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

Edited By Satyajeet Kumar | Updated on Jan 20, 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.

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

## Differentiation Excercise: 10.4

Differentiation exercise 10.4 question 1

$\left(\frac{-y}{x}\right)$
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

$\frac{(x+y)^{2}}{y^{2}-2 x y-x^{2}}$
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
$\frac{d}{d x}\left(y^{3}-3 x y^{2}\right)=\frac{d}{d x}\left(x^{3}+3 x^{2} y\right)$
$\frac{d}{d x}\left(y^{3}\right)-\frac{d}{d x}\left(3 x y^{2}\right)=\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}\left(3 x^{2} y\right)$

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

### Differentiation exercise 10.4 question 3

$-\left(\frac{y}{x}\right)^{\frac{1}{3}}$
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

$\frac{4}{3}\left(\frac{1-4 x+3 y}{4 x-3 y+1}\right)$
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

$\frac{-b^{2} x}{a^{2} y}$
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

$\frac{d y}{d x}=\frac{y-x^{4}}{y^{4}-x}$
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

$\left(\frac{a y-x-y}{x+y-a x}\right)$
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

$\frac{y-4 x^{3}-4 x y^{2}}{4 y x^{2}+4 y^{3}-x}$
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:$\frac{y}{x}\left[\frac{x e^{(x-y)}-1}{y e^{(x-y)}-1}\right]$
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

$\left[\frac{\sin (x+y)-y \cos x y}{x \cos x y-\sin (x+y)}\right]$
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

$\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
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

$-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
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

$\frac{d y}{d x}+y^{2}=0$
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

$2 \frac{d y}{d x}+y^{3}=0$
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

$(1+x)^{2} \frac{d y}{d x}+1=0$
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

$\frac{d y}{d x}=\frac{x+y}{x-y}$
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

$\frac{d y}{d x}=\frac{y}{x}$
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

$\frac{d y}{d x}=\frac{x}{y}\left(\frac{1-\tan a}{1+\tan a}\right)$
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

$\frac{d y}{d x}=-\frac{y\left(x^{2} y+x+y\right)}{x\left(x y^{2}+x+y\right)}$
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

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin (a+y)-y \cos (a+y)}$
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

$\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$
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

$\frac{d y}{d x}=\frac{\sin y}{1-x \cos y}$
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

$\left(x^{2}+1\right) \frac{d y}{d x}+x y+1=0$
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

$\frac{d y}{d x}=\frac{2 x^{3}+y-x^{2} y \cos x y}{x\left(x^{2} \cos x y+1+2 x y\right)}$
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

$\frac{d y}{d x}=\frac{\sec ^{2}(x+y)+\sec ^{2}(x-y)}{\sec ^{2}(x-y)-\sec ^{2}(x+y)}$
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

$\frac{d y}{d x}=\frac{-e^{x}\left(e^{y}-1\right)}{e^{y}\left(e^{x}-1\right)}$
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

$\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}$
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

$\frac{d y}{d x}=\frac{\pi}{4}(\sqrt{2}+1)$
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

$\frac{d y}{d x}=8\left[\frac{4}{16+\pi^{2}}-\frac{1}{\log 2}\right]_{\mathrm{At}} x=\frac{\pi}{4}$
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

$\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$ 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.

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## RD Sharma Chapter-wise Solutions

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