned} &\frac{d}{d x}\left(e^{x}\right)=e^{x} \\\\ &\frac{d y}{d x}=e^{x}+10^{x} \log 10+x^{x}(\log x+1) \end{aligned}$ 0

Differentiation exercise 10.5 question 20

Answer: $\frac{d y}{d x}=n \cdot x^{n-1}+n^{x} \cdot \log n+x^{x}(\log x+1)$
Hint: Differentiate the equation taking log on both sides
Given: $y=x^{x}+n^{x}+x^{x}+n^{n}$
Solution: $y=x^{x}+n^{x}+x^{x}+n^{n}$
As we know
$\left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$
$\left[\frac{d}{d x}(\log x)=\frac{1}{x}\right]$
According to question
$\frac{d y}{d x}=\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(n^{x}\right)+\frac{d u}{d x}+\frac{d}{d x}\left(x^{n}\right)$ .......(1)
Take $u=x^{x}$
$\begin{aligned} \frac{d}{d x}(\log u) &=\log x^{x} \\ &=\frac{d}{d x}(x \log x) \end{aligned}$
$\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=\frac{d x}{d x} \cdot \log x+x \frac{d}{d x}(\log x) \\\\ &\frac{d u}{d x}=u\left[\log +\not{x} \cdot \frac{1}{\not{x}}\right] \\\\ &\frac{d u}{d x}=x^{x}[\log x+1] \end{aligned}\$
Put in eq. (1)
$\frac{d y}{d x}=n \cdot x^{n-1}+n^{x} \cdot \log n+x^{x}(\log x+1)$

Differentiation exercise 10.5 question 21

Answer: $\frac{d y}{d x}=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}\left[\frac{6 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2(x-3)}-\frac{2 x}{(4 x-1)}\right]$
Hint: Differentiate the equation taking log on both sides
Given: $y=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}$
Solution:
$y=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}$
Taking log on both sides,
$\log y=\log \left[\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{(x-3)^{\frac{1}{2}}(4 x-1)^{\frac{1}{2}}}\right]$
w.r.t x
$\frac{1}{y} \frac{d y}{d x}=3 \log \left(x^{2}-1\right)+\log (2 x-1)-\frac{1}{2} \log (x-3)-\frac{1}{2} \log (4 x-1)$
$\left[\begin{array}{l} \because \log A B=\log A+\log B \\\\ \log \frac{a}{B}=\log A-\log B \\\\ \log A^{n}=n \log A \end{array}\right]$
$\frac{1}{y} \frac{d y}{d x}=\frac{3.2 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2} \cdot \frac{1}{(x-3)}-\frac{1}{2} \cdot \frac{1}{(4 x-1)} 4 x$
$\frac{d y}{d x}=y\left[\frac{6 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2(x-3)}-\frac{2 x}{(4 x-1)}\right]$
$\frac{d y}{d x}=\frac{\left(x^{2}-1\right)^{3}(2 x-1)}{\sqrt{(x-3)(4 x-1)}}\left[\frac{6 x}{x^{2}-1}+\frac{2}{2 x-1}-\frac{1}{2(x-3)}-\frac{2 x}{(4 x-1)}\right]$

Differentiation exercise 10.5 question 22

Answer: $\frac{d y}{d x}=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}\left[a+\tan x+\frac{1}{x \log x}+\frac{1}{1-2 x}\right]$
Hint: Differentiate the equation taking log on both sides
Given: $y=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}$
Solution:
$y=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}$
Taking log on both sides,
$\log y=\log \left[\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}\right]$
$=\log \left(e^{a x}\right)+\log (\sec x)+\log (\log x)-\log (1-2 x)^{\frac{1}{2}}$
$\log y=a x+\log (\sec x)+\log (\log x)-\frac{1}{2} \log (1-2 x)$
$\frac{1}{y} \frac{d y}{d x}=a+\frac{\sec x \cdot \tan x}{\sec x}+\frac{1}{\log x} \cdot \frac{1}{x}-\frac{1}{2} \cdot \frac{1(-2)}{(1-2 x)}$
$\frac{d y}{d x}=y\left[a+\tan x+\frac{1}{x \log x}+\frac{1}{1-2 x}\right]$
$\frac{d y}{d x}=\frac{e^{a x} \sec x \log x}{\sqrt{1-2 x}}\left[a+\tan x+\frac{1}{x \log x}+\frac{1}{1-2 x}\right]$

Differentiation exercise 10.5 question 23

Answer: $\frac{d y}{d x}=e^{3 x} \cdot \sin 4 x .2 x\left[e^{3 x} \cdot \sin 4 x \cdot 2^{x}(3+4 \cot 4 x+\ln 2)\right]$
Hint: Differentiate the equation taking log on both sides
Given: $y=e^{3 x} \cdot \sin 4 x \cdot 2^{x}$
Solution:
$[\because \log A B=\log A+\log B]$
$\ln y=\ln e^{3 x}+\ln \sin 4 x+\ln 2^{x}$
Diff w.r.t x
$\frac{1}{y} \frac{d y}{d x}=\frac{1}{e^{3 x}} \cdot \cdot e^{3 x} \cdot 3+\frac{\cos 4 x}{\sin 4 x} \cdot 4+\ln 2$
$\frac{d y}{d x}=e^{3 x} \cdot \sin 4 x .2 x\left[e^{3 x} \cdot \sin 4 x .2 x(3+4 \cot 4 x+\ln 2)\right]$

Differentiation exercise 10.5 question 24

Answer: $\frac{d y}{d x}=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x[\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x]$
Hint: Differentiate the equation taking log on both sides
Given: $y=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x$
Solution:
$y=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x$
Taking log on both sides,
$\log y=\log (\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x)$
$=\log \sin x+\log \sin 2 x+\log \sin 3 x+\log \sin 4 x$
$\frac{1}{y} \frac{d y}{d x}=\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x\left[\because \frac{d}{d x}(\log \sin x)=\frac{1}{\sin x} \cdot \cos x=\cot x\right]$
$\frac{d y}{d x}=y[\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x]$
$\frac{d y}{d x}=\sin x \cdot \sin 2 x \cdot \sin 3 x \cdot \sin 4 x[\cot x+\cot 2 x .2+3 \cot 3 x+4 \cot 4 x]$

Differentiation exercise 10.5 question 27

Answer: $\frac{d y}{d x}=(\tan x)^{\cot x}\left[\operatorname{cosec}^{2} x(1-\log \tan x)\right]+(\cot x)^{\tan x}\left[\sec ^{2} x \cdot \log (\cot x)-1\right]$
Hint: Differentiate the equation taking log on both sides
Given: $y=(\tan x)^{\cot x}+(\cot x)^{\tan x}$
Solution: $y=(\tan x)^{\cot x}+(\cot x)^{\tan x}$
Let's assume $y_{1}=(\tan x)^{\cot x} \text { and }$
$\begin{aligned} &y_{2}=(\cot x)^{\tan x} \\\\ &\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x} \end{aligned}$ .......(1)
Now $y_{1}=(\tan x)^{\cot x}$
$\log y_{1}=\cot x \cdot \log (\tan x)$
On diff both sides w.r.t. x
$\frac{1}{y_{1}} \frac{d y_{1}}{d x}=\left[\cot x \cdot \frac{1}{\tan x} \cdot \sec ^{2} x+\log (\tan x) x-\cos e c^{2} x\right]$
$\frac{d y_{1}}{d x}=(\tan x)^{\cot x}\left[\operatorname{cosec}^{2} x(1-\log (\tan x)]\right.$ .......(2)
$\begin{aligned} &y_{2}=(\cot x)^{\tan x} \\\\ &\log y_{2}=\tan x \log (\cot x) \end{aligned}$
On diff both sides w.r.t. x
$\frac{1}{y_{2}} \frac{d y_{2}}{d x}=\left[\tan x \cdot \frac{1}{\cot x} \cdot-\left(\cos e c^{2 x}\right)+\log \cot x \cdot \sec ^{2} x\right]$
$\frac{d y_{2}}{d x}=y_{2}\left[-\sec ^{2} x+\log (\cot x) \cdot \sec ^{2} x\right]$
$\frac{d y_{2}}{d x}=(\cot x)^{\tan x}\left[\sec ^{2} x \log (\cot x-1)\right]$ ........(3)
$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x}$

Differentiation exercise 10.5 question 28 (i)

Answer: $\frac{d y}{d x}=(\sin x)^{x}[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}$
Hint: Differentiate the equation taking log on both sides
Given:
Solution:
$\begin{aligned} &y=(\sin x)^{x}+\sin ^{-1}(\sqrt{x}) \\\\ &y=y_{1}+y_{2} \end{aligned}$
Diff w.r.t x
$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x}$ .......(1)
$y_{1}=(\sin x)^{x}$
Taking log on both sides
$\log y_{1}=\log (\sin x)^{x} \quad\left[\because \log m^{n}=n \log m\right]$
$\log y_{1}=x \log \sin x$
$\frac{1}{y_{1}} \frac{d y}{d x}=x \cdot \frac{1}{\sin x} \cdot \cos x+\log \sin x \cdot 1$ $\left[\begin{array}{l} \because \frac{d}{d x} \log x=\frac{1}{x} \\\\ \frac{d}{d x} I . I I=I \frac{d}{d x} I I+I I \frac{d}{d x} I \end{array}\right]$
$\begin{aligned} &\frac{d y_{1}}{d x}=y_{1}[x \cdot \cot x+\log \sin x] \\\\ &\frac{d y_{1}}{d x}=(\sin x)^{n}[x . \cot x+\log \sin x] \end{aligned}$ .......(2)
$\begin{aligned} &y_{2}=\sin ^{-1}(\sqrt{x}) \\\\ &\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x}) \end{aligned}$
$\begin{aligned} &\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}} \\\\ &\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x(1-x)}} \end{aligned}$
$\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$ .........(3)
Put (2) and (3) in eq(1)
$\frac{d y}{d x}=(\sin x)^{x}[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}$

Differentiation exercise 10.5 question 28 (ii)

Answer: $\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)+\frac{1}{2 \sqrt{x-x^{2}}}$
Hint: Differentiate the equation taking log on both sides
Given: $y=x^{\sin x}+\sin ^{-1}(\sqrt{x})$
Solution: $y=x^{\sin x}+\sin ^{-1}(\sqrt{x})$
$y=y_{1}+y_{2}$
Diff w.r.t x
$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x}$ .......(1)
$\begin{aligned} &\text { Let } y_{1}=x^{\sin x} \\\\ &\log y_{1}=\log x(\sin x)^{x} \end{aligned}$
On diff both side with respect to x we get
$\frac{1}{y} \frac{d y}{d x}=\sin x \frac{d}{d x} \log x+\log x \frac{d}{d x} \sin x$
$\frac{1}{y} \frac{d y}{d x}=\sin x \frac{1}{x}+\log x \cos x$
$\frac{d y}{d x}=y\left(\frac{\sin x}{x}+\log x \cos x\right)$
$\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)$ from 1
$y_{2}=\sin ^{-1}(\sqrt{x})$
$\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x}) \quad\left[\because \frac{d}{d x} \sin ^{-1}(\sqrt{x})=\frac{1}{\sqrt{1-x^{2}}}\right]$
$\begin{aligned} &\frac{d y_{2}}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}} \\\\ &\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x(1-x)}} \end{aligned}$
$\frac{d y_{2}}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$ ........(3)
Put (2) and (3) in eq(1)
$\frac{d y}{d x}=x^{\sin x}\left(\frac{\sin x}{x}+\log x \cos x\right)+\frac{1}{2 \sqrt{x-x^{2}}}$

Differentiation exercise 10.5 question 29 (i)

Answer: $\frac{d y}{d x}=x \cos x\left(\frac{\cos x}{x}-\sin x \log x\right)+\sin ^{\tan x}\left(1+\sec ^{2} x \log \sin x\right)$
Hint: To solve this equation we use $f(x)^{f(x)}$ formula
Given:
Solution:
$\begin{gathered} y=x^{\cos x}+\sin x^{\tan x} \\\\ y=(f(x))^{f(x)} \end{gathered}$
$\begin{aligned} &\text { Let } u=x^{\cos x} \\\\ &\log u=x^{\cos x} \\\\ &\log u=\cos x \log x \end{aligned}$
$\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=\cos x \cdot \frac{1}{x}+\log x(-\sin x) \\\\ &\frac{d u}{d x}=u\left(\frac{\cos x}{x}-\sin x \log x\right) \end{aligned}$
$\begin{aligned} &=x \cos x\left[\left(\frac{\cos x}{x}-\sin x \log x\right)\right] \\\\ &v=\sin x^{\tan x} \\\\ &\log v=\tan x \log \sin x \end{aligned}$
$\frac{1}{v} \frac{d v}{d x}=\tan x\left[\frac{1}{\sin x} \frac{d}{d x}(\sin x)\right]+\log \sin x \sec ^{2} x$
$\frac{d v}{d x}=v\left(\tan x \times \frac{\cos x}{\sin x}+\sec ^{2} x \log \sin x\right)$
$\frac{d v}{d x}=v\left(\frac{\sin x}{\cos x} \times \frac{\cos x}{\sin x}+\sec \sin ^{2} x \log \sin x\right)$
$\frac{d v}{d x}=\sin x^{\tan x}\left(1+\sec ^{2} x \log \sin x\right)$
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
$\frac{d v}{d x}=x \cos x\left[\left(\frac{\cos x}{x}-\sin x \log x\right)\right]+\sin x^{\tan x}\left ( 1+\sec ^{2} x\log \sin x\right )$

Differentiation exercise 10.5 question 29 (ii)

Answer: $\frac{d y}{d x}=x^{x}(1+\log x)+(\sin x)^{x}[x \cot x+\log (\sin x)]$
Hint: Differentiate the equation taking log on both sides
Given: $y=x^{x}+(\sin x)^{x}$
Solution: $y=x^{x}+(\sin x)^{x}$
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$ .............(1)
$\begin{aligned} &\text { Let } u=x^{x} \\ &\log u=\log x^{x} \\ &\log u=x \log x \end{aligned}$
Diff w.r.t x
$\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{x}+\log x \\\\ &\frac{d u}{d x}=u(1+\log x) \\\\ &\frac{d u}{d x}=x^{x}(1+\log x) \end{aligned}$ ..............(2)
$\begin{aligned} &\text { Again let } v=(\sin x)^{x} \\\\ &\log v=\log (\sin x)^{x} \\\\ &\log v=x \log (\sin x) \end{aligned}$
Diff w.r.t x
$\begin{aligned} &\frac{1}{v} \frac{d v}{d x}=x \cdot \frac{\cos x}{\sin x}+\log (\sin x) \\\\ &\frac{d v}{d x}=v[x \cdot \cot x+\log (\sin x)] \\\\ &\frac{d v}{d x}=(\sin x)^{x}[x \cdot \cot x+\log (\sin x)] \end{aligned}$ ...........(3)
Put (2) and (3) in eq (1)
$\frac{d y}{d x}=x^{x}(1+\log x)+(\sin x)^{x}[x \cot x+\log (\sin x)]$

Differentiation exercise 10.5 question 30

Answer: $\frac{d y}{d x}=(\tan x)^{\log x}\left[\frac{\log x(\tan x)}{x}+\frac{\log x \sec ^{2} x}{\tan x}\right]$
Hint: Differentiate the equation taking log on both sides
Given: $y=(\tan x)^{\log x}+\cos ^{2}\left(\frac{\pi}{4}\right)$
Solution:
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(u+\cos ^{2}\left(\frac{\pi}{4}\right)\right) \\\\ &\frac{d y}{d x}=\frac{d u}{d x} \end{aligned}$
Let $u=(\tan x)^{\log x}$
$\begin{aligned} &\log u=\log (\tan x)^{\log x} \\\\ &\frac{d}{d x}(\log u)=\log (\tan x)^{\log x} \end{aligned}$
$\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(\log x) \log (\tan x)+\log x \cdot \frac{d}{d x} \log (\tan x)$
$\frac{d u}{d x}=u\left[\frac{\log (\tan x)}{x}+\log x \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\right]$
$\begin{aligned} &\frac{d u}{d x}=(\tan x)^{\log x}\left[\frac{\log x(\tan x)}{x}+\frac{\log x \cdot \sec ^{2} x}{\tan x}\right] \\\\ &\frac{d y}{d x}=(\tan x)^{\log x}\left[\frac{\log x(\tan x)}{x}+\frac{\log x \sec ^{2} x}{\tan x}\right] \end{aligned}$

Differentiation exercise 10.5 question 31

Answer: $\frac{d y}{d x}=x^{x}(\log x+1)+x^{\frac{1}{x-2}}(1-\log x)$
Hint: Differentiate the equation taking log on both sides
Given: $y=x^{x}+x^{\frac{1}{x}}$
Solution: $y=x^{x}+x^{\frac{1}{x}}$
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$
Let $u=x^{x}$
$\begin{aligned} &\log u=\log x^{x} \\\\ &\frac{d}{d x}(\log u)=\log x^{x} \\\\ &\frac{d}{d x}(\log u)=\frac{d}{d x}(x \log x) \end{aligned}$
$\begin{aligned} &\frac{1}{u} \cdot \frac{d u}{d x}=\frac{d x}{d x} \cdot \log x+x \frac{d}{d x}(\log x) \\\\ &\left.\frac{d u}{d x}=u[\log x+x] \cdot \frac{1}{x}\right] \\\\ &\frac{d u}{d x}=x^{x}[\log x+1] \end{aligned}$
Let $v=x^{\frac{1}{x}}$
$\log v=\log x^{\frac{1}{x}}$
$\begin{aligned} &\frac{d}{d x}(\log v)=\frac{d}{d x}\left(\frac{1}{x} \log x\right) \\\\ &\frac{1}{v} \cdot \frac{d v}{d x}=\frac{1}{x} \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}\left(\frac{1}{x}\right) \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=v\left[\frac{1}{x^{2}}+\log x \cdot\left(\frac{-1}{x^{2}}\right)\right] \\\\ &=x^{\frac{1}{x}}\left[\frac{1-\log x}{x^{2}}\right] \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\\\ &=x^{x}(\log x+1)+x^{\frac{1}{x-2}}(1-\log x) \end{aligned}$

Differentiation exercise 10.5 question 32

Answer: $\frac{d y}{d x}=x^{\log x}\left[\frac{2 \log x}{x}\right]+(\log x)^{x}\left(\frac{1}{\log x}+\log (\log x)\right)$
Hint: Differentiate the equation taking log on both sides
Given: $y=x^{\log x}+(\log x)^{x}$
Solution:
Let’s assume $y_{1}=x^{\log x} \text { and }$
$\begin{aligned} &y_{2}=(\log x)^{x} \\\\ &\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x} \end{aligned}$
Now $y_{1}=x^{\log x}$
$\log y_{1}=\log x \cdot \log x$
Diff w.r.t x
$\frac{1}{y_{1}} \times \frac{d y_{1}}{d x}=\left[\log x \cdot \frac{1}{x}+\log x \cdot \frac{1}{x}\right]$
$\frac{d y_{1}}{d x}=x^{\log x}\left[\frac{2 \log x}{x}\right]$ .............(1)
$\begin{aligned} &y_{2}=(\log x)^{x} \\\\ &\log y_{2}=x \log (\log x) \end{aligned}$
Diff both sides w.r.t x
$\frac{1}{y_{2}} \frac{d y_{2}}{d x}=\left[x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \cdot 1+\log (\log x)\right]$
$\frac{d y_{2}}{d x}=(\log x)^{x}\left[\frac{1}{\log x}+\log (\log x)\right]$ ...............(2)
$\begin{aligned} &\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d y_{2}}{d x} \\\\ &x^{\log x}\left[\frac{2 \log x}{x}\right]+(\log x)^{x}\left(\frac{1}{\log x}+\log (\log x)\right) \end{aligned}$

Differentiation exercise 10.5 question 33

Answer: $\frac{d y}{d x}=\frac{y}{x}$
Hint: To differentiate the equation take log on both the sides
Given: $x^{13} y^{7}=(x+y)^{20}$
Solution:
$\log \left(x^{13} y^{7}\right)=\log (x+y)^{20}$
$\log \left(x^{13}\right)+\log \left(y^{7}\right)=\log (x+y)^{20}$
$\frac{d}{d x}(13 \log x+7 \log y)=\frac{d}{d x}(20 \log (x+y))$
$13 \cdot \frac{1}{x}+7 \cdot \frac{1}{y} \cdot \frac{d y}{d x}=20 \cdot \frac{1}{x+y} \cdot \frac{d}{d x}(x+y)$
$\frac{13}{x}+\frac{7}{y} \frac{d y}{d x}=\frac{20}{x+y} \cdot\left(1+\frac{d y}{d x}\right)$
$\left(\frac{7}{y}-\frac{20}{x+y}\right) \cdot \frac{d y}{d x}=\frac{20}{x+y}-\frac{13}{x}$
$\frac{7 x+7 y-20 y}{(x+y) y} \frac{d y}{d x}=\frac{20 x-13 x-13 y}{x(x+y)}$
$\frac{7 x-13 y}{y(x+y)} \frac{d y}{d x}=\frac{7 x-13 y}{x(x+y)}$
$\begin{aligned} &\frac{d y}{d x}=\frac{7 x-13 y}{\sqrt{(x+y)}} \times \frac{y[(x+y)}{\frac{d y}{7 x-13 y}} \\\\ &\frac{d y}{d x}=\frac{y}{x} \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 34

Answer: $x \cdot \frac{d y}{d x}=2 y$
Hint: Differentiate the equation by taking log on both the sides
Given: $x^{16} y^{9}=\left(x^{2}+y\right)^{17}$
Solution:
$x^{16} y^{9}=\left(x^{2}+y\right)^{17}$
Taking log on both sides,
$\begin{aligned} &\log \left(x^{16} y^{9}\right)=\log \left(x^{2}+y\right)^{17} \\\\ &\log \left(x^{16}\right)+\log \left(y^{9}\right)=\log \left(x^{2}+y\right)^{17} \end{aligned}$
$\frac{d}{d x} 16 \log (x)+\frac{d}{d x} 9 \log (y)=\frac{d}{d x}\left(17 \log \left(x^{2}+y\right)\right)$
$\frac{16}{x}+\frac{9}{y} \cdot \frac{d y}{d x}=17 \cdot \frac{1}{x+y} \cdot\left[\left(x^{2}+y\right) \frac{d y}{d x}\right]$
$\frac{16}{x}-\frac{34}{x^{2}+y}=\left(\frac{17}{x^{2}+y}-\frac{9}{y}\right) \frac{d y}{d x}$
$\begin{aligned} &\frac{16 x^{2}+16 y-34 x}{x\left(x^{2}+y\right)}=\left(\frac{17 y-9 x^{2}-9 y}{\left(x^{2}+y\right) y} \cdot \frac{d y}{d x}\right. \\\\ &\frac{-18 x^{2}+16 y}{x\left(x^{2}+y\right)}=\frac{-9 x^{2}+8 y}{\left(x^{2}+y\right) y} \cdot \frac{d y}{d x} \end{aligned}$ $x \cdot \frac{d y}{d x}=2 y$

Differentiation exercise 10.5 question 35

Answer: $\frac{d y}{d x}=\cos \left(x^{x}\right) \cdot x^{x}(1+\log x)$
Hint: Differentiate the equation by taking log on both the sides
Given: $y=\sin (x)^{x}$
Solution:
Let take $u=(x)^{x}$
$\begin{aligned} &y=\sin u \\\\ &\frac{d y}{d x}=\cos u \cdot \frac{d u}{d x} \end{aligned}$
$\frac{d}{d x}(\log u)=\frac{d}{d x}(x \log x)$
$\begin{aligned} &\frac{1}{u} \cdot \frac{d u}{d x}=\log x \cdot \frac{d x}{d x}+x \cdot \frac{d}{d x}(\log x) \\\\ &\frac{d u}{d x}=u\left[\log x+x \cdot \frac{1}{x}\right] \\\\ &\frac{d u}{d x}=x^{x}[\log x+1] \end{aligned}$
Put the value of $du/dx$ we get
$\frac{d y}{d x}=\cos \left(x^{x}\right) \cdot x^{x} \cdot \log (x+1)$

Differentiation exercise 10.5 question 36

Answer: $\frac{d y}{d x}=-\left\{\frac{\left.x^{x}(1+\log x)+y^{x} \log y\right)}{x \cdot y^{(x-1)}}\right\}$
Hint: To solve this equation we denote both term as u and v
Given: $x^{x}+y^{x}=1$
Solution:
$\frac{d u}{d x}+\frac{d v}{d x}=0$
$u=x^{x}$
Taking log on both sides,
$\log u=x \log x$
on diff. we get
$\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x \frac{d}{d x}(\log x)+\log x \cdot \frac{d x}{d x} \\\\ &\frac{d u}{d x}=u\left(x \cdot \frac{1}{x}+\log x\right) \end{aligned}$
$\begin{aligned} &\frac{d u}{d x}=x^{x}(\log x+1) \\\\ &v=y^{x} \end{aligned}$
Taking log both side
$\log v=x \log y$
on diff. we get
$\frac{1}{v} \cdot \frac{d v}{d x}=x \cdot \frac{d}{d x}(\log y)+\log y \cdot \frac{d x}{d x}$
$\begin{aligned} &\frac{d v}{d x}=v\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right] \\\\ &\frac{d v}{d x}=y^{x} \cdot\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right] \end{aligned}$
Hence $\frac{d y}{d x}=-\left\{\frac{\left.x^{x}(1+\log x)+y^{x} \log y\right)}{x \cdot y^{(x-1)}}\right\}$

Differentiation exercise 10.5 question 37

Answer: $\frac{d y}{d x}=\frac{-(y(y+x \log y))}{x(y \log x+x)}$
Hint: Adding log on both sides of equation
Given: $-x^{y} \cdot y^{x}=1$
Solution:
Taking log on both sides,
$\begin{aligned} &\log x^{y}+\log y^{x}=0 \\\\ &\mathrm{y} \log x+x \log y=0 \end{aligned}$
$\log x \frac{d y}{d x}+y \cdot \frac{d}{d x}(\log x)+\log y \frac{d}{d x} x+x \frac{d}{d x}(\log y)=0$
$\begin{aligned} &\log x \frac{d y}{d x}+\frac{y}{x}+\log y+\frac{x}{y} \frac{d y}{d x}=0 \\\\ &\frac{d y}{d x}\left(\log x+\frac{x}{y}\right)=-\left(\frac{y}{x}+\log y\right) \end{aligned}$
$\frac{d y}{d x}=\frac{-\left(\frac{y}{x}+\log y\right)}{\left(\log x+\cdot \frac{x}{y}\right)} \frac{d y}{d x}$
$\begin{aligned} &\frac{d y}{d x}=\frac{-(y(y+x \log y)}{x(y \log x+x)} \\\\ &\frac{d y}{d x}=\frac{-(y(y+x \log y)}{x(y \log x+x)} \end{aligned}$

Differentiation exercise 10.5 question 38

Answer: $\frac{d y}{d x}=\frac{x(1+\log (x+y)-y x \log y}{y \log x+x-y(1+\log x+y)}$
Hint: To solve this we add log on both sides
Given: $x^{y}+y^{x}=(x+y)^{x+y}$
Solution:
Taking log on both sides,
$\begin{aligned} &\log x^{y}+\log y^{x}=\log (x+y)^{x+y} \\\\ &y \log x+x \log y=(x+y)+\log (x+y) \end{aligned}$ $\left[\because \frac{d}{d x}(u-v)=u . d v+v \cdot d u\right]$
$y \cdot \log x+x \log y^{\prime}+\frac{x}{y} y^{\prime}+\log y=\frac{(x+y)}{(x+y)}\left(1+y^{\prime}\right)+\log (x+y)\left(1+y^{\prime}\right)$
$\begin{aligned} &=\left(1+y^{\prime}\right)(1+\log (x+y)) \\\\ &\frac{y}{x}+\log y+y^{\prime}\left(\log x+\frac{x}{y}\right)=1+\log (x+y)+y^{\prime}(1+\log (x+y)) \end{aligned}$
$y^{\prime}\left(\log x+\frac{x}{y}-(1+\log (x+y))=1+\log (x+y)-\left(\frac{y}{x}+\log y\right)\right.$
$y^{\prime}=\frac{x\left(1+\log (x+y)^{\prime-y}+x \log y\right)}{y \log x+x-y(1+\log (x+y))}$

Differentiation exercise 10.5 question 39

Answer: $\frac{d y}{d x}=-\frac{m y}{n x}$
Hint: To solve this we add log on both sides
Given: $x^{m} y^{n}=1$
Solution:
$\log \left(x^{m} y^{n}\right)=\log 1=0$
$\begin{aligned} &\log x^{m}+\log y^{n}=0 \\\\ &\frac{d}{d x} m \log x+n \log y=\frac{d}{d x} 0 \end{aligned}$
$\begin{aligned} &m \cdot \frac{1}{x}+n \frac{1}{y} \frac{d y}{d x}=0 \\\\ &\frac{d y}{d x}=\frac{-m y}{n x} \end{aligned}$

Differentiation exercise 10.5 question 40

Answer: $\frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y}$
Hint: To solve this we differentiate method
Given: $y^{x}=e^{y-x}$
Solution:
$\frac{d}{d x} x \log y=(y-x) \log e=\frac{d}{d x}(y-x)$
$\begin{aligned} &\frac{d x}{d x} \log y+x \frac{d}{d x}(\log y)=\frac{d y}{d x}-\frac{d x}{d x} \\\\ &\log y+x \frac{1}{y} \frac{d y}{d x}=\frac{d y}{d x}-1 \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}\left(\frac{x}{y}-1\right)=-(1+\log y) \\\\ &\frac{d y}{d x}=\frac{-y(1+\log y)}{x-y} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{y(1+\log y)}{y-x} \\\\ &\frac{d y}{d x}=\frac{y(1+\log y)}{x \cdot \log y} \\\\ &\frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y} \end{aligned}$

Differentiation exercise 10.5 question 41

Answer: $\frac{d y}{d x}=\frac{\log (\operatorname{cosy})-y \cot x}{\log (\sin x)+x \tan y}$
Hint: To solve this equation we will convert log function
Given: $(\sin x)^{y}=(\cos y)^{x}$
Solution:
$\begin{aligned} &\log (\sin x)^{y}=\log (\cos y)^{x} \\\\ &y\log (\sin x)=x \log (\cos y) \end{aligned}$
$\begin{aligned} &y \frac{d}{d x} \log (\sin x)+\log \sin x \frac{d y}{d x}=x \frac{d}{d x} \log \cos y+\log \cos y \frac{d x}{d x} \\\\ &y \frac{1}{\sin x} \cos x+\log \sin x \frac{d y}{d x}=x \frac{1}{\cos y}(-\sin x) \frac{d y}{d x}+\log \cos y \cdot 1 \end{aligned}$
$\begin{aligned} &y \cot x+\log \sin x \frac{d y}{d x}=-x \tan \frac{d y}{d x}+\log \cos y \\\\ &\frac{d x}{d x} \log (\sin x)+y \frac{\cos x}{\sin x}=\log (\cos y)+x \frac{-\sin y}{\cos y} \end{aligned}$
$\begin{aligned} &\frac{d x}{d x} \log (\sin x)+x \tan y=\log (\cos y)-y \cot x \\\\ &\frac{d y}{d x}=\frac{\log (\cos y)-y \cot x}{\log (\sin x)+x \tan y} \end{aligned}$
Hence it is proved

Differentiation exercise 10.5 question 42

Answer: $\frac{d y}{d x}=\frac{\log (\tan y+y \tan x)}{\log (\cos x)-x \sec y \; \cos ec \; y}$
Hint: To solve this equation, we will convert them in log
Given: $(\cos x)^{y}=(\tan y)^{x}$
Solution:
Taking log on both sides,
$\begin{aligned} &\log (\cos x)^{y}=\log (\tan y)^{x} \\\\ &y \log (\cos x)=x \log (\tan x) \end{aligned}$
$\frac{d}{d x}(y \log (\cos x))=\frac{d}{d x}(x \log (\tan x))$ $\left[\begin{array}{l} \because \frac{d}{d x}(\log x)=\frac{1}{x} \\\\ \frac{d}{d x} \cos x=-\sin x \\\\ \frac{d}{d x}(\tan x)=\sec ^{2} x \end{array}\right]$
$\begin{aligned} &\frac{d y}{d x} \cdot \log (\cos x)+y \frac{d}{d x}(\log (\cos x)) =\frac{d x}{d x} \cdot \log (\tan x)+x \cdot \frac{d}{d x}(\log (\tan x)) \end{aligned}$

$\frac{d y}{d x} \log (\cos x)+y\left(\frac{-\sin x}{\cos x}\right)=\log (\tan x)+x \cdot \frac{\sec ^{2} y}{\tan y}-\frac{d y}{d x}$
$\frac{d y}{d x} \log (\cos x)-\tan x \cdot y=\log (\tan x)+x \cdot \frac{\frac{1}{\cos ^{2} y}}{\frac{\sin y}{\cos y}} \cdot \frac{d y}{d x}$
$\frac{d y}{d x}\left[\log (\cos x)-\frac{x}{\sin y \cos y}\right]=\log (\tan x)+y \tan x$
$\frac{d y}{d x}=\frac{\log (\tan x)+y \tan x}{\log (\cos x)-x \cos ec\; y \cdot \sec y}$

Differentiation exercise 10.5 question 43

Answer: $\frac{d y}{d x}+e^{y-x}=0$
Hint: To solve this equation we will do differentiate differently
Given: $e^{x}+e^{y}=e^{x+y}$
Solution:
$e^{x}+e^{y}=e^{x+y}$
Diff w.r.t x $\left[\because \frac{d}{d x} e^{x}=e^{x}\right]$
$\frac{d}{d x} e^{x}+\frac{d}{d x} e^{y}=\frac{d}{d x} e^{x+y}$
$\begin{aligned} &e^{x}+e^{y} \frac{d y}{d x}=e^{x+y} \frac{d}{d x}(x+y) \\\\ &e^{x}+e^{y} \frac{d y}{d x}=e^{x+y}\left(1+\frac{d y}{d x}\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}\left(e^{x+y}-e^{y}\right)=e^{x}-e^{x+y} \\\\ &\frac{d y}{d x}=\frac{\left(e^{x}-e^{x} e^{y}\right)}{e^{x} e^{y}-e^{y}} \end{aligned}$
$\frac{d y}{d x}=\frac{e^{x}\left(1-e^{y}\right)}{e^{y}\left(e^{x}-1\right)}$
$\begin{aligned} &\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)} \\\\ &\frac{d y}{d x}=\frac{e^{x}-e^{x+y}}{e^{x+y}-e^{y}}=\frac{e^{x}-\left(e^{x}+e^{y}\right)}{\left(e^{x}+e^{y}\right)-e^{y}} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{-e^{y}}{e^{x}}=-e^{y-x} \\\\ &\frac{d y}{d x}+e^{y-x}=0 \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 44

Answer: $\frac{d y}{d x}=\frac{(\log y)^{2}}{\log y-1}$
Hint: To solve this equation we convert terms in log
Given: $e^{y}=y^{x}$
Solution:
$e^{y}=y^{x}$
$y \log e=x \log y \quad\left[\because \log a^{b}=b \log a\right]$
$y=x \log y$ ...................(1)
$\begin{aligned} &\frac{d y}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y \\\\ &\frac{d y}{d x}-\frac{x}{y} \frac{d y}{d x}=\log y \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}\left(1-\frac{x}{y}\right)=\log y \\\\ &\frac{d y}{d x}=\frac{\log y}{\left(1-\frac{x}{y}\right)} \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{\log y}{\left(1-\frac{1}{\log y}\right)} \\\\ &\frac{d y}{d x}=\frac{(\log y)^{2}}{(\log y-1)} \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 45

Answer: $\frac{d y}{d x}=\frac{1-x}{x}$
Hint: To solve this equation we will convert term into log
Given: $e^{x+y}-x=0$
Solution:
$e^{x+y}-x=0$
$\begin{aligned} &\log e^{x+y}=\log x \\\\ &(x+y) \log e=\log x \\\\ &\frac{d}{d x}(x+y)=\frac{d}{d x}(\log x) \end{aligned}$
$\begin{aligned} &1+\frac{d y}{d x}=\frac{1}{x} \\\\ &\frac{d y}{d x}=\frac{1}{x}-1=\frac{1-x}{x} \\\\ &\frac{d y}{d x}=\frac{1-x}{x} \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 46

Answer: $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$
Hint: To solve this equation we use uv' form
Given: $y=x \sin (a+y)$
Solution:
$(u v)^{\prime}=u^{\prime} v+v^{\prime} u$
$\begin{aligned} &\frac{d y}{d x}=1 \times \sin (a+y)+x \cos (a+y)\left[0+\frac{d y}{d x}\right] \\\\ &y^{\prime}=\sin (a+y)+x \cos (a+y) y^{\prime} \end{aligned}$
$\begin{aligned} &y^{\prime}=\frac{\sin (a+y)}{1-x \cos (a+y)} \cdot \frac{\sin (a+y)}{\sin (a+y)} \\\\ &\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cdot \cos (a+y)} \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 47

Answer: $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$
Hint: To solve this equation we use uv' form
Given: $x \sin (a+y)+\sin a \cdot \cos (a+y)=0$
Solution: $(u v)^{\prime}=u^{\prime} v+v^{\prime} u$
$\sin (a+y)+x \cdot \cos (a+y) y^{\prime}+\sin a\left(-\sin (a+y)\left(0+y^{\prime}\right)=0\right.$
$y^{\prime}[x \cdot \cos (a+y)+(-\sin a) \cdot \sin (a+y)]=-\sin (a+y)$
$y^{\prime}=\frac{-\sin (a+y)}{x \cdot \cos (a+y)+(-\sin a) \cdot(\sin a+y)} \cdot \frac{\sin (a+y)}{\sin (a+y)}$
$\frac{d y}{d x}=\frac{-\sin ^{2}(a+y)}{-\sin a\left(\cos ^{2}(a+y)+\sin ^{2}(a+y)\right.} \quad\left(\sin ^{2} x+\cos ^{2} x=1\right)$
$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$
Hence proved

Differentiation exercise 10.5 question 48

Answer: $\frac{d y}{d x}=\frac{1-(x+y) y \cot u}{(x+y) \log \sin x-1}$
Hint: To solve this we convert terms into log
Given: $(\sin x)^{y}=x+y$
Solution:
$(\sin x)^{y}=x+y$ $\left[\because \frac{d y}{d x} \int \log \sin x=y\right]$
$\log (\sin x)^{y}=\log (x+y)$ $\left[y=\frac{1}{\sin x} \times \cos x\right]$
$y \log (\sin x)=\log (x+y)$ $\left[\frac{\cos x}{\sin x}=\cot x\right]$
$y \frac{d}{d x} \log (\sin x)+\log \sin x \cdot \frac{d y}{d x}=\frac{1}{x+y} \frac{d}{d x}(x+y)$
$y \cot x+\log \sin x \cdot \frac{d y}{d x}=\frac{1}{x+y}+\frac{1}{x+y} \frac{d y}{d x}$
$\frac{d y}{d x}\left(\frac{1}{x+y}-\log \sin x\right)=y \cot x \cdot \frac{-1}{x+y}$
$\frac{d y}{d x}\left(\frac{1-(x+y) \log \sin x}{x+y}\right)=\frac{(x+y) y \cot x-1}{x+y}$
$\begin{aligned} &\frac{d y}{d x}=\frac{(x+y) y \cot x-1}{1-(x+y) \log \sin x} \\\\ &\frac{d y}{d x}=\frac{-(1-y \cot x(x+y)}{-(x+y) \log \sin x-1} \end{aligned}$
$d y=\frac{1-(x+y) \cdot y \cot x}{(x+y) \log \sin x-1}$
Hence proved

Differentiation exercise 10.5 question 49

Answer: $\frac{d y}{d x}=\frac{-y\left(x^{2} y+x+y\right)}{x\left(x y^{2}+x+y\right)}$
Hint: To solve this equation we solve the log term firstly
Given: $x y \log (x+y)=1$
Solution:
$x y \log (x+y)=1$
$\frac{d}{d x}(x) \cdot y \cdot \log (x+y)+x \cdot \frac{d}{d x}(y) \cdot \log (x+y)+x-y$
$\begin{aligned} &\frac{d}{d x} \log (x+y)=\frac{d}{d x}(1) \\\\ &y \cdot \log (x+y)+\frac{d y}{d x} \cdot x \cdot \log (x+y)+x \cdot y \cdot \frac{1}{x+y}\left(1+\frac{d y}{d x}\right)=0 \end{aligned}$
$\frac{d y}{d x}\left[x \cdot \log (x+y)+\frac{x y}{x+y}\right]=-\left(y \log (x+y)+\frac{x y}{x+y}\right)$
$\frac{d y}{d x}\left[\frac{x \cdot(x+y) \log (x+y)+x y}{x+y}\right]=\frac{-y(x+y) \log (x+y)}{x+y}$
$\frac{d y}{d x}=\frac{-y(x+y) \log (x+y)+x}{x \cdot(x+y) \cdot \log (x+y)+y}$
Hence proved

Differentiation exercise 10.5 question 50

Answer: $\frac{d y}{d x}=\frac{y}{x(1-x \cos y)}$
Hint: To solve this equation we differentiate it separately
Given: $y=x \sin y$
Solution:
$y=x \sin y$
$\begin{aligned} &u=x-1 \\ &v=\sin y \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=x \cos y \frac{d y}{d x}+\sin y \cdot 1 \\\\ &\frac{d y}{d x}=x \cos y \frac{d y}{d x}+\sin y \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}-x \cos y \frac{d y}{d x}=\sin y \\\\ &\frac{d y}{d x}(1-x \cos y)=\sin y \\\\ &\frac{d y}{d x}=\frac{y}{x(1-x \cos y)} \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 51

Answer: 120
Hint: To solve this equation we use $\frac{d}{dx}uvwz$ form
Given: $\begin{aligned} &f(x)=(1+x) \cdot\left(1+x^{2}\right) \cdot\left(1+x^{4}\right)\left(1+x^{8}\right)\\ \end{aligned}$
u v w z
Solution:
$\frac{d}{d x}(u v w z)=v w z \cdot \frac{d u}{d x}+u w z \cdot \frac{d v}{d x}+u v z \cdot \frac{d w}{d x}+u v w \frac{d z}{d x}$
$\begin{array}{r} f^{\prime}(x)=1 \times\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)+(1+x)\left(1+x^{4}\right)\left(1+x^{8}\right) 2 x+ \\ (1+x)\left(1+x^{2}\right)\left(1+x^{8}\right) 4 x^{3}+8 x^{7}(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \end{array}$
$\begin{aligned} &f^{\prime}(1)=1 \times 2 \times 2 \times 2+2 \times 2 \times 2 \times 2+2 \times 2 \times 2 \times 4+8 \times 1 \times 2 \times 2 \times 2 \\\\ &=8+16+36+64 \\\\ &f(1)^{\prime}=120 \end{aligned}$

Differentiation exercise 10.5 question 52

Answer: $\frac{4}{1+x^{2}+x^{4}}$
Hint: To solve this equation we use $\left ( \frac{u}{v} \right )^{'}$ form
Given: $y=\frac{\log \left(x^{2}+x+1\right)}{x^{2}-x+1}+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{1-x^{2}}\right)$
Solution:
$\frac{d y}{d x}=\left(\frac{(2 x+1)\left(x^{2}-x+1\right)-(2 x-1)\left(x^{2}+x+1\right.}{\left(x^{2}-x+1\right)^{2}}\right)+\frac{2}{\sqrt{3}}\left(\frac{1}{\frac{1+3 x^{2}}{\left(1-x^{2}\right)^{2}}}\right)\left(\frac{\sqrt{3}\left(1-x^{2}\right)-\sqrt{3} x(-2 x)}{\left(1-x^{2}\right)^{2}}\right)$

$=\frac{2 x^{3}+x^{2}-2 x^{2}-x+2 x+1-2 x^{3}-2 x^{2}-2 x+x^{2}+x+1}{\left(x^{2}-x+1\right)\left(x^{2}+x+1\right)}\; \; +$ $\frac{2}{\sqrt{x}} \cdot \frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}+x^{4}\right)}\left[\frac{\sqrt{3}-\sqrt{3} x^{2}+2 \sqrt{3} x^{2}}{\left(1-x^{2}\right)^{2}}\right]$
$\begin{aligned} &=\frac{2-2 x^{2}+2+2 x^{2}}{1+x^{2}+x^{4}} \\\\ &=\frac{4}{1+x^{2}+x^{4}} \end{aligned}$

Differentiation exercise 10.5 question 53

Answer: $\frac{d y}{d x}=(\sin x-\cos x)^{\sin x-\cos x}(\sin x+\cos x) \cdot(\log \sin x-\cos x)+1$
Hint: To solve this we take log both sides
Given: $y=(\sin x-\cos x)^{\sin x-\cos x}$
Solution:
$y=(\sin x-\cos x)^{\sin x-\cos x}$
Taking log on both sides,
$\log y=\log (\sin x-\cos x)^{\sin x-\cos x}$ $\left[\because \log x^{n}=n \log x\right]$

$\log y=(\sin x-\cos x) \log (\sin x-\cos x)$

$\frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x)(\cos x+\sin x)+(\sin x-\cos x)\left(\frac{1}{(\sin x-\cos x)}(\cos x+\sin x)\right)$

$\begin{array}{r} \frac{1}{y} \frac{d y}{d x}=(\cos x+\sin x)[1+\log (\sin x-\cos x)] \frac{d y}{d x}=y(\sin x+\cos x)[\log (\sin x- \cos x)+1 \end{array}$

$\frac{d y}{d x}=(\sin x-\cos x)^{\sin x-\cos x}(\sin x+\cos x) \cdot(\log \sin x-\cos x)+1$

Differentiation exercise 10.5 question 54

Answer: $\frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}$
Hint: To solve this equation we solve side by side
Given: $x y=e^{(x-y)}$
Solution:
$\log (x y)=\log \left(e^{(x-y)}\right)$
$\left[\because \frac{d u v}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}\right]$
Diff w.r.t $x^{u}$ $\left[\because \frac{d e^{x}}{d x}=e^{x}\right]$
$\begin{aligned} &x \frac{d y}{d x}+1(y)=e^{(x-y)} \frac{d}{d x}(x-y) \\\\ &x \frac{d y}{d x}+y=e^{(x-y)}\left(1-\frac{d y}{d x}\right) \end{aligned}$
$\begin{aligned} &x \frac{d y}{d x}+y=e^{(x-y)}-e^{(x-y)} \frac{d y}{d x} \\\\ &x \frac{d y}{d x}+e^{(x-y)} \frac{d y}{d x}=e^{(x-y)}-y \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}\left(x+e^{(x-y)}\right)=e^{(x-y)}-y \\\\ &\frac{d y}{d x}=\frac{y(x-1)}{x(y+1)} \end{aligned}$

Differentiation exercise 10.5 question 55

Answer: $\frac{d y}{d x}=\frac{-\left(y^{x} \log y+x^{y-1} \cdot y+x^{x}(\log x+1)\right)}{\left(x^{y} \log y+y^{x-1} \cdot x\right)}$
Hint: To solve this equation we solve this differently
Given: $y^{x}+x^{y}+x^{x}=a^{b}$
Solution: Let uvw
$u+v+w=a^{b}$
Diff w.r.t. x
$\begin{aligned} &\frac{d}{d x}(u+v+w)=\frac{d}{d x} a^{b} \\\\ &\frac{d u}{d x}+\frac{d v}{d u}+\frac{d w}{d v}=0 \\\\ &u=y^{x} \end{aligned}$ ...........(1)
Taking log on both sides,
$\log u=\log y^{x} \quad\left[\because \log a^{b}=b \log a\right]$
Diff w.r.t. u $\left[\because \frac{d}{d x}(u v)=u v^{\prime}+v u^{\prime}\right]$
$\frac{d}{d x}(\log u)=\frac{d}{d x}(x \log y) \quad\left[\because \frac{d}{d x} \log x=\frac{1}{x}\right]$
$\begin{aligned} &\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y(1) \\\\ &\frac{d u}{d x}=u\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \end{aligned}$

$v=x^{y}$ ..............(2)
Taking log on both sides,
$\begin{aligned} &\log v=\log x^{y} \\\\ &\log v=y \log x \end{aligned}$
Diff w.r.t
$\begin{aligned} &\frac{1}{v} \frac{d v}{d x}=y \cdot \frac{1}{x}+\log x \frac{d y}{d x} \\\\ &\frac{d v}{d u}=v\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\ &w=x^{x} \end{aligned}$ ..........(3)
Log on b.s.
$\begin{aligned} &\log w=\log x^{x} \\\\ &\log w=x \log x \end{aligned}$
Diff w.r.t
$\frac{1}{w} \frac{d w}{d x}=x \cdot \frac{1}{x}+\log x(1)$
From (1)
$\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}=0$
$y^{x}\left(\frac{x}{y} d y-\log y\right)+x^{y}\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+x^{x}(1+\log x)=0$
$y^{x} \log y+y^{x-1} x \frac{d y}{d x}+x^{y} \log x \frac{d y}{d x}+x^{y-1} \frac{y}{x}+x^{x}(1+\log x)=0$
$y^{x} \log y+x^{y} \cdot \frac{y}{x}+x^{x}(\log x+1)+y^{x-1} x \frac{d y}{d x}+x^{y} \log x \frac{d y}{d x}=0$
$y^{x-1} \frac{d y}{d x}+x^{y} \log y \frac{d y}{d x}=-\left(y^{x} \log x+x^{y} \cdot \frac{y}{x}+x^{x}(\log x+1)\right.$
$\frac{d y}{d x}=\frac{-\left(y^{x} \log y+x^{y-1} y+x^{x}(\log x+1)\right)}{\left(x^{y} \log y+y^{x-1} x\right)}$

Differentiation exercise 10.5 question 56

Answer: $\frac{d y}{d x}=\frac{y \tan x+\log \cos y}{\log (\cos x)+x \tan y}$
Hint: To solve this equation we add log on both sides
Given: $(\cos x)^{y}=(\cos y)^{x}$
Solution:
$(\cos x)^{y}=(\cos y)^{x}$
Taking log on both sides,
$\log (\cos x)^{y}=\log (\cos y)^{x}$
$y \log (\cos x)=x \log (\cos y) \quad\left[\because \log m^{n}=n \log \mathrm{m}\right]$
$\frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}$
$y \frac{d}{d x} \log (\cos x)+\log (\cos x) \frac{d y}{d x}=x \frac{d}{d x} \log (\cos y)+\log \cos y \frac{d y}{d x}$
$y \cdot\left(\frac{-\sin x}{\cos x}\right)+\log (\cos x) \frac{d y}{d x}=x \frac{d}{d x} \log (\cos y) \frac{d y}{d x}+\log \cos y$ ..........(1)
$-y \tan x+\log (\cos x) \frac{d y}{d x}=\frac{x(-\sin y)}{\cos y} \frac{d y}{d x}+\log \cos y$
$\log (\cos x) \frac{d y}{d x}+x \tan y \frac{d y}{d x}=y \tan x+\log \cos x$
$\frac{d y}{d x}(\log (\cos x)+x \tan y)=y \tan x+\log \cos x$
$\frac{d y}{d x}=\frac{y \tan x+\log \cos y}{\log (\cos x)+x \tan y}$

Differentiation exercise 10.5 question 57

Answer: $\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$
Hint: To solve this equation we make statement $\frac{\cos (a+y)-a}{\cos (a+y)}=x$
Given: $\cos y=x \cdot \cos (a+y)$
Solution: we have
$\cos y=x \cdot \cos (a+y)$
$\frac{\cos ((a+y)-a)}{\cos (a+y)}=x$
$\frac{\cos (a+y) \cos a+\sin (a+y) \sin a}{\cos (a+y)}=x$
$\frac{\cos (a+y) \cos a}{\cos (a+y)}+\frac{\sin (a+y) \cdot \sin a}{\cos (a+y)}$
$\frac{d}{d x}(\cos a+\tan (a+y) \sin a)=\frac{d x}{d x}$
$\sin a \cdot \frac{d}{d x}(\tan (a+y))=1 \quad\left[\because \frac{d}{d x} \tan x=\sec ^{2} x\right]$
$\sin a \cdot \sec ^{2}(a+y) \frac{d}{d x}(a+y)=1$
$\begin{aligned} &\frac{\sin a}{\cos ^{2}(a+y)} \cdot \frac{d y}{d x}=1 \\\\ &\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a} \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 58

Answer: $y \cdot \frac{d y}{d x}+x=2 y$
Hint: To solve this we add log on both side
Given: $(x-y)^{e \frac{x}{x-y}}=\mathrm{a}$
Solution:
$e^{\frac{x}{x-y}}=\frac{a}{x-y}$
Taking log on both sides,
$\log e^{\frac{x}{x-y}}=\log \left(\frac{a}{x-y}\right)$
$\frac{x}{x-y} \log e=\log a-\log (x-y) \quad\left[\because \log \frac{p}{q}=\log p-\log q\right]$
$\frac{x}{x-y}=\log a-\log (x-y)$
Differentiate w.r.t x we get
$\frac{d}{d x}\left(\frac{x}{x-y}\right)=0-\frac{d y}{d x} \log (x-y)$
$\frac{\left(\frac{x}{x-y}\right) \frac{d x}{d x}-x^{\frac{d y}{d x}(x-y)}}{(x-y)^{2}}=\frac{-d}{d x} \log t$ $\left[\because \frac{d}{d x} \frac{p(x)}{q(x)}=q(x) \frac{d p}{d x}-\frac{p \frac{d q}{d x}}{q^{2}}\right]$
$\frac{(x-y) 1-x\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}}=\frac{-d}{d x} \log \frac{d t}{d x}$ $\mathrm{x}-\mathrm{y}=\mathrm{t}=>\frac{d t}{d x}=\frac{d x}{d x}-\frac{d y}{d x}$
$\frac{d t}{d x}=1-\frac{d y}{d x}$
$\frac{x^{2}-y-x+x \frac{d y}{d x}}{(x-y)^{2}}=\frac{-1}{t}$
$\begin{aligned} &\frac{-y+x \cdot d y}{(x-y)^{2}}=\frac{-1}{(x-y)}\left(1-\frac{d y}{d x}\right) \\\\ &-y+x \cdot \frac{d y}{d x}=\left(\frac{d y}{d x}-1\right)(x-y) \end{aligned}$
$\begin{aligned} &=x \cdot \frac{d y}{d x}-x-y \frac{d y}{d x}+y \\\\ &-y=x+y-y \frac{d y}{d x} \\\\ &y \frac{d y}{d x}+x=2 y \end{aligned}$
Hence proved

Differentiation exercise 10.5 question 59

Answer: $\frac{d y}{d x}=\frac{x-y}{x \log x}$
Hint: To solve this equation we use chain rule and quotient rule
Given: $x=e^{\frac{x}{y}}$
Solution: we have $x=e^{\frac{x}{y}}$
Diff w.r.t x we get $\left[\because \frac{d}{d x} e^{x}=e^{x}\right]$
$1=e^{\frac{x}{y}} \frac{d}{d x}\left(\frac{x}{y}\right)$ [Chain rule]
$1=e^{\frac{x}{y}} \times \frac{y \frac{d}{d x} x-x \frac{d}{d x} y}{y^{2}} \quad\left[\because\left(\frac{u}{v}\right)^{\prime}=\frac{u r v-v r u}{v^{2}}\right]$
$\begin{aligned} &y^{2}=e^{\frac{x}{y}}\left(y-x \frac{d y}{d x}\right) \\\\ &y^{2}=y \cdot e^{\frac{x}{y}}-x \frac{d y}{d x} e^{\frac{x}{y}} \end{aligned}$
$\begin{aligned} &x \frac{d y}{d x} e^{\frac{x}{y}}=y \cdot e^{\frac{x}{y}}-y^{2} \\\\ &\frac{d y}{d x}=\frac{y\left(e^{\frac{x}{y}}-y\right)}{x \cdot e^{\frac{x}{y}}} \end{aligned}$
$x. \frac{d y}{d x}=\frac{1}{\log x}(x-y) \quad\left[\because x=e^{\frac{x}{y}}, \log x=\frac{x}{y}\right]$
$\frac{d y}{d x}=\frac{x-y}{x(\log x)}$

Differentiation exercise 10.5 question 60

Answer: $\frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{x}{\sqrt{2\left(x^{2}+1\right)}}$
Hint: To solve this equation we use log on both side
Given: $y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$
Solution: $\frac{d}{d x}(\log u)=\log (x \tan x)=\tan x-\log x$
$\text { Let } \mathrm{u}=x^{\tan x}$
$\frac{1}{u} \frac{d u}{d x}=\tan x \frac{d}{d x} \log x+\log x \cdot \frac{d}{d x}(\tan x)$
$\begin{aligned} &\frac{d u}{d x}=u\left(\frac{\tan x}{x}+\left(\log x \cdot \sec ^{2} x\right)\right) \\\\ &\frac{d u}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\left(\log x \cdot \sec ^{2} x\right)\right) \end{aligned}$
$\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(u+\sqrt{\frac{x^{2}+1}{2}}\right) \\\\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d}{d x}\left(\sqrt{\frac{x^{2}+1}{2}}\right) \end{aligned}$
$\frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x\right)$ $+\frac{1}{2 \sqrt{\frac{x^{2}+1}{2}}} \frac{d}{d x}\left(\frac{x^{2}+1}{2}\right)$ $\begin{aligned} &{\left[\because \frac{d}{d x} \sqrt{x}=\frac{1}{2 \sqrt{x}}\right]} \\ &{\left[\because \frac{d}{d x} x^{2}=2 x\right]} \end{aligned}$
$=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{1}{\sqrt{2\left(x^{2}+1\right)}} \cdot \frac{2 x}{2}$
$\frac{d y}{d x}=x^{\tan x}\left(\frac{\tan x}{x}+\log x \sec ^{2} x\right)+\frac{x}{\sqrt{2\left(x^{2}+1\right)}}$

Differentiation exercise 10.5 question 61

Answer: $\frac{d y}{d x} \frac{\alpha}{(1-\alpha x)^{2}}+\frac{\beta\left(1-\beta x^{2}\right)}{(1-x)^{2}\left(1-\beta x^{2}\right)}+\frac{2 \alpha \beta y^{2} x^{3}-\alpha \beta y x-\alpha y^{2} x^{2}-y^{3} \beta x^{3}}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-y x)^{2}}$
Hint: To solve this equation we differentiate it differently
Given: $y=1+\frac{\alpha}{\left(\frac{1}{x}-\alpha\right)}+\frac{\frac{\beta}{x}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)}+\frac{\frac{\gamma}{x^{2}}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)\left(\frac{1}{x}-\gamma\right)}$
Solution:
$y=1+\frac{\alpha}{\left(\frac{1}{x}-\alpha\right)}+\frac{\frac{\beta}{x}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)}+\frac{\frac{\gamma}{x^{2}}}{\left(\frac{1}{x}-\alpha\right)\left(\frac{1}{x}-\beta\right)\left(\frac{1}{x}-\gamma\right)}$
$y=1+A+B+C$
$\frac{d y}{d x}=0+\frac{d A}{d x}+\frac{d B}{d x}+\frac{d C}{d x}$
$\begin{aligned} &A=\frac{\alpha x}{(1-\alpha x)} \\\\ &\frac{d A}{d x}=\frac{(1-\alpha x) \alpha-\alpha x(-\alpha)}{(1-\alpha x)^{2}} \end{aligned}$
$\begin{aligned} &\frac{d A}{d x}=\frac{\alpha-\alpha^{2} x+\alpha^{2} x}{(1-\alpha x)^{2}} \\\\ &\frac{d A}{d x}=\frac{\alpha}{(1-\alpha x)^{2}} \end{aligned}$ .............(1)
$\frac{d B}{d x}=\frac{d}{d x}\left(\frac{\beta x}{(1-x)(1-\beta x)}\right)$
$\frac{d B}{d x}=\frac{(1-x)(1-\beta x) \beta-\beta x(1-x)(-\beta)+(1-\beta x)(-1)}{(1-x)^{2}(1-\beta x)^{2}}$
$\frac{d B}{d x}=\frac{\beta\left(1-x-\beta x+\beta x^{2}\right)-\beta x(-\beta+\beta x-1+\beta x)}{(1-x)^{2}(1-\beta x)^{2}}$
$\frac{d B}{d x}=\frac{\left.\beta-\beta x-\beta^{2} x+\beta^{2} x^{2}+\beta^{2} x-\beta^{2} x^{2}+\beta x-\beta^{2} x^{2}\right)}{(1-x)^{2}(1-\beta x)^{2}}$
$\frac{d B}{d x}=\frac{\beta\left(1-\beta x^{2}\right)}{(1-x)^{2}(1-\beta x)^{2}}$ .................(2)
$c=\frac{y x}{(1-\alpha x)(1-\beta x)(1-\gamma x)}$
$\frac{d c}{d x}=\frac{(1-\alpha x)(1-\beta x)(1-\gamma x) y-y x(1-\alpha x)(1-\beta x)-\gamma(1-\beta)(1-\gamma x)(-\alpha)+(1-\alpha x)(1-\gamma x)(-\beta)}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-\gamma x)^{2}}$
$\frac{dc}{dx}=\frac{\begin{aligned} &\left(1-\alpha x-\beta x+\alpha \beta x^{2}\right)(1-\gamma x) \gamma-\gamma x\{(-\gamma)(1-\alpha x-\beta x+\alpha \beta x)+(-\alpha)+(1-\beta x-\gamma x+\beta \gamma x)+ &\left.(-\beta)\left(1-\alpha x-\gamma x+\alpha \gamma x^{2}\right)\right\} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}$

$\frac{dc}{dx}=\frac{\begin{aligned} &\left(1-\alpha x-\beta x+\alpha \beta x^{2}\right)(1-\gamma x) \gamma- &\gamma x\left\{(-\gamma)(1-\alpha x-\beta x+\alpha \beta x)+(-\alpha)+\left(1-\beta x-\gamma x+\beta \gamma x^{2}\right)+\right. &\left.(-\beta)\left(1-\alpha x-\gamma x+\alpha \gamma x^{2}\right)\right\} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}$
$\frac{dc}{dx}=\frac{\begin{aligned} &\gamma\left(1-\alpha x-\beta x+\alpha \beta x^{2}-\gamma x+\alpha y x^{2}+\beta \gamma x^{2}-\alpha \beta \gamma x^{2}\right. &-\gamma x\left(-\gamma+\alpha \gamma x+\beta \gamma x-\alpha \beta \gamma^{2} x^{2}-\alpha+\alpha \beta x+\alpha \gamma x-\alpha \beta y x^{2}-\right. &\left.\left.\beta+\alpha \beta x+\gamma \beta x+\alpha \beta \gamma x^{2}\right)\right\} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}$
$\frac{dc}{dx}=\frac{\begin{aligned} &\gamma-\alpha \gamma x-\beta \gamma x+\alpha \beta \gamma x^{2}-\gamma^{2} x+\alpha \gamma^{2} x^{2}+\beta \gamma^{2} x^{2} &-\alpha \beta \gamma^{2} x^{2}-\gamma^{2} x-\alpha \gamma^{2} x^{2}-\beta \gamma^{2} x^{2}+\alpha \beta \gamma^{2} x^{3}+ &\alpha \gamma x-\alpha \beta \gamma x-\alpha \gamma^{2} x^{2}+\alpha \beta y^{2} x^{3}+\beta \gamma x- &\alpha \beta \gamma x^{2}+\gamma^{2} \beta x^{2}+\alpha \beta \gamma^{2} x^{3} \end{aligned}}{\left ( 1-ax \right )^{2}\left ( 1-\beta x \right )^{2}\left ( 1-\gamma x \right )^{2}}$
$\frac{d c}{d x}=\frac{-\alpha \beta \gamma x-\alpha \gamma^{2} x^{2}-\alpha \beta \gamma^{2} x^{3}-\beta \gamma^{2} x^{2}}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-\gamma x)^{2}}$ $\frac{d y}{d x}=\frac{\alpha}{(1-\alpha x)^{2}}+\frac{\beta\left(1-\beta x^{2}\right)}{(1-x)^{2}(1-\beta x)^{2}}+\frac{+2 \alpha \beta \gamma^{2} x^{3}-\alpha \beta \gamma x-\alpha \gamma^{2} x^{2}-\beta \gamma^{3} x^{3}}{(1-\alpha x)^{2}(1-\beta x)^{2}(1-\gamma x)^{2}}$

Differentiation exercise 10.5 question 62

Answer: $\frac{d y}{d x}=y \times\left(\log y+\frac{x}{y}\right)$
Hint: To solve this equation we use log on both side
Given: $x^{y}-y^{x}=a^{b}$
Solution:
$x^{y}=z$
Taking log on both side
$\begin{aligned} &\log x^{y}=\log z \\\\ &y \log x=\log z \\\\ &\frac{d y}{d x} \log x+\frac{y}{x}=\frac{1}{z} \frac{d z}{d x} \end{aligned}$
$\frac{d z}{d x}=x^{y}\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)$ ............(1)
$\begin{aligned} &y^{x}=p \\\\ &\log y^{x}=\log p \\\\ &x \log y=\log p \end{aligned}$
$\begin{aligned} &\log y+\frac{x}{y} \frac{d y}{d x}=\frac{1}{p} \frac{d p}{d x} \\\\ &\frac{d p}{d x}=y^{x}\left(\log y+\frac{x}{y} \frac{d y}{d x}\right) \\\\ &\frac{d y}{d x}=y^{x}\left(\log y+\frac{x}{y}\right) \end{aligned}$

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.5 question 62
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RD Sharma Class 12th Exercise 10.5 deals with the Chapter Differentiation and is a lengthy exercise consisting of 71 questions. As solving all of them at once is a tedious task, students should divide the sums and practice accordingly. Although these are basic level one questions they can be quite tricky unless students know their basics right.

In this exercise you will learn topics like:

1. Differentiation of Trigonometric equations

2. Differentiation involving exponents

3. Finding first derivatives of complex Algebraic equations

4. Proof sums involving derivatives

These are the following traits that make RD Sharma Class 12th Exercise 10.5 material the best choice for students:

1. Detailed questions

The solutions provided by Brinley contain detailed questions that cover the entire syllabus. This means that students don't have to worry about any missing topics.

2. Step-by-step solutions

These solutions are interpreted with step-by-step explanations that are extremely helpful for students to understand the questions properly. Students who find RD Sharma material confusing can refer to these solutions for a clear understanding. Class 12 RD Sharma Chapter 10 Exercise 10.5 Solution material is made in such a way that it is easy for both a class topper as well as an average student to study effectively.

3. Helpful for class lectures

As differentiation is a vast topic it is not necessary that teachers might cover all the questions from the book. This is why RD Sharma class 12 Chapter 10 Exercise 10.5 material is helpful for students to stay in line with their class lectures and side-by-side prepare for their exams.

4. Exam-oriented material

The sole purpose of this material is to enable students to score good marks in exams by merging all the solutions and concepts in a singular material. The questions from RD Sharma class 12 Chapter 10 Exercise 10.5 material are exam-oriented and as RD Sharma books are widely used the questions from this material can also show up in your exam.

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. What are the charges for this material?

RD Sharma Class 12th Exercise 10.5 solutions are provided on Careers360’s website and are accessible for free. Students can refer to this material through their browser on the go without any hassle.

2. Which material is best suited for maths, RD Sharma or NCERT?

When compared to NCERT RD Sharma materials are exam oriented and best suited for maths. They contain detailed concepts with step-by-step explanations that help students better understand the subject.

3. Can I find solutions for all chapters on the Careers360’s website?

Students can search their relevant chapter name on Branly’s website to get access to their materials. To find solutions for Differentiation, refer to RD Sharma Class 12th Exercise 10.5.

4. What is differentiation?

The process of finding the rate of change of a function is called Differentiation. To learn more, check RD Sharma Class 12 solutions Differentiation Ex 10.5.

5. What are the real life applications of differentiation?

Given below are some of the applications: