What is the benefit of this material from an exam standpoint?

Class 12 RD Sharma Chapter 10 Exercise 10.4 Solution material has solutions that cover all important topics and are essential for exam preparation.

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What is differentiation?

Differentiation is the process of finding derivatives of a function that helps determine its rate of change.

What are the applications of differentiation?

Differentiation and the maxima and minima concept is widely used in economics and finding the rate of growth of businesses. You can refer to RD Sharma Class 12 Solutions Chapter 10 Ex 10.4 to learn more about this topic.

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RD Sharma Class 12th Exercise 10.4 material will help students get a better understanding of the chapter and score gold marks in exams.

RD Sharma Class 12 Exercise 10.4 Differentiation Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 10.4 Differentiation Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 10.4 Differentiation Solutions Maths - Download PDF Free Online

Updated on 20 Jan 2022, 05:24 PM IST

RD Sharma is considered the best material for CBSE students. They are used all over the country and are extremely helpful for exam preparation. They are preferred over NCERT due to their syllabus coverage and detail of answers.
RD Sharma Class 12th Exercise 10.4 deals with the chapter ' Differentiation.' This is a fundamental chapter with its applications across many subjects. This is why students need to familiarise themselves with the basics of this chapter.

This Story also Contains

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
Differentiation Excercise: 10.4
Differentiation exercise 10.4 question 31
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: 10.4

Differentiation exercise 10.4 question 1

Answer:

$\left(\frac{-y}{x}\right)$
Hint:
Use product rule to find $\frac{dy}{dx}$
Given:
$x y=C^{2}$
Solution:
Differentiate the given equation $x y=C^{2}$ w.r.t x
$\begin{aligned} &x \cdot \frac{d y}{d x}+y=0 \\ &x \frac{d y}{d x}=-y \\ &\frac{d y}{d x}=\frac{-y}{x} \end{aligned}$ $\left[\begin{array}{c} \because \frac{d(\operatorname{cons} \tan t)}{d x}=0 \\ \frac{d(u \cdot v)}{d x}=u \cdot \frac{d v}{d x}+v \frac{d u}{d x} \end{array}\right]$

Hence $\frac{d y}{d x}=\frac{-y}{x}$ is the required answer.

Differentiation exercise 10.4 question 2

Answer:
$\frac{(x+y)^{2}}{y^{2}-2 x y-x^{2}}$
Hint:
Use product rule to find $\frac{d y}{d x}$
Given:
$y^{3}-3 x y^{2}=x^{3}+3 x^{2} y$
Solution:
Differentiating the given equation w.r.t x
$\frac{d}{d x}\left(y^{3}-3 x y^{2}\right)=\frac{d}{d x}\left(x^{3}+3 x^{2} y\right)$
$\frac{d}{d x}\left(y^{3}\right)-\frac{d}{d x}\left(3 x y^{2}\right)=\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}\left(3 x^{2} y\right)$

$3 y^{2} \frac{d y}{d x}-3\left[x \frac{d y^{2}}{d x}+y^{2} \frac{d x}{d x}\right]=3 x^{2}+3\left[x^{2} \frac{d y}{d x}+y \frac{d x^{2}}{d x}\right]$ $\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$3 y^{2} \frac{d y}{d x}-3\left[x \cdot \frac{d y^{2}}{d y} \times \frac{d y}{d x}+y^{2}\right]=3 x^{2}+3\left[x^{2} \frac{d y}{d x}+y(2 x)\right]$
$3 y^{2} \frac{d y}{d x}-3\left[x(2 y) \cdot \frac{d y}{d x}+y^{2}\right]=3 x^{2}+3 x^{2} \frac{d y}{d x}+6 x y$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$3 y^{2} \frac{d y}{d x}-6 x y \frac{d y}{d x}-3 y^{2}=3 x^{2}+3 x^{2} \frac{d y}{d x}+6 x y$
$3 y^{2} \frac{d y}{d x}-6 x y \frac{d y}{d x}-3 x^{2} \frac{d y}{d x}=3 x^{2}+6 x y+3 y^{2}$
$3 \frac{d y}{d x}\left[y^{2}-2 x y-x^{2}\right]=3\left(x^{2}+2 x y+y^{2}\right)$
$\frac{d y}{d x}=\frac{3\left(x^{2}+2 x y+y^{2}\right)}{3\left(y^{2}-2 x y-x^{2}\right)}$
$\frac{d y}{d x}=\frac{(x+y)^{2}}{y^{2}-2 x y-x^{2}}$ $\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$
Is the required answer.
Hence $\frac{d y}{d x}=\frac{(x+y)^{2}}{y^{2}-2 x y-x^{2}}$

Differentiation exercise 10.4 question 3

Answer:
$-\left(\frac{y}{x}\right)^{\frac{1}{3}}$
Hint:
Use the differentiation formula of $\left(x^{n}\right)$
i.e. $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$
Given:
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)=\frac{d}{d x}\left(a^{\frac{2}{3}}\right)$
$\frac{d}{d x}\left(x^{\frac{2}{3}}\right)+\frac{d}{d x}\left(y^{\frac{2}{3}}\right)=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$\frac{2}{3}(x)^{\frac{2}{3}-1}+\frac{d\left(y^{\frac{2}{3}}\right)}{d y} \times \frac{d y}{d x}=0 \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$\frac{2}{3}(x)^{\frac{-1}{3}}+\frac{2}{3}(y)^{\frac{2}{3}-1} \frac{d y}{d x}=0$
$\frac{2}{3}(x)^{\frac{-1}{3}}+\frac{2}{3}(y)^{\frac{-1}{3}} \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{\frac{-2}{3}(x)^{\frac{-1}{3}}}{\frac{2}{3}(y)^{\frac{-1}{3}}}=\frac{-(x)^{\frac{-1}{3}}}{(y)^{\frac{-1}{3}}}$
$\frac{d y}{d x}=-\frac{(y)^{\frac{1}{3}}}{(x)^{\frac{1}{3}}} \quad\left[\because(x)^{-n}=\frac{1}{(x)^{n}}\right]$
$\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$
Hence $\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$is required answer

Differentiation exercise 10.4 question 4

Answer:
$\frac{4}{3}\left(\frac{1-4 x+3 y}{4 x-3 y+1}\right)$
Hint:
Use chain rule to find the differentiation
Given:
$4 x+3 y=\log (4 x-3 y)$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}(4 x+3 y)=\frac{d}{d x}(\log (4 x-3 y))$
$\frac{d}{d x}(4 x)+\frac{d}{d x}(3 y)=\frac{d(\log (4 x-3 y))}{d(4 x-3 y)} \times \frac{d(4 x-3 y)}{d x}$
[Using the chain $\frac{d(f(x+y))}{d x}=\frac{d(f(x+y))}{d(x+y)} \times \frac{d(x+y)}{d x}$ ]

$4 \frac{d x}{d x}+3 \frac{d y}{d x}=\frac{1}{(4 x-3 y)} \times\left[\frac{d(4 x)}{d x}-\frac{d(3 y)}{d x}\right] \quad\left[\because \frac{d(\log x)}{d x}=\frac{1}{x}\right]$
$4+3 \frac{d y}{d x}=\frac{1}{4 x-3 y} \cdot\left[4-\frac{d(3 y)}{d y} \times \frac{d y}{d x}\right]$
$4+3 \frac{d y}{d x}=\frac{1}{4 x-3 y} \cdot\left[4-3 \times \frac{d y}{d x}\right]$
$4+3 \frac{d y}{d x}=\frac{4}{4 x-3 y}-\frac{3}{4 x-3 y} \frac{d y}{d x}$
$3 \frac{d y}{d x}+\frac{3}{4 x-3 y} \cdot \frac{d y}{d x}=\frac{4}{4 x-3 y}-4$
$3 \frac{d y}{d x}\left(1+\frac{1}{4 x-3 y}\right)=\frac{4-4(4 x-3 y)}{4 x-3 y}$
$3 \frac{d y}{d x} \times\left(\frac{4 x-3 y+1}{4 x-3 y}\right)=\frac{4-16 x+12 y}{4 x-3 y}$
$\frac{d y}{d x}=\frac{4-16 x+12 y}{(4 x-3 y)} \times \frac{4 x-3 y}{(4 x-3 y+1)} \times \frac{1}{3}$
$\frac{d y}{d x}=\frac{1}{3} \cdot \frac{(4-16 x+12 y)}{(4 x-3 y+1)}$
$\frac{d y}{d x}=\frac{4}{3} \cdot\left(\frac{1-4 x+3 y}{4 x-3 y+1}\right)$
Hence
$\frac{d y}{d x}=\frac{4}{3}\left(\frac{1-4 x+3 y}{4 x-3 y+1}\right)$ Is the required answer

Differentiation exercise 10.4 question 5

Answer:
$\frac{-b^{2} x}{a^{2} y}$
Hint:
Use chain rule and differentiation formulas like $\frac{d x^{n}}{d x}=n x^{n-1}$
Given:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=\frac{d(1)}{d x}$
$\frac{d}{d x}\left(\frac{x^{2}}{a^{2}}\right)+\frac{d}{d x}\left(\frac{y^{2}}{b^{2}}\right)=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$\frac{1}{a^{2}} \cdot \frac{d\left(x^{2}\right)}{d x}+\frac{1}{b^{2}} \cdot \frac{d\left(y^{2}\right)}{d x}=0$
$\frac{1}{a^{2}} \times 2 x+\frac{1}{b^{2}} \times\left(\frac{d y^{2}}{d y} \times \frac{d y}{d x}\right)=0 \quad\left[\because \frac{d x^{n}}{d x}=n x^{n-1}\right]$
$\frac{2 x}{a^{2}}+\frac{1}{b^{2}}\left(2 y \cdot \frac{d y}{d x}\right)=0$
$\frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{\frac{-2 x}{a^{2}}}{\frac{2 y}{b^{2}}}=\frac{-2 x \times b^{2}}{2 y \times a^{2}}$
$\frac{d y}{d x}=\frac{-x b^{2}}{y a^{2}}$
$\frac{d y}{d x}=-\frac{b^{2} x}{a^{2} y}$
Hence $\frac{d y}{d x}=-\frac{b^{2} x}{a^{2} y}$ is the required answer

Differentiation exercise 10.4 question 6

Answer:
$\frac{d y}{d x}=\frac{y-x^{4}}{y^{4}-x}$
Hint:
Use chain rule to find the differentiation formula like $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$
Given:
$x^{5}+y^{5}=5 x y$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left(x^{5}+y^{5}\right)=\frac{d}{d x}(5 x y)$
$\frac{d}{d x}\left(x^{5}\right)+\frac{d}{d x}\left(y^{5}\right)=5 \frac{d(x y)}{d x}$
$5 x^{4}+\left(\frac{d y^{5}}{d y} \times \frac{d y}{d x}\right)=5\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}, \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$5 x^{4}+5 y^{4} \frac{d y}{d x}=5\left[x \frac{d y}{d x}+y\right]$
$5 x^{4}+5 y^{4} \frac{d y}{d x}=5 x \frac{d y}{d x}+5 y$
$5 y^{4} \frac{d y}{d x}-5 x \frac{d y}{d x}=5 y-5 x^{4}$
$\frac{d y}{d x}\left(5 y^{4}-5 x\right)=5 y-5 x^{4}$
$\frac{d y}{d x}=\frac{5 y-5 x^{4}}{5 y^{4}-5 x}$
$\begin{aligned} &\frac{d y}{d x}=\frac{5\left(y-x^{4}\right)}{5\left(y^{4}-x\right)} \\ &\frac{d y}{d x}=\frac{y-x^{4}}{y^{4}-x} \end{aligned}$
Hence $\frac{d y}{d x}=\frac{y-x^{4}}{y^{4}-x}$ is the required answer

Differentiation exercise 10.4 question 7

Answer:
$\left(\frac{a y-x-y}{x+y-a x}\right)$
Hint:
Use chain rule and product rule
Given:
$(x+y)^{2}=2 a x y$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left[(x+y)^{2}\right]=\frac{d}{d x}[2 a x y]$
$\frac{d}{d x}\left[x^{2}+2 x y+y^{2}\right]=2 a \frac{d}{d x}(x y) \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$
$\frac{d\left(x^{2}\right)}{d x}+\frac{d(2 x y)}{d x}+\frac{d y^{2}}{d x}=2 a\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right] \quad\left[\because \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$2 x+2 \frac{d(x y)}{d x}+\frac{d y^{2}}{d y} \times \frac{d y}{d x}=2 a\left[x \frac{d y}{d x}+y\right]$
$2 x+2\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]+2 y \frac{d y}{d x}=2 a x \frac{d y}{d x}+2 a y \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$2 x+2 x \frac{d y}{d x}+2 y+2 y \frac{d y}{d x}=2 a x \frac{d y}{d x}+2 a y$
$2 x \frac{d y}{d x}+2 y \frac{d y}{d x}-2 a x \frac{d y}{d x}=2 a y-2 x-2 y$
$\frac{d y}{d x}[2 x+2 y-2 a x]=2 a y-2 x-2 y$
$\frac{d y}{d x}=\frac{2 a y-2 x-2 y}{2 x+2 y-2 a x}$
$\frac{d y}{d x}=\frac{2(a y-x-y)}{2(x+y-a x)}$
$\frac{d y}{d x}=\frac{a y-x-y}{x+y-a x}$
Hence $\frac{d y}{d x}=\frac{a y-x-y}{x+y-a x}$ is the required answer

Differentiation exercise 10.4 question 8

Answer:
$\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$
Hint:
Use chain rule and the product rule of differentiation
Given:
$\left(x^{2}+y^{2}\right)^{2}=x y$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left[\left(x^{2}+y^{2}\right)^{2}\right]=\frac{d}{d x}(x y)$
$\frac{d}{d x}\left[\left(x^{2}\right)^{2}+\left(y^{2}\right)^{2}+2 x^{2} \times y^{2}\right]=x \frac{d y}{d x}+y \frac{d x}{d x} \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$
[Product Rule $\frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}$]

$\frac{d}{d x}\left[x^{4}+y^{4}+2 x^{2} y^{2}\right]=x \frac{d y}{d x}+y$
$\frac{d}{d x}\left(x^{4}\right)+\frac{d}{d x}\left(y^{4}\right)+\frac{d}{d x}\left(2 x^{2} y^{2}\right)=x \frac{d y}{d x}+y$
$4 x^{3}+\left(\frac{d y^{4}}{d y} \times \frac{d y}{d x}\right)+2\left(\frac{d\left(x^{2} y^{2}\right)}{d x}\right)=x \frac{d y}{d x}+y$
$4 x^{3}+4 y^{3} \frac{d y}{d x}+2\left[x^{2} \times \frac{d y^{2}}{d x}+y^{2} \frac{d x^{2}}{d x}\right]=x \frac{d y}{d x}+y \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$4 x^{3}+4 y^{3} \frac{d y}{d x}+2\left[x^{2} \times \frac{d y^{2}}{d y} \times \frac{d y}{d x}+y^{2}(2 x)\right]=x \frac{d y}{d x}+y$
$4 x^{3}+4 y^{3} \frac{d y}{d x}+2\left[x^{2} \times(2 y) \frac{d y}{d x}+2 x y^{2}\right]=x \frac{d y}{d x}+y$
$4 x^{3}+4 y^{3} \frac{d y}{d x}+4 x^{2} y \frac{d y}{d x}+4 x y^{2}=x \frac{d y}{d x}+y$
$4 y^{3} \frac{d y}{d x}+4 x^{2} y \frac{d y}{d x}-x \frac{d y}{d x}=y-4 x^{3}-4 x y^{2}$
$\frac{d y}{d x}=\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$
Hence $\frac{d y}{d x}=\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$ is the required answer.

Differentiation exercise 10.4 question 9

Answer:$\left[-\left(\frac{x}{y}\right)\right]$
Hint:
Use chain rule and $\frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}}$
Given:
$\tan ^{-1}\left(x^{2}+y^{2}\right)=a$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left(\tan ^{-1}\left(x^{2}+y^{2}\right)\right)=\frac{d(a)}{d x}$
$\frac{d\left(\tan ^{-1}\left(x^{2}+y^{2}\right)\right)}{d\left(x^{2}+y^{2}\right)} \times \frac{d\left(x^{2}+y^{2}\right)}{d x}=0$ [Using chain rule] $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(\frac{d x^{2}}{d x}+\frac{d y^{2}}{d x}\right)=0$ $\left[\because \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}}\right]$
$\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(2 x+\frac{d y^{2}}{d y} \times \frac{d y}{d x}\right)=0$
$\frac{1}{1+\left(x^{2}+y^{2}\right)^{2}} \times\left(2 x+2 y \frac{d y}{d x}\right)=0$
$\frac{2 x}{1+\left(x^{2}+y^{2}\right)^{2}}+\frac{2 y}{1+\left(x^{2}+y^{2}\right)^{2}} \frac{d y}{d x}=0$
$\frac{2 y}{1+\left(x^{2}+y^{2}\right)^{2}} \frac{d y}{d x}=-\frac{2 x}{1+\left(x^{2}+y^{2}\right)^{2}}$
$\frac{d y}{d x}=-\frac{2 x}{\left[1+\left(x^{2}+y^{2}\right)^{2}\right]} \times \frac{1+\left(x^{2}+y^{2}\right)^{2}}{2 y}$
$\frac{d y}{d x}=\frac{-2 x}{2 y}=-\frac{x}{y}$
$\frac{d y}{d x}=-\frac{x}{y}$
Hence $\frac{d y}{d x}=-\frac{x}{y}$ is the required answer

Differentiation exercise 10.4 question 10

Answer:

Answer:$\frac{y}{x}\left[\frac{x e^{(x-y)}-1}{y e^{(x-y)}-1}\right]$
Hint:
Use chain rule and quotient rule
Given:
$e^{x-y}=\log \left(\frac{x}{y}\right)$
Solution:
Differentiate the given equation w.r.t x
$\frac{d\left(e^{x-y}\right)}{d x}=\frac{d\left(\log \left(\frac{x}{y}\right)\right)}{d x}$
$\frac{d\left(e^{x-y}\right)}{d(x-y)} \times \frac{d(x-y)}{d x}=\frac{d\left(\log \left(\frac{x}{y}\right)\right)}{d\left(\frac{x}{y}\right)} \times \frac{d\left(\frac{x}{y}\right)}{d x}$
$\left(e^{x-y}\right) \times\left[\frac{d x}{d x}-\frac{d y}{d x}\right]=\frac{1}{\left(\frac{x}{y}\right)} \times\left[\frac{y \cdot \frac{d x}{d x}-x \cdot \frac{d y}{d x}}{y^{2}}\right]$ $\left[\because \frac{d\left(e^{x}\right)}{d x}=e^{x}\right]$
[Using quotient rule $\frac{d\left(\frac{u}{v}\right)}{d x}=\frac{v \cdot \frac{d u}{d x}-u \cdot \frac{d v}{d x}}{v^{2}}$ ]
$e^{x-y} \times\left[1-\frac{d y}{d x}\right]=\frac{y}{x} \times\left[\frac{y-x \frac{d y}{d x}}{y^{2}}\right]$
$e^{x-y}-e^{x-y} \frac{d y}{d x}=\frac{1}{x y}\left(y-x \frac{d y}{d x}\right)$
$e^{x-y}-e^{x-y} \frac{d y}{d x}=\frac{y}{x y}-\frac{x}{x y} \cdot \frac{d y}{d x}$
$\frac{x}{x y} \frac{d y}{d x}-e^{x-y} \frac{d y}{d x}=\frac{y}{x y}-e^{x-y}$
$\frac{d y}{d x}\left(\frac{x}{x y}-e^{x-y}\right)=\frac{y}{x y}-e^{x-y}$
$\frac{d y}{d x}=\left[\frac{\frac{y}{x y}-e^{x-y}}{\frac{x}{x y}-e^{x-y}}\right]$
$\frac{d y}{d x}=\left[\frac{\frac{1}{x}-e^{x-y}}{\frac{1}{y}-e^{x-y}}\right]$
$=\frac{\left(\frac{1-x e^{x-y}}{x}\right)}{\left(\frac{1-y e^{x-y}}{y}\right)}$
$=\frac{y\left(1-x e^{x-y}\right)}{x\left(1-y e^{x-y}\right)}$
$=\frac{-y\left(x e^{x-y}-1\right)}{-x\left(y e^{x-y}-1\right)}$
$\frac{d y}{d x}=\frac{y}{x} \cdot\left(\frac{x e^{x-y}-1}{y e^{x-y}-1}\right)$
Hence $\frac{d y}{d x}=\frac{y}{x} \left[\frac{x e^{x-y}-1}{y e^{x-y}-1}\right]$ is the required differentiation.

Differentiation exercise 10.4 question 11

Answer:
$\left[\frac{\sin (x+y)-y \cos x y}{x \cos x y-\sin (x+y)}\right]$
Hint:
Use chain rule and product rule
Given:
$\sin (x y)+\cos (x+y)=1$
Solution:
Differentiate the given equation w.r.t x
$\frac{d}{d x}[\sin x y+\cos (x+y)]=\frac{d(1)}{d x}$
$\frac{d}{d x}[\sin x y]+\frac{d}{d x}[\cos (x+y)]=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$\frac{d(\sin x y)}{d x y} \times \frac{d x y}{d x}+\frac{d \cos (x+y)}{d(x+y)} \times \frac{d(x+y)}{d x}=0$ [Using chain rule]
$\cos (x y) \times\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]+(-\sin (x+y))\left[\frac{d x}{d x}+\frac{d y}{d x}\right]=0$ $\left[\begin{array}{c} \frac{d(\sin x)}{d x}=\cos x \\ \frac{d(\cos x)}{d x}=-\sin x \\ \frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x} \end{array}\right]$
$\cos x y\left[x \frac{d y}{d x}+y\right]+(-\sin (x+y))\left[1+\frac{d y}{d x}\right]=0$
$x \cos x y \cdot \frac{d y}{d x}+y \cos x y-\sin (x+y)-\sin (x+y) \frac{d y}{d x}=0$
$\frac{d y}{d x}(x \cos x y-\sin (x+y))=\sin (x+y)-y \cos x y$
$\frac{d y}{d x}=\frac{\sin (x+y)-y \cos x y}{x \cos x y-\sin (x+y)}$
Hence, $\frac{d y}{d x}=\frac{\sin (x+y)-y \cos x y}{x \cos x y-\sin (x+y)}$ is the required answer.

Differentiation exercise 10.4 question 12

Answer:

$\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Hint:
Use $(\sin A-\sin B) \text { and }(\cos A+\cos B)$ to formulas and use $\frac{d\left(\sin ^{-1} x\right)}{d x}=\frac{1}{\sqrt{1-x^{2}}}$
Given:
$\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$
Solution:
Let $x=\sin A, y=\sin B$
So, the equation will become
$\sqrt{1-\sin ^{2} A}+\sqrt{1-\sin ^{2} B}=a(\sin A-\sin B)$
$\sqrt{\cos ^{2} A}+\sqrt{\cos ^{2} B}=a(\sin A-\sin B) \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$\begin{aligned} &\cos A+\cos B=a(\sin A-\sin B) \\ &a=\frac{\cos A+\cos B}{\sin A-\sin B} \end{aligned}$
$a=\frac{2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)}$ $\left[\begin{array}{l} \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right) \\\\ \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right) \end{array}\right]$
$a=\cot \left(\frac{A-B}{2}\right)$
$\begin{aligned} &\cot ^{-1} a=\frac{A-B}{2} \\ &2 \cot ^{-1} a=A-B \end{aligned}$ $\left[\because \frac{\cos \theta}{\sin \theta}=\cot \theta\right]$
$2 \cot ^{-1} a=\sin ^{-1} x-\sin ^{-1} y \quad[x=\sin A, y=\sin B]$
Differentiate the above equation w.r.t x
$\frac{d\left(2 \cot ^{-1} a\right)}{d x}=\frac{d\left(\sin ^{-1} x-\sin ^{-1} y\right)}{d x}$
$2 \frac{d\left(\cot ^{-1} a\right)}{d x}=\frac{d\left(\sin ^{-1} x\right)}{d x}-\frac{d\left(\sin ^{-1} y\right)}{d x}$
$2 \times 0=\frac{1}{\sqrt{1-x^{2}}}-\frac{d\left(\sin ^{-1} y\right)}{d y} \times \frac{d y}{d x}$ $\left[\because \frac{d\left(\sin ^{-1} x\right)}{d x}=\frac{1}{\sqrt{1-x^{2}}}, \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$0=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}$
$\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}$
$\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$
Hence, if $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$
Then,$\frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$ is the required answer
Hence proved

Differentiation exercise 10.4 question 13

Answer:
$-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Hint:
Use trigonometric identities and differentiation formula of inverse trigonometric functions
Given:
$y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$
Solution:
Let $x=\sin A, y=\sin B$
$\therefore$ The given equation becomes
$\sin B \sqrt{1-\sin ^{2} A}+\sin A \sqrt{1-\sin ^{2} B}=1$
$\sin B \sqrt{\cos ^{2} A}+\sin A \sqrt{\cos ^{2} B}=1 \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$\sin B \cos A+\sin A \cos B=1$
$\sin (A+B)=1 \quad[\because \sin (A+B)=\sin A \cos B+\cos A \sin B]$
$\begin{aligned} &A+B=\sin ^{-1}(1) \\ &A+B=\frac{\pi}{2} \end{aligned}$
$\therefore \sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2} \quad[\because x=\sin A, y=\sin B]$
Differentiate $\left(\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}\right)$ w.r.t x
$\frac{d\left(\sin ^{-1} x+\sin ^{-1} y\right)}{d x}=\frac{d\left(\frac{\pi}{2}\right)}{d x}$
$\frac{d\left(\sin ^{-1} x\right)}{d x}+\frac{d\left(\sin ^{-1} y\right)}{d x}=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$\frac{1}{\sqrt{1-x^{2}}}+\frac{d\left(\sin ^{-1} y\right)}{d y} \times \frac{d y}{d x}=0 \quad\left[\because \frac{d\left(\sin ^{-1} x\right)}{d x}=\frac{1}{\sqrt{1-x^{2}}}\right]$
$\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=0$
$\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=-\frac{1}{\sqrt{1-x^{2}}}$
$\frac{d y}{d x}=\frac{-\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$
$\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Hence if $y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1$
Then, $\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$ Is the required answer

Differentiation exercise 10.4 question 14

Answer:
$\frac{d y}{d x}+y^{2}=0$
Hint:
Use product rule
Given:
$x y=1$
Solution:
Differentiate $x y=1$ w.r.t x
$\frac{d(x y)}{d x}=\frac{d(1)}{d x}$
$x \frac{d y}{d x}+y \frac{d x}{d x}=0$ $\left[\begin{array}{l} \because \frac{d(\operatorname{cons} \tan t)}{d x}=0 \\ \frac{d(u \cdot v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x} \end{array}\right]$
$x \frac{d y}{d x}+y=0$
$\begin{aligned} &x \frac{d y}{d x}=-y \\ &\frac{d y}{d x}=-\frac{y}{x} \end{aligned}$
$\frac{d y}{d x}=-\frac{y}{\left(\frac{1}{y}\right)} \; \; \; \; \; \; \; \quad\left[\begin{array}{c} x y=1 \\ x=\frac{1}{y} \end{array}\right]$
$\frac{d y}{d x}=-y^{2} \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{d y}{d x}+y^{2}=0\right]$
Hence, if $x y=1$
Then $\frac{d y}{d x}+y^{2}=0$ hence proved

Differentiation exercise 10.4 question 15

Answer:
$2 \frac{d y}{d x}+y^{3}=0$
Hint:
Use product rule and $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$
Given:
$x y^{2}=1$
Solution:
Differentiate $\left[x y^{2}=1\right]$ w.r.t x
$\frac{d\left(x y^{2}\right)}{d x}=\frac{d(1)}{d x}$
$x \frac{d y^{2}}{d x}+y^{2} \frac{d x}{d x}=0$ $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$ [Product rule $\frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}$]
$x \frac{d y^{2}}{d y} \times \frac{d y}{d x}+y^{2}=0$
$x(2 y) \times \frac{d y}{d x}+y^{2}=0$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$2 x y \frac{d y}{d x}+y^{2}=0$
$\begin{aligned} &\frac{d y}{d x}=-\frac{y^{2}}{2 x y}=\frac{-y}{2 x} \\ &2 \frac{d y}{d x}=-\frac{y}{x} \end{aligned}$
$2 \frac{d y}{d x}=-\frac{y}{\left(\frac{1}{y^{2}}\right)}$ $\left[\begin{array}{c} x y^{2}=1 \\ x=\frac{1}{y^{2}} \end{array}\right]$
$2 \frac{d y}{d x}+y^{3}=0$
Hence, if $x y^{2}=1$ then $2 \frac{d y}{d x}+y^{3}=0$ is the required answer
Hence proved

Differentiation exercise 10.4 question 16

Answer:
$(1+x)^{2} \frac{d y}{d x}+1=0$
Hint:
Use quotient rule and algebraic identities
Given:
$x \sqrt{1+y}+y \sqrt{1+x}=0$
Solution:
$\begin{aligned} &x \sqrt{1+y}+y \sqrt{1+x}=0 \\ &x \sqrt{1+y}=-y \sqrt{1+x} \end{aligned}$
Squaring both the side,
$(x \sqrt{1+y})^{2}=(-y \sqrt{1+x})^{2}$
$\begin{aligned} &x^{2}(1+y)=y^{2}(1+x) \\ &x^{2}+x^{2} y=y^{2}+x y^{2} \end{aligned}$
$x^{2}-y^{2}=x y^{2}-x^{2} y$
$(x+y)(x-y)=x y(y-x)$ $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$
$\begin{aligned} &(x+y)=-x y \\ &y+x y=-x \\ &y(1+x)=-x \end{aligned}$
$y=\frac{-x}{1+x}$
Differentiate this above equation w.r.t x
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{-x}{1+x}\right)$
$\frac{d y}{d x}=\frac{(1+x) \cdot \frac{d(-x)}{d x}-(-x) \cdot \frac{d(1+x)}{d x}}{(1+x)^{2}}$ [Using quotient rule]
$\frac{d y}{d x}=\frac{-(1+x)+x(0+1)}{(1+x)^{2}}$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$\begin{aligned} &\frac{d y}{d x}=\frac{-(1+x)+x}{(1+x)^{2}} \\ &\frac{d y}{d x}=\frac{-1-x+x}{(1+x)^{2}} \end{aligned}$
$\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}$
$(1+x)^{2} \frac{d y}{d x}=-1$
$(1+x)^{2} \frac{d y}{d x}+1=0$
Hence proved

Differentiation exercise 10.4 question 17

Answer:
$\frac{d y}{d x}=\frac{x+y}{x-y}$
Hint:
Use quotient rule and properties of logarithm
Given:
$\log \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)$
Solution:
$\log \sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)$
$\log \left(x^{2}+y^{2}\right)^{\frac{1}{2}}=\tan ^{-1}\left(\frac{y}{x}\right)$
$\frac{1}{2} \log \left(x^{2}+y^{2}\right)=\tan ^{-1}\left(\frac{y}{x}\right) \quad\left[\because \log a^{m}=m \log a\right]$
Differentiate this above equation w.r.t x
$\frac{1}{2} \cdot \frac{d}{d x}\left(\log \left(x^{2}+y^{2}\right)\right)=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{y}{x}\right)\right)$
$\frac{1}{2} \cdot \frac{d \log \left(x^{2}+y^{2}\right)}{d\left(x^{2}+y^{2}\right)} \times\left(\frac{d x^{2}}{d x}+\frac{d y^{2}}{d x}\right)=\frac{1}{1+\left(\frac{y}{x}\right)^{2}} \times \frac{x \frac{d y}{d x}-y \frac{d x}{d x}}{x^{2}}$
$\left[\because \frac{d \log x}{d x}=\frac{1}{x}, \frac{d \tan ^{-1} x}{d x}=\frac{1}{1+x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\right]$
$\frac{1}{2} \times\left(\frac{1}{x^{2}+y^{2}}\right) \times\left(2 x+\frac{d y^{2}}{d y} \times \frac{d y}{d x}\right)=\frac{1}{1+\left(\frac{y^{2}}{x^{2}}\right)} \times \frac{x \frac{d y}{d x}-y}{x^{2}}$
$\begin{aligned} &\text { - }\\ &\frac{1}{2\left(x^{2}+y^{2}\right)} \times\left(2 x+2 y \frac{d y}{d x}\right)=\frac{1}{\left(\frac{x^{2}+y^{2}}{x^{2}}\right)} \cdot\left(\frac{x}{x^{2}} \cdot \frac{d y}{d x}-\frac{y}{x^{2}}\right) \end{aligned}$
$\frac{1}{2\left(x^{2}+y^{2}\right)} \times 2\left(x+y \frac{d y}{d x}\right)=\frac{x^{2}}{x^{2}+y^{2}} \times\left(\frac{1}{x} \cdot \frac{d y}{d x}-\frac{y}{x^{2}}\right)$

$\frac{x}{x^{2}+y^{2}}+\frac{y}{x^{2}+y^{2}} \cdot \frac{d y}{d x}=\left[\frac{x^{2}}{x^{2}+y^{2}} \times \frac{1}{x} \cdot \frac{d y}{d x}\right]-\frac{x^{2}}{x^{2}+y^{2}} \times \frac{y}{x^{2}}$
$\frac{x}{x^{2}+y^{2}}+\frac{y}{x^{2}+y^{2}} \cdot \frac{d y}{d x}=\left[\frac{x}{x^{2}+y^{2}} \cdot \frac{d y}{d x}\right]-\frac{y}{x^{2}+y^{2}}$
$\frac{x}{x^{2}+y^{2}} \cdot \frac{d y}{d x}-\frac{y}{x^{2}+y^{2}} \cdot \frac{d y}{d x}=\frac{x}{x^{2}+y^{2}}+\frac{y}{x^{2}+y^{2}}$
$\frac{d y}{d x}\left(\frac{x-y}{x^{2}+y^{2}}\right)=\frac{x+y}{\left(x^{2}+y^{2}\right)}$
$\frac{d y}{d x}=\frac{x+y}{x-y}$
Hence proved

Differentiation exercise 10.4 question 18

Answer:
$\frac{d y}{d x}=\frac{y}{x}$
Hint:
Use quotient rule
Given:
$\sec \left(\frac{x+y}{x-y}\right)=a$
Solution:
$\begin{aligned} &\sec \left(\frac{x+y}{x-y}\right)=a \\ &\sec ^{-1} a=\frac{x+y}{x-y} \end{aligned}$
Differentiating $\left[\sec ^{-1} a=\frac{x+y}{x-y}\right]$ w.r.t x
$\frac{d\left(\sec ^{-1} a\right)}{d x}=\frac{d}{d x}\left(\frac{x+y}{x-y}\right)$
$0=\frac{(x-y) \frac{d(x+y)}{d x}-(x+y) \frac{d(x-y)}{d x}}{(x-y)^{2}}$ [Use quotient rule]
$\frac{(x-y)\left(\frac{d x}{d x}+\frac{d y}{d x}\right)-(x+y)\left(\frac{d x}{d x}-\frac{d y}{d x}\right)}{(x-y)^{2}}=0$
$(x-y)\left(1+\frac{d y}{d x}\right)-(x+y)\left(1-\frac{d y}{d x}\right)=0$
$(x-y)+(x-y) \frac{d y}{d x}-(x+y)+(x+y) \frac{d y}{d x}=0$
$\frac{d y}{d x}[(x-y)+(x+y)]=x+y-(x-y)$
$\frac{d y}{d x}[x-y+x+y]=x+y-x+y$
$\frac{d y}{d x}(2 x)=2 y$
$\begin{aligned} &\frac{d y}{d x}=\frac{2 y}{2 x}=\frac{y}{x} \\ &\frac{d y}{d x}=\frac{y}{x} \end{aligned}$
Thus, proved

Differentiation exercise 10.4 question 19

Answer:
$\frac{d y}{d x}=\frac{x}{y}\left(\frac{1-\tan a}{1+\tan a}\right)$
Hint:
Use differentiation formulas
Given:
$\tan ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=a$
Solution:
$\tan ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=a$
$\tan a=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$
$\begin{aligned} &\left(x^{2}+y^{2}\right) \tan a=x^{2}-y^{2} \\ &x^{2}-y^{2}=(\tan a)\left(x^{2}+y^{2}\right) \end{aligned}$
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left(x^{2}-y^{2}\right)=\tan a \cdot \frac{d}{d x}\left(x^{2}+y^{2}\right)$
$\frac{d\left(x^{2}\right)}{d x}-\frac{d y^{2}}{d x}=\tan a \cdot\left[\frac{d x^{2}}{d x}+\frac{d y^{2}}{d x}\right]$
$2 x-\frac{d y^{2}}{d y} \times \frac{d y}{d x}=\tan a \cdot\left[2 x+\frac{d y^{2}}{d y} \times \frac{d y}{d x}\right]$
$2 x-2 y \frac{d y}{d x}=\tan a\left[2 x+2 y \cdot \frac{d y}{d x}\right]$
$2 x-2 y \frac{d y}{d x}=2 x \tan a+2 y \tan a \frac{d y}{d x}$
$2 y \tan a \frac{d y}{d x}+2 y \frac{d y}{d x}=2 x-2 x \tan a$
$2 y \frac{d y}{d x}(\tan a+1)=2 x(1-\tan a)$
$\frac{d y}{d x}=\frac{(1-\tan a)}{(1+\tan a)} \times \frac{2 x}{2 y}$
$\frac{d y}{d x}=\left(\frac{1-\tan a}{1+\tan a}\right) \times \frac{x}{y}$
Thus, proved

Differentiation exercise 10.4 question 20

Answer:
$\frac{d y}{d x}=-\frac{y\left(x^{2} y+x+y\right)}{x\left(x y^{2}+x+y\right)}$
Hint:
Use product rule and chain rule
Given:
$x y \log (x+y)=1$
Solution:
$x y \log (x+y)=1$
Differentiate the given equation w.r.t x
$\frac{d}{d x}[x y \log (x+y)]=\frac{d(1)}{d x}$
$x y \cdot \frac{d}{d x}(\log (x+y))+\log (x+y) \frac{d(x y)}{d x}=0$ $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$x y\left[\frac{d \log (x+y)}{d(x+y)} \times \frac{d(x+y)}{d x}\right]+\log (x+y)\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]=0$ [Use product rule and chain rule]
$x y\left[\frac{1}{x+y} \times\left(\frac{d x}{d x}+\frac{d y}{d x}\right)\right]+\log (x+y)\left[x \frac{d y}{d x}+y\right]=0$
$x y\left[\frac{1}{x+y}+\frac{1}{x+y} \frac{d y}{d x}\right]+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0$
$\frac{x y}{x+y}+\frac{x y}{x+y} \frac{d y}{d x}+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0$
$\frac{d y}{d x}\left[\frac{x y}{x+y}+x \log (x+y)\right]=-\frac{x y}{x+y}-y \log (x+y)$
Also $x y \log (x+y)=1$ $\left[\log (x+y)=\frac{1}{x y}\right]$
Put $\log (x+y)=\frac{1}{x y}$ in the above equation
$\frac{d y}{d x}\left[\frac{x y}{x+y}+x \cdot \frac{1}{x y}\right]=-\frac{x y}{x+y}-y \cdot \frac{1}{x y}$
$\frac{d y}{d x}\left[\frac{x y}{x+y}+\frac{1}{y}\right]=-\left(\frac{x y}{x+y}+\frac{1}{x}\right)$
$\frac{d y}{d x}\left[\frac{x y^{2}+(x+y)}{y(x+y)}\right]=-\left(\frac{x^{2} y+(x+y)}{x(x+y)}\right)$
$\frac{d y}{d x}\left[\frac{\left(x y^{2}+(x+y)\right)}{y}\right]=-\left[\frac{\left(x^{2} y+(x+y)\right)}{x}\right]$
$\begin{aligned} &\frac{d y}{d x}=\frac{-\left(x^{2} y+x+y\right)}{\left(x y^{2}+x+y\right)} \times \frac{y}{x} \\ &\frac{d y}{d x}=\frac{-y}{x} \cdot\left(\frac{x^{2} y+x+y}{x y^{2}+x+y}\right) \end{aligned}$
Thus, proved

Differentiation exercise 10.4 question 21

Answer:
$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$
Hint:
Use chain rule
Given:
$y=x \sin (a+y)$
Solution:
$y=x \sin (a+y)$
Differentiate the given equation w.r.t x
$\frac{d y}{d x}=\frac{d(x \sin (a+y))}{d x}$
$\frac{d y}{d x}=x \cdot \frac{d \sin (a+y)}{d x}+\sin (a+y) \cdot \frac{d x}{d x}$
$\frac{d y}{d x}=x\left[\frac{d \sin (a+y)}{d(a+y)} \times \frac{d(a+y)}{d y} \times \frac{d y}{d x}\right]+\sin (a+y)$ [Using chain rule]

$=x\left[\cos (a+y) \times\left(\frac{d a}{d y}+\frac{d y}{d y}\right) \times \frac{d y}{d x}\right]+\sin (a+y)$
$=x\left[\cos (a+y) \times(0+1) \times \frac{d y}{d x}\right]+\sin (a+y)$
$\frac{d y}{d x}=x \cos (a+y) \frac{d y}{d x}+\sin (a+y)$
$\frac{d y}{d x}-x \cos (a+y) \frac{d y}{d x}=\sin (a+y)$
$\frac{d y}{d x}(1-x \cos (a+y))=\sin (a+y)$
$\frac{d y}{d x}=\frac{\sin (a+y)}{1-x \cos (a+y)}$
$y=x \sin (a+y)$
$x=\frac{y}{\sin (a+y)}$
Put $x=\frac{y}{\sin (a+y)}$ in the above equation
$\frac{d y}{d x}=\frac{\sin (a+y)}{1-\frac{y}{\sin (a+y)} \times \cos (a+y)}$
$\frac{d y}{d x}=\frac{\sin (a+y)}{\left[\frac{\sin (a+y)-y \cos (a+y)}{\sin (a+y)}\right]}$
$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$
Thus, proved

Differentiation exercise 10.4 question 22

Answer:
$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$
Hint:
Use chain rule
Given:
$x \sin (a+y)+\sin a \cos (a+y)=0$
Solution:
$\begin{aligned} &x \sin (a+y)+\sin a \cos (a+y)=0 \\ &x \sin (a+y)=-\sin a \cos (a+y) \end{aligned}$
$-\frac{x}{\sin a}=\frac{\cos (a+y)}{\sin (a+y)}$
$-\frac{x}{\sin a}=\cot (a+y)$
Differentiating equation w.r.t x
$\frac{d}{d x}\left(\frac{-x}{\sin a}\right)=\frac{d}{d x} \cot (a+y)$
$\frac{-1}{\sin a} \cdot \frac{d x}{d x}=\frac{d \cot (a+y)}{d(a+y)} \times \frac{d(a+y)}{d x}$ [Use chain rule]
$\frac{-1}{\sin a}=-\cos e c^{2}(a+y) \times\left(\frac{d a}{d x}+\frac{d y}{d x}\right)$ $\left[\because \frac{d(\cot \theta)}{d \theta}=-\cos e c^{2} \theta\right]$
$-\frac{1}{\sin a}=-\cos e c^{2}(a+y)\left[0+\frac{d y}{d x}\right]$
$-\frac{1}{\sin a}=-\operatorname{cosec}^{2}(a+y) \cdot \frac{d y}{d x}$
$\frac{d y}{d x}=\frac{1}{\sin a \cdot \cos e c^{2}(a+y)}$
$\frac{d y}{d x}=\frac{1}{\sin a \cdot\left(\frac{1}{\sin ^{2}(a+y)}\right)}$ $\left[\because \cos e c \theta=\frac{1}{\sin \theta}\right]$
$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$
Thus, proved

Differentiation exercise 10.4 question 23

Answer:
$\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}$
Hint:
Use product rule and chain rule
Given:
$y=x \sin y$
Solution:
$y=x \sin y$
Differentiate w.r.t x
$\frac{d y}{d x}=\frac{d(x \sin y)}{d x}$
$\frac{d y}{d x}=x \cdot \frac{d(\sin y)}{d x}+\sin y \cdot \frac{d x}{d x}$ [Using product rule]
$\frac{d y}{d x}=x \cdot \frac{d \sin y}{d y} \cdot \frac{d y}{d x}+\sin y$ [Using chain rule]
$\frac{d y}{d x}=x \cdot \cos y \cdot \frac{d y}{d x}+\sin y$
$\frac{d y}{d x}-x \cos y \frac{d y}{d x}=\sin y$
$\frac{d y}{d x}(1-x \cos y)=\sin y$
$\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}$
Thus, proved

Differentiation exercise 10.4 question 24

Answer:
$\left(x^{2}+1\right) \frac{d y}{d x}+x y+1=0$
Hint:
Use product rule and chain rule
Given:
$y \sqrt{x^{2}+1}=\log \left(\sqrt{x^{2}+1}-x\right)$
Solution:
$y \sqrt{x^{2}+1}=\log \left(\sqrt{x^{2}+1}-x\right)$
Differentiate w.r.t x
$\frac{d}{d x}\left(y \sqrt{x^{2}+1}\right)=\frac{d}{d x}\left(\log \left(\sqrt{x^{2}+1}-x\right)\right)$
$y \frac{d}{d x}\left(\sqrt{x^{2}+1}\right)+\left(\sqrt{x^{2}+1}\right) \frac{d y}{d x}=\frac{d\left(\log \left(\sqrt{x^{2}+1}-x\right)\right)}{d\left(\sqrt{x^{2}+1}-x\right)} \times \frac{d\left(\sqrt{x^{2}+1}-x\right)}{d x}$
[Use product rule and chain rule]
$\Rightarrow y \cdot\left[\frac{d\left(\sqrt{x^{2}+1}\right)}{d\left(x^{2}+1\right)} \times \frac{d\left(x^{2}+1\right)}{d x}\right]+\sqrt{x^{2}+1} \frac{d y}{d x}=\frac{1}{\left(\sqrt{x^{2}+1}-x\right)} \times\left(\frac{\left(\sqrt{x^{2}+1}\right)}{d x}-\frac{d x}{d x}\right)$
$y \cdot \frac{1}{2 \sqrt{x^{2}+1}} \cdot 2 x+\sqrt{x^{2}+1} \frac{d y}{d x}=\frac{1}{\left(\sqrt{x^{2}+1}-x\right)} \times\left[\frac{d \sqrt{x^{2}+1}}{d\left(x^{2}+1\right)} \times \frac{d\left(x^{2}+1\right)}{d x}-1\right]$
$\frac{x y}{\sqrt{x^{2}+1}}+\sqrt{x^{2}+1} \frac{d y}{d x}=\frac{1}{\left(\sqrt{x^{2}+1}-x\right)} \cdot\left[\frac{1}{2 \sqrt{x^{2}+1}} \times 2 x-1\right]$
$\frac{x y}{\sqrt{x^{2}+1}}+\sqrt{x^{2}+1} \cdot \frac{d y}{d x}=\frac{1}{\left(\sqrt{x^{2}+1}-x\right)} \cdot\left[\frac{x-\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}\right]$
$\frac{x y}{\sqrt{x^{2}+1}}+\sqrt{x^{2}+1} \cdot \frac{d y}{d x}=-\frac{1}{\sqrt{x^{2}+1}}$
$\sqrt{x^{2}+1} \cdot \frac{d y}{d x}=\frac{-1}{\sqrt{x^{2}+1}}-\frac{x y}{\sqrt{x^{2}+1}}$
$\sqrt{x^{2}+1} \cdot \frac{d y}{d x}=\frac{-1-x y}{\sqrt{x^{2}+1}}$
$\left(x^{2}+1\right) \frac{d y}{d x}=-(1+x y)$
$\left(x^{2}+1\right) \frac{d y}{d x}+1+x y=0$
Thus, proved

Differentiation exercise 10.4 question 25

Answer:
$\frac{d y}{d x}=\frac{2 x^{3}+y-x^{2} y \cos x y}{x\left(x^{2} \cos x y+1+2 x y\right)}$
Hint:
Use chain rule and $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$
Given:
$\sin (x y)+\frac{y}{x}=x^{2}-y^{2}$
Solution:
$\sin (x y)+\frac{y}{x}=x^{2}-y^{2}$
Differentiate w.r.t x
$\frac{d}{d x}(\sin x y)+\frac{d}{d x}\left(\frac{y}{x}\right)=\frac{d\left(x^{2}\right)}{d x}-\frac{d\left(y^{2}\right)}{d x}$
$\frac{d(\sin x y)}{d(x y)} \times \frac{d(x y)}{d x}+\frac{d}{d x}\left(\frac{y}{x}\right)=2 x-\frac{d y^{2}}{d y} \times \frac{d y}{d x}$ [Using chain rule and $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$]
$\cos x y \times\left(x \frac{d y}{d x}+y \frac{d x}{d x}\right)+\frac{x \frac{d y}{d x}-y \frac{d x}{d x}}{x^{2}}=2 x-2 y \frac{d y}{d x}$
$x \cos x y \frac{d y}{d x}+y \cos x y+\frac{x \frac{d y}{d x}-y}{x^{2}}=2 x-2 y \frac{d y}{d x}$
$x \cos x y \frac{d y}{d x}+y \cos x y+\frac{1}{x} \frac{d y}{d x}-\frac{y}{x^{2}}=2 x-2 y \frac{d y}{d x}$
$x \cos x y \frac{d y}{d x}+\frac{1}{x} \frac{d y}{d x}+2 y \frac{d y}{d x}=2 x+\frac{y}{x^{2}}-y \cos x y$
$\frac{d y}{d x}\left(x \cos x y+\frac{1}{x}+2 y\right)=2 x+\frac{y}{x^{2}}-y \cos x y$
$\frac{d y}{d x}\left(\frac{x^{2} \cos x y+1+2 x y}{x}\right)=\frac{2 x^{3}+y-x^{2} y \cos x y}{x^{2}}$
$\frac{d y}{d x}\left(x^{2} \cos x y+2 x y+1\right)=\frac{1}{x}\left(2 x^{3}+y-x^{2} y \cos x y\right)$
$\frac{d y}{d x}=\frac{2 x^{3}+y-x^{2} y \cos x y}{x\left(x^{2} \cos x y+1+2 x y\right)}$
Hence proved

Differentiation exercise 10.4 question 26

Answer:
$\frac{d y}{d x}=\frac{\sec ^{2}(x+y)+\sec ^{2}(x-y)}{\sec ^{2}(x-y)-\sec ^{2}(x+y)}$
Hint:
Use chain rule
Given:
$\tan (x+y)+\tan (x-y)=1$
Solution:
$\tan (x+y)+\tan (x-y)=1$
Differentiate w.r.t x
$\frac{d(\tan (x+y))}{d x}+\frac{d(\tan (x-y))}{d x}=\frac{d(1)}{d x}$
$\frac{d(\tan x+y)}{d(x+y)} \times \frac{d(x+y)}{d x}+\frac{d(\tan (x-y))}{d(x-y)} \times \frac{d(x-y)}{d x}=0$
[Using chain rule and $\frac{d(\operatorname{cons} \tan t)}{d x}=0$]
$\sec ^{2}(x+y) \times\left(\frac{d x}{d x}+\frac{d y}{d x}\right)+\sec ^{2}(x-y)\left(\frac{d x}{d x}-\frac{d y}{d x}\right)=0$
$\sec ^{2}(x+y)+\sec ^{2}(x+y) \frac{d y}{d x}+\sec ^{2}(x-y)-\sec ^{2}(x-y) \frac{d y}{d x}=0$
$\frac{d y}{d x}\left(\sec ^{2}(x+y)-\sec ^{2}(x-y)\right)=-\left(\sec ^{2}(x+y)+\sec ^{2}(x-y)\right)$
$\frac{d y}{d x}=\frac{\sec ^{2}(x+y)+\sec ^{2}(x-y)}{\sec ^{2}(x-y)-\sec ^{2}(x+y)}$

Differentiation exercise 10.4 question 27

Answer:
$\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$
Hint:
Use chain rule
Given:
$e^{x}+e^{y}=e^{x+y}$
Solution:
$e^{x}+e^{y}=e^{x+y}$
Differentiate w.r.t x
$\frac{d}{d x}\left(e^{x}+e^{y}\right)=\frac{d\left(e^{x+y}\right)}{d x}$
$\frac{d\left(e^{x}\right)}{d x}+\frac{d\left(e^{y}\right)}{d x}=\frac{d\left(e^{x+y}\right)}{d(x+y)} \times \frac{d(x+y)}{d x}$ [Using chain rule]
$e^{x}+\frac{d\left(e^{y}\right)}{d y} \times \frac{d y}{d x}=e^{x+y} \times\left(\frac{d x}{d x}+\frac{d y}{d x}\right)$ $\left[\because \frac{d\left(e^{x}\right)}{d x}=e^{x}\right]$
$e^{x}+e^{y} \frac{d y}{d x}=e^{x+y}+e^{x+y} \frac{d y}{d x}$
$e^{y} \frac{d y}{d x}-e^{x+y} \frac{d y}{d x}=e^{x+y}-e^{x}$
$\frac{d y}{d x}\left(e^{y}-e^{x+y}\right)=\left(e^{x+y}-e^{x}\right)$
$\frac{d y}{d x}=\frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}}$
$\frac{d y}{d x}=\frac{e^{x} \cdot e^{y}-e^{x}}{e^{y}-e^{x} \cdot e^{y}}$
$=\frac{e^{x}\left(e^{y}-1\right)}{e^{y}\left(1-e^{x}\right)}$
$=\frac{e^{x}\left(e^{y}-1\right)}{e^{y}\left(-\left(e^{x}-1\right)\right)}$
$\frac{d y}{d x}=-\frac{e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$
Thus proved

Differentiation exercise 10.4 question 28

Answer:
$\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$
Hint:
Use chain rule and product rule
Given:
$\cos y=x \cos (a+y)$
Solution:
Differentiate the given equation w.r.t x
$\frac{d \cos y}{d x}=\frac{d(x \cos (a+y))}{d x}$
$\frac{d(\cos y)}{d y} \times \frac{d y}{d x}=x \frac{d(\cos (a+y))}{d x}+\cos (a+y) \cdot \frac{d x}{d x}$ [Use chain rule and product rule]
$(-\sin y) \times \frac{d y}{d x}=x \times\left[\frac{d \cos (a+y)}{d(a+y)} \times \frac{d(a+y)}{d x}\right]+\cos (a+y)$
$-\sin y \frac{d y}{d x}=x \cdot\left(-\sin (a+y) \times\left(\frac{d a}{d x}+\frac{d y}{d x}\right)\right)+\cos (a+y)$
$-\sin y \frac{d y}{d x}=-x \sin (a+y)\left(0+\frac{d y}{d x}\right)+\cos (a+y)$ $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$
$-\sin y \frac{d y}{d x}=-x \sin (a+y) \frac{d y}{d x}+\cos (a+y)$
$x \sin (a+y) \frac{d y}{d x}-\sin y \frac{d y}{d x}=\cos (a+y)$
$\frac{d y}{d x}(x \sin (a+y)-\sin y)=\cos (a+y)$
Also $\cos y=x \cos (a+y)$
$x=\frac{\cos y}{\cos (a+y)}$
Put this values of x in the above equation
$\frac{d y}{d x}\left[\frac{\cos y}{\cos (a+y)} \sin (a+y)-\sin y\right]=\cos (a+y)$
$\frac{d y}{d x}\left[\frac{\cos y \cdot \sin (a+y)-\sin y \cdot \cos (a+y)}{\cos (a+y)}\right]=\cos (a+y)$
$\frac{d y}{d x}\left[\frac{\sin ((a+y)-y)}{\cos (a+y)}\right]=\cos (a+y)$ $[\because \sin (A-B)=\sin A \cos B-\cos A \sin B]$
$\frac{d y}{d x} \times\left[\frac{\sin a}{\cos (a+y)}\right]=\cos (a+y)$
$\begin{aligned} &\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a} \end{aligned}$
Thus, proved

Differentiation exercise 10.4 question 29

Answer:
$\frac{d y}{d x}=\frac{\pi}{4}(\sqrt{2}+1)$
Hint:
Use chain rule and product rule
Given:
$\sin ^{2} y+\cos x y=K$
Solution:
$\sin ^{2} y+\cos x y=K$
Differentiate the given equation w.r.t x
$\frac{d}{d x}\left(\sin ^{2} y+\cos x y\right)=\frac{d(k)}{d x}$
$\frac{d \sin ^{2} y}{d x}+\frac{d \cos x y}{d x}=0$ $\left[\because \frac{d(\operatorname{con} s \tan t)}{d x}=0\right]$
$\left(\frac{d \sin ^{2} y}{d \sin y} \times \frac{d \sin y}{d y} \times \frac{d y}{d x}\right)+\left(\frac{d \cos x y}{d x y}\right) \times \frac{d x y}{d x}=0$ [Using chain rule]
$2 \sin y \times \cos y \times \frac{d y}{d x}+(-\sin x y) \times\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]=0$ [Using product rule]
$\sin 2 y \frac{d y}{d x}-\sin x y\left(x \frac{d y}{d x}\right)-y \sin x y=0$
$\frac{d y}{d x}(\sin 2 y-x \sin x y)=y \sin x y$
$\frac{d y}{d x}=\frac{y \sin x y}{\sin 2 y-x \sin x y}$
At $x=1, y=\frac{\pi}{4}$
$\frac{d y}{d x}=\frac{\left(\frac{\pi}{4}\right) \sin \left(1 \times \frac{\pi}{4}\right)}{\sin \left(2 \times \frac{\pi}{4}\right)-1 \times \sin \left(1 \times \frac{\pi}{4}\right)}$
$=\frac{\frac{\pi}{4} \cdot\left(\frac{1}{\sqrt{2}}\right)}{\sin \frac{\pi}{2}-\frac{1}{\sqrt{2}}}$ $\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin \frac{\pi}{2}=1\right]$
$=\frac{\frac{\pi}{4 \sqrt{2}}}{1-\frac{1}{\sqrt{2}}}$
$=\frac{\frac{\pi}{4 \sqrt{2}}}{\frac{(\sqrt{2}-1)}{\sqrt{2}}}$
$=\frac{\pi \times \sqrt{2}}{(\sqrt{2}-1) 4 \sqrt{2}}=\frac{\pi}{4(\sqrt{2}-1)} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\pi(\sqrt{2}+1)}{4}$
Hence at $x=1, y=\frac{\pi}{4}$
$\frac{d y}{d x}=\frac{\pi}{4}(\sqrt{2}+1)$

Differentiation exercise 10.4 question 30

Answer:
$\frac{d y}{d x}=8\left[\frac{4}{16+\pi^{2}}-\frac{1}{\log 2}\right]_{\mathrm{At}} x=\frac{\pi}{4}$
Hint:
Use chain rule and differentiation of inverse trigonometric function
Given:
$y=\left[\log _{\cos x} \sin x\right]\left[\log _{\sin x} \cos x\right]^{-1}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
Solution:
$y=\left[\log _{\cos x} \sin x\right]\left[\log _{\sin x} \cos x\right]^{-1}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$y=\left[\log _{\cos x} \sin x\right]\left[\log _{\cos x} \sin x\right]+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ $\left[\because \log _{b} a=\left(\log _{a} b\right)^{-1}\right]$
$y=\left[\log _{\cos x} \sin x\right]^{2}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$y=\left(\frac{\log \sin x}{\log \cos x}\right)^{2}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ $\left[\because \log _{a} b=\frac{\log b}{\log a}\right]$
$y=\left(\frac{\log \sin x}{\log \cos x}\right)^{2}+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

Differentiate w.r.t x
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\log \sin x}{\log \cos x}\right)^{2}+\frac{d}{d x}\left(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right)$
$=\left[\frac{d\left(\frac{\log \sin x}{\log \cos x}\right)^{2}}{d\left(\frac{\log \sin x}{\log \cos x}\right)} \times \frac{d\left(\frac{\log \sin x}{\log \cos x}\right)}{d x}\right]+\frac{d \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)}{d\left(\frac{2 x}{1+x^{2}}\right)} \times \frac{d\left(\frac{2 x}{1+x^{2}}\right)}{d x}$ [Using chain rule]
$\frac{d y}{d x}=2\left(\frac{\log \sin x}{\log \cos x}\right) \times \frac{d}{d x}\left(\frac{\log \sin x}{\log \cos x}\right)+\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)$
$=\frac{2 \log \sin x}{\log \cos x} \times \frac{\log \cos x \cdot \frac{d(\log \sin x)}{d x}-\log \sin x \cdot \frac{d(\log \cos x)}{d x}}{(\log \cos x)^{2}}$$+\frac{1}{\sqrt{1-\frac{4 x^{2}}{\left(1+x^{2}\right)^{2}}}} \times \frac{\left(1+x^{2}\right) \frac{d(2 x)}{d x}-2 x \frac{d\left(1+x^{2}\right)}{d x}}{\left(1+x^{2}\right)^{2}}$
$=\frac{2 \log \sin x}{\log \cos x} \times \frac{\log \cos x \cdot \frac{d(\log (\sin x))}{d \sin x} \times \frac{d(\sin x)}{d x}-\log \sin x \cdot \frac{d(\log \cos x)}{d x}}{(\log \cos x)^{2}}$$+\frac{1}{\sqrt{\frac{1+x^{4}+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}}} \times \frac{\left(1+x^{2}\right) \times 2-(2 x)(2 x)}{\left(1+x^{2}\right)^{2}}$
$\frac{d y}{d x}=2 \frac{\log \sin x}{\log \cos x} \times \frac{\log \cos x \times \frac{1}{\sin x} \cos x-\log \sin x \cdot \frac{d(\log \cos x)}{d \cos x} \times \frac{d \cos x}{d x}}{(\log (\cos x))^{2}}$$+\frac{1+x^{2}}{\sqrt{\left(1-x^{2}\right)^{2}}} \times\left[\frac{2+2 x^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}\right]$
$\frac{d y}{d x}=\frac{2 \log \sin x}{\log \cos x} \times \frac{(\log \cos x) \cot x+(\log \sin x) \times \tan x}{(\log \cos x)^{2}}+\frac{1+x^{2}}{1-x^{2}} \times \frac{2-2 x^{2}}{\left(1+x^{2}\right)^{2}}$
$\frac{d y}{d x}=\frac{2 \log \sin x}{(\log \cos x)^{3}} \times \frac{[\cot x \times(\log \cos x)+\tan x \times(\log \sin x)]}{1}+\frac{2}{1+x^{2}}$
$\frac{d y}{d x}=\frac{2 \log \sin x}{(\log \cos x)^{3}} \times(\cot x \log \cos x+\tan x \log \sin x)+\frac{2}{1+x^{2}}$

Put $x=\frac{\pi}{4}$

$\frac{d y}{d x}=2\left(\frac{\log \left(\sin \frac{\pi}{4}\right)}{\left(\log \left(\cos \frac{\pi}{4}\right)\right)^{3}}\right) \times\left(\cot \frac{\pi}{4} \times \log \cos \frac{\pi}{4}+\tan \frac{\pi}{4} \cdot \log \sin \frac{\pi}{4}\right)+\frac{2}{\left[1+\left(\frac{\pi}{4}\right)^{2}\right]}$ $\left[\because \sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] ;\left[\tan \frac{\pi}{4}=\cot \frac{\pi}{4}=1\right]$
$\frac{d y}{d x}=2\left[\frac{\log \left(\frac{1}{\sqrt{2}}\right)}{\left(\log \left(\frac{1}{\sqrt{2}}\right)\right)^{3}}\right]\left[1 \times \log \frac{1}{\sqrt{2}}+1 \times \log \frac{1}{\sqrt{2}}\right]+\frac{2}{\left(1+\frac{\pi^{2}}{16}\right)}$$\frac{d y}{d x}=\frac{2}{\left(\log \frac{1}{\sqrt{2}}\right)^{2}}\left[2 \log \frac{1}{\sqrt{2}}\right]+\frac{32}{16+\pi^{2}}$
$=\frac{4}{\left(\log \frac{1}{\sqrt{2}}\right)}+\frac{32}{16+\pi^{2}}$
$=\frac{4}{\log (2)^{\frac{-1}{2}}}+\frac{32}{16+\pi^{2}}$
$=\frac{4}{\frac{-1}{2} \log 2}+\frac{32}{16+\pi^{2}}$
$=\frac{-8}{\log 2}+\frac{32}{16+\pi^{2}}$
$\therefore\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{4}}=8\left[\frac{4}{16+\pi^{2}}-\frac{1}{\log 2}\right]$

Differentiation exercise 10.4 question 31

Answer:
$\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$ At $x=\frac{\pi}{4}$
Hint:
Use chain rule
Given:
$\sqrt{y+x}+\sqrt{y-x}=c$
Solution:
$\sqrt{y+x}+\sqrt{y-x}=c$
Differentiate the given equation w.r.t x
$\frac{d}{d x}(\sqrt{y+x}+\sqrt{y-x})=\frac{d(c)}{d x}$
$\frac{d(\sqrt{y+x})}{d x}+\frac{d(\sqrt{y-x})}{d x}=0$ $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$\frac{d(\sqrt{y+x})}{d(y+x)} \times \frac{d(y+x)}{d x}+\frac{d(\sqrt{y-x})}{d(y-x)} \times \frac{d(y-x)}{d x}=0$ [Using chain rule]

$\frac{1}{2 \sqrt{y+x}} \times\left(\frac{d y}{d x}+\frac{d x}{d x}\right)+\frac{1}{2 \sqrt{y-x}} \times\left(\frac{d y}{d x}-\frac{d x}{d x}\right)=0$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$\frac{1}{2 \sqrt{y+x}}\left(\frac{d y}{d x}+1\right)+\frac{1}{2 \sqrt{y-x}}\left(\frac{d y}{d x}-1\right)=0$
$\frac{d y}{d x}\left(\frac{1}{2 \sqrt{y+x}}+\frac{1}{2 \sqrt{y-x}}\right)=\frac{1}{2 \sqrt{y-x}}-\frac{1}{2 \sqrt{y+x}}$
$\frac{d y}{d x}\left(\frac{\sqrt{y-x}+\sqrt{y+x}}{2 \sqrt{y+x} \cdot \sqrt{y-x}}\right)=\frac{\sqrt{y+x}-\sqrt{y-x}}{(2 \sqrt{y-x} \cdot \sqrt{y+x})}$
$\frac{d y}{d x} \times[\sqrt{y-x}+\sqrt{y+x}]=[\sqrt{y+x}-\sqrt{y-x}]$
$\frac{d y}{d x}=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}}$

Rationalizing the denominator

$\frac{d y}{d x}=\frac{(\sqrt{y+x}-\sqrt{y-x})(\sqrt{y+x}-\sqrt{y-x})}{(\sqrt{y+x}+\sqrt{y-x})(\sqrt{y+x}-\sqrt{y-x})}$
$=\frac{(\sqrt{y+x}-\sqrt{y-x})^{2}}{(\sqrt{y+x})^{2}-(\sqrt{y-x})^{2}}$ $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$
$\frac{d y}{d x}=\frac{(\sqrt{y+x})^{2}+(\sqrt{y-x})^{2}-2 \sqrt{y+x} \cdot \sqrt{y-x}}{(y+x)-(y-x)}$
$=\frac{y+x+y-x-2 \sqrt{y+x} \cdot \sqrt{y-x}}{y+x-y+x}$
$=\frac{2 y-2 \sqrt{y+x} \cdot \sqrt{y-x}}{2 x}$
$=\frac{2(y-\sqrt{y+x} \cdot \sqrt{y-x})}{2 x}$
$=\frac{y-\sqrt{y+x} \cdot \sqrt{y-x}}{x}$
$=\frac{y-\sqrt{y^{2}-x^{2}}}{x}$
$=\frac{y}{x}-\frac{\sqrt{y^{2}-x^{2}}}{\sqrt{x^{2}}}$
$\therefore \frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$
Thus proved

RD Sharma Class 12th Exercise 10.4 has a total of 31 questions including subparts, that are divided into two levels. Level 1 sums are less complex and can be solved using the fundamentals. In contrast, Level 2 sums are lengthy and require additional knowledge. Therefore, students should divide their work and study accordingly because they cannot cover the entire topic in one go.

The Level 1 questions in example 10.4 cover topics like finding dy/dx of quadratic and trigonometric equations, prove that sums, evaluation sums, etc. Level 2, On the other hand, contains questions mainly consisting of finding dy/dx at given parameters.

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