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RD Sharma Class 12 Exercise 10.7 Differentiation Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 10.7 Differentiation Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 10.7 Differentiation Solutions Maths - Download PDF Free Online

Q: What are the concepts covered in chapter 10 of the class 12 maths book?

The 10th chapter of the NCERT maths book, class 12, contains concepts under the topic Differentiation. The chapter includes differentiation of inverse trigonometric functions, differentiation of a function, differentiation by using trigonometric substitutions, logarithmic differentiation, etc.

Updated on 20 Jan 2022, 05:41 PM IST

School students may find it difficult to solve their board exam papers if they don't practice beforehand. This insufficient practice may cause students to panic and lose confidence before their exams. Therefore, high school students should use the RD Sharma class 12th exercise 10.7 for their exam preparations. This book will help them practice at home and clear their doubts well before the exam commences.
The RD Sharma class 12 chapter 10 exercise 10.7 is a trusted NCERT solution that has already helped hundreds of students in their exams. Be it school exams or boards, Chapter 10 of the Class 12 maths book is a crucial part of the syllabus, which needs to be understood well. The 10th chapter of the book is titled Differentiations. RD Sharma class 12th exercise 10.7 covers the concepts of differentiation of inverse trigonometric functions, Differentiation of a function, Differentiation by using trigonometric substitutions, Logarithmic differentiation, etc. Exercise 10.7 includes 29 questions on finding the value of variables in linear trigonometric equations.

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RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
Differentiation Excercise: 10.7
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: 10.7

Differentiation Exercise 10.7 Question 2

Answer:
$\frac{d y}{d x}=\tan \left(\frac{\theta}{2}\right)$
Hint:
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
Given:
$x=a(\theta+\sin x) \$ , $y=a(1-\cos \theta) \\$
Solution:
$x=a(\theta+\sin x) \$
$\ \frac{d x}{d \theta}=a(1+\cos \theta) \\$
$y=a(1-\cos \theta) \\$
$\frac{d y}{d \theta}=a(-\sin \theta)=a \sin \theta \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{dx}{d\theta}}=\frac{a \sin \theta}{a(1+\cos \theta)} \end{aligned}$
$\frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}$
$\begin{aligned} & \\ &=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \end{aligned}$ $\left[\begin{array}{l} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1+\cos 2 \theta=2 \cos ^{2} \theta \end{array}\right]$
$=\frac{\sin \left(\frac{\theta}{2}\right)}{\cos \left(\frac{\theta}{2}\right)}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$\frac{d y}{d x}=\tan \left(\frac{\theta}{2}\right)$

Differentiation Exercise 10.7 Question 3

Answer:
$\frac{d y}{d x}=\frac{-b}{a} \cot \theta$
Hint:
$\frac{d(\sin x)}{d x}=\cos x, \frac{d(\cos x)}{d x}=-\sin x$
Given:
$x=a \cos \theta$ , $y=b \sin \theta \\$
Solution:
$x=a \cos \theta$
$\\ \frac{d x}{d \theta}=-a \sin \theta \\$
$y=b \sin \theta \\$
$\begin{aligned} & &\frac{d y}{d \theta}=b \cos \theta \end{aligned}$
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{b \cos \theta}{-a \sin \theta}=\frac{-b}{a} \cot \theta$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

Differentiation Exercise 10.7 Question 4

Answer:
$\frac{dy}{dx}=\cot \theta$
Hint:
Use product rule
Given:
$x=a e^{\theta}(\sin \theta-\cos \theta)$
$\begin{aligned}&y=a e^{\theta}(\sin \theta+\cos \theta) \end{aligned}$
Solution:
$x=a e^{\theta}(\sin \theta-\cos \theta)$
$\begin{aligned} & \\ &\frac{d x}{d \theta}=a e^{\theta} \frac{d(\sin \theta-\cos \theta)}{d \theta}+(\sin \theta-\cos \theta) \frac{d\left(a e^{\theta}\right)}{d \theta} \end{aligned}$ [Use product rule]
$\begin{aligned} &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) a \cdot \frac{d e^{\theta}}{d \theta} \\ &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) \cdot a \cdot e^{\theta} \\ &\frac{d x}{d \theta}=a e^{\theta}(\cos \theta+\sin \theta+\sin \theta-\cos \theta) \end{aligned}$
$\begin{aligned} &\frac{d x}{d \theta}=2 a e^{\theta} \sin \theta \\ &y=a e^{\theta}(\sin \theta+\cos \theta) \end{aligned}$
$\frac{d y}{d \theta}=a e^{\theta} \cdot \frac{d(\sin \theta+\cos \theta)}{d \theta}+(\sin \theta+\cos \theta) \cdot \frac{d\left(a e^{\theta}\right)}{d \theta}$ [Using product rule]
$\begin{aligned} &=a e^{\theta}(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) a \cdot \frac{d e^{\theta}}{d \theta} \\ &=a e^{\theta}(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) a e^{\theta} \\ &=a e^{\theta}(\cos \theta-\sin \theta+\sin \theta+\cos \theta) \\ &\frac{d y}{d \theta}=2 a e^{\theta} \cos \theta \end{aligned}$
So,
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2 a e^{\theta} \cos \theta}{2 a e^{\theta} \sin \theta}=\frac{\cos \theta}{\sin \theta}=\cot \theta$

Differentiation Excercise 10.7 Question 5

Answer:
$\frac{dy}{dx}=-\frac{a}{b}$
Hint:
Use chain rule
Given:
$x=b \sin ^{2} \theta \\$ , $y=a \cos ^{2} \theta \\$
Solution:
$x=b \sin ^{2} \theta \\$
$\frac{d x}{d \theta}=b \frac{d\left(\sin ^{2} \theta\right)}{d \theta} \\$
$\frac{d x}{d \theta}=b \times\left[\frac{d \sin ^{2} \theta}{d \sin \theta} \times \frac{d \sin \theta}{d \theta}\right] \$ [Using chain rule]
$\ =b \times[2 \sin \theta \times \cos \theta] \\$
$\begin{aligned} &&\frac{d x}{d \theta}=2 b \sin \theta \cos \theta \end{aligned}$
$y=a \cos ^{2} \theta \\$
$\frac{d y}{d \theta}=a \cdot \frac{d\left(\cos ^{2} \theta\right)}{d \theta}$
$\begin{aligned} &\\ &=a \cdot \frac{d \cos ^{2} \theta}{d \cos \theta} \times \frac{d \cos \theta}{d \theta} \end{aligned}$ [Using chain rule]
$=a(2 \cos \theta)(-\sin \theta) \\$
$\frac{d y}{d \theta}=-2 a \sin \theta \cos \theta$
$\begin{aligned} & \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-2 a \sin \theta \cos \theta}{2 b \sin \theta \cos \theta}=-\frac{a}{b} \end{aligned}$

Differentiation Excercise 10.7 Question 6

Answer:
$\frac{dy}{dx}=1$ At $\theta =\frac{\pi}{2}$
Hint:
Use differentiation formula
Given:
$x=a(1-\cos \theta) \\$ , $\begin{aligned} &y=a(\theta+\sin \theta) \end{aligned}$
Solution:
$x=a(1-\cos \theta) \\$
$\frac{d x}{d \theta}=a \frac{d(1-\cos \theta)}{d \theta} \\$
$\begin{aligned} & &=a\left[0-\frac{d \cos \theta}{d \theta}\right] \end{aligned}$ $\left(\frac{d(\operatorname{CONSTANT})}{d \theta}=0\right)$
$=a(-(-\sin \theta)) \\$
$\frac{d x}{d \theta}=a \sin \theta \\$
$\begin{aligned} &y=a(\theta+\sin \theta) \end{aligned}$
$\frac{d y}{d \theta}=a\left(\frac{d \theta}{d \theta}+\frac{d \sin \theta}{d \theta}\right) \\$
$\begin{aligned} & &=a(1+\cos \theta) \end{aligned}$
$\frac{d x}{d \theta}$ At $\quad \theta=\frac{\pi}{2} \\$
$=a\left(1+\cos \frac{\pi}{2}\right) \\$
$=a(1+0) \\$
$\begin{aligned} &=a \end{aligned}$
$\frac{d x}{d \theta}$ At $\quad \theta=\frac{\pi}{2} \\$
$=a \sin \frac{\pi}{2} \\$
$\begin{aligned} &=a \end{aligned}$
$\frac{d y}{d \theta}$ At $\quad \theta=\frac{\pi}{2} \$
$\begin{aligned} &\ &\frac{\left(\frac{d y}{d \theta}\right) a t \theta=\frac{\pi}{2}}{\left(\frac{d x}{d \theta}\right) \operatorname{at} \theta=\frac{\pi}{2}}=\frac{a}{a}=1 \end{aligned}$

Differentiation Excercise 10.7 Question 7

Answer:
$\frac{dy}{dx}=\frac{x}{y}$
Hint:
Use $\frac{d\left(e^{x}\right)}{d x}=e^{x}$
Given:
$x=\frac{e^{t}+e^{-t}}{2} \\$ , $y=\frac{e^{t}-e^{-t}}{2} \$
Solution:
$x=\frac{e^{t}+e^{-t}}{2} \\$
$\begin{aligned} & &\frac{d x}{d y}=\frac{1}{2} \frac{d\left(e^{t}+e^{-t}\right)}{d t} \end{aligned}$
$=\frac{1}{2}\left(e^{t}-e^{-t}\right) \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{1}{2} \frac{d\left(e^{t}-e^{-t}\right)}{d t} \end{aligned}$
$=\frac{1}{2}\left(e^{t}+e^{-t}\right) \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{1}{2}\left(e^{t}+e^{-t}\right)}{\frac{1}{2}\left(e^{t}-e^{-t}\right)}=\frac{e^{t}+e^{-t}}{e^{t}-e^{-t}} \end{aligned}$
Now
$x=\frac{e^{t}+e^{-t}}{2} \\$
$2 x=e^{t}+e^{-t} \\$
$\begin{aligned} & &e^{t}+e^{-t}=2 x \end{aligned}$
Also
$y=\frac{e^{t}-e^{-t}}{2} \$
$\begin{aligned} &\ &e^{t}-e^{-t}=2 y \end{aligned}$
Put these values of $\left(e^{t}+e^{-t}\right) \text { and }\left(e^{t}-e^{-t}\right)$ in $\frac{dy}{dx}$ expression
$\frac{d y}{d x}==\frac{e^{t}+e^{-t}}{e^{t}-e^{-t}}=\frac{2 x}{2 y}=\frac{x}{y}$

Differentiation Exercise 10.7 question 8

Answer:
$\frac{d y}{d x}=\frac{2 t}{1-t^{2}}$
Hint:
Use quotient rule
Given:
$x=\frac{3 a t}{1+t^{2}} \\$ , $y=\frac{3 a t^{2}}{1+t^{2}} \\$
Solution:
$x=\frac{3 a t}{1+t^{2}} \\$
$\frac{d x}{d y}=\frac{d}{d t}\left(\frac{3 a t}{1+t^{2}}\right) \\$
$\begin{aligned} & &=\frac{\left(1+t^{2}\right) \frac{d(3 a t)}{d t}-3 a t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}$ [Using quotient rule]
$=\frac{\left(1+t^{2}\right)(3 a)-(3 a t)(2 t)}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} & &=\frac{3 a\left(1+t^{2}\right)-6 a t^{2}}{\left(1+t^{2}\right)^{2}} \end{aligned}$
$=\frac{3 a+3 a t^{2}-6 a t^{2}}{\left(1+t^{2}\right)^{2}} \\$
$=\frac{3 a-3 a t^{2}}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \end{aligned}$ (1)
$y=\frac{3 a t^{2}}{1+t^{2}} \\$
$\frac{d y}{d t}=\frac{d}{d t}\left(\frac{3 a t^{2}}{1+t^{2}}\right)$
$\begin{aligned} &\\ &=\frac{(1+t) \frac{d\left(3 a t^{2}\right)}{d t}-3 a t^{2} \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}$ [Using quotient rule]
$=\frac{\left(1+t^{2}\right)\left(3 a \times \frac{d t^{2}}{d t}\right)-3 a t^{2}\left(\frac{d(1)}{d t}+\frac{d\left(t^{2}\right)}{d t}\right)}{\left(1+t^{2}\right)^{2}}$

$=\frac{(1+t) \cdot 3 a \cdot(2 t)-3 a t^{2}(0+2 t)}{\left(1+t^{2}\right)}$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$=\frac{6 a t\left(1+t^{2}\right)-6 a t^{3}}{\left(1+t^{2}\right)^{2}} \\$
$=\frac{6 a t+6 a t^{3}-6 a t^{3}}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{6 a t}{(1+t)^{2}} \end{aligned}$ (2)
Now
$\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d t}}$
Put the value of $\frac{dy}{dt}\:and\: \frac{dx}{dt}$ from the equation (2) and (1)
In $\frac{dy}{dx}$
$\frac{d y}{d x}=\frac{\frac{6 a t}{\left(1+t^{2}\right)^{2}}}{\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}}$
$=\frac{6 a t}{3 a\left(1-t^{2}\right)}$
$\begin{aligned} & \\ &\frac{d y}{d x}=\frac{2 t}{1-t^{2}} \end{aligned}$

Differentiation Exercise 10.7 question 9

Answer:
$\frac{d y}{d x}=\tan \theta$
Hint:
Use product rule
Given:
$\begin{aligned} &x=a(\cos \theta+\theta \sin \theta) \\ &y=a(\sin \theta-\theta \cos \theta) \end{aligned}$
Solution:
$x=a(\cos \theta+\theta \sin \theta) \\$
$\frac{d x}{d \theta}=a \frac{d}{d \theta}(\cos \theta+\theta \sin \theta)$
$\begin{aligned} &\\ &=a\left[\frac{d \cos \theta}{d \theta}+\frac{d}{d \theta}(\theta \sin \theta)\right] \end{aligned}$
$=a\left[-\sin \theta+\left(\theta \frac{d \sin \theta}{d \theta}+\sin \theta \cdot \frac{d \theta}{d \theta}\right)\right]$ [Using product rule]
$=a[-\sin \theta+(\theta \cdot \cos \theta+\sin \theta)]$ $\left[\because \frac{d \sin \theta}{d \theta}=\cos \theta\right]$
$=a(\theta \cos \theta) \\$
$\begin{aligned} & &\frac{d x}{d \theta}=a \theta \cos \theta \end{aligned}$ (1)
$y=a(\sin \theta-\theta \cos \theta) \\$
$\frac{d y}{d \theta}=a \frac{d}{d \theta}(\sin \theta-\theta \cos \theta) \\$
$=a\left[\frac{d \sin \theta}{d \theta}-\frac{d(\theta \cos \theta)}{d \theta}\right] \\$
$\begin{aligned} & &=a\left[\cos \theta-\left(\theta \cdot \frac{d \cos \theta}{d \theta}+\cos \theta \cdot \frac{d \theta}{d \theta}\right)\right] \end{aligned}$ [Using product rule]
$=a[\cos \theta-(\theta(-\sin \theta)+\cos \theta)] \\$ $\left[\begin{array}{l} \because \frac{d \cos \theta}{d \theta}=-\sin \theta \\ \frac{d x}{d x}=1 \end{array}\right]$
$=a[\cos \theta+\theta \sin \theta-\cos \theta] \\$
$\begin{aligned} & &\frac{d y}{d \theta}=a \theta \sin \theta \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Put the value of $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$ from equations (2) and (1) respectively
$\frac{d y}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta \\$
$\begin{aligned} & &\frac{d y}{d x}=\tan \theta \end{aligned}$

Differentiation Exercise 10.7 question 10

Answer:
$\frac{d y}{d x}=e^{2 \theta}\left[\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{3}+\theta^{2}-1}\right]$
Hint:
Use product rule
Given:
$\begin{aligned} &x=e^{\theta}\left(\theta+\frac{1}{\theta}\right) \\ &y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right) \end{aligned}$
Solution:
$x=e^{\theta}\left(\theta+\frac{1}{\theta}\right) \\$
$\frac{d x}{d \theta}=\frac{d}{d \theta}\left[e^{\theta}\left(\theta+\frac{1}{\theta}\right)\right] \\$
$\begin{aligned} & &=e^{\theta} \frac{d\left(\theta+\frac{1}{\theta}\right)}{d \theta}+\left(\theta+\frac{1}{\theta}\right) \cdot \frac{d\left(e^{\theta}\right)}{d \theta} \end{aligned}$ [Using product rule]
$=e^{\theta}\left[\frac{d \theta}{d \theta}+\frac{d}{d \theta}\left(\frac{1}{\theta}\right)\right]+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta}$ $\left[\because \frac{d\left(e^{\theta}\right)}{d \theta}=e^{\theta}\right]$
$=e^{\theta}\left[1+\left(-\frac{1}{\theta^{2}}\right)\right]+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta}$ $\left[\because \frac{d\left(\frac{1}{x}\right)}{d x}=\frac{-1}{x^{2}}\right]$
$=e^{\theta}\left[1-\frac{1}{\theta^{2}}\right]+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta} \\$
$=e^{\theta}\left[\left(1-\frac{1}{\theta^{2}}\right)+\left(\theta+\frac{1}{\theta}\right)\right] \\$
$=e^{\theta}\left[\frac{\theta^{2}-1-\theta^{3}+\theta}{\theta^{2}}\right]$
$\begin{aligned} &\\ &\frac{d y}{d \theta}=e^{\theta}\left(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}}\right) \end{aligned}$ (1)
$y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right) \\$
$\frac{d y}{d \theta}=\frac{d}{d \theta}\left(e^{-\theta} \cdot\left(\theta-\frac{1}{\theta}\right)\right) \\$
$\begin{aligned} & &=e^{-\theta}\left(\frac{d \theta}{d \theta}-\frac{d\left(\frac{1}{\theta}\right)}{d \theta}\right]+\left(\theta-\frac{1}{\theta}\right)\left(-e^{-\theta}\right) \end{aligned}$
$=e^{-\theta}\left[1-\left(-\frac{1}{\theta^{2}}\right)\right]-e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$
$=e^{-\theta}\left[1+\frac{1}{\theta^{2}}\right]-e^{-\theta}\left[\theta-\frac{1}{\theta}\right] \\$
$\begin{aligned} & &=e^{-\theta}\left[1+\frac{1}{\theta^{2}}-\theta+\frac{1}{\theta}\right] \end{aligned}$
$=e^{-\theta}\left[\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right]$ $\left[\because L C M\left(\theta, \theta^{2}\right)=\theta^{2}\right]$
$=e^{-\theta}\left[\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right]$
$\frac{d y}{d \theta}=e^{-\theta}\left[\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right]$ (2)
Put the values of $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$ from equation (2) and (1)
$\frac{d y}{d x}=\frac{e^{\theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right)}{e^{-\theta}\left(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}}\right)} \\$
$\begin{aligned} & &=e^{\theta} \cdot e^{\theta} \frac{\left(-\theta^{3}+\theta^{2}+\theta+1\right)}{\left(\theta^{3}+\theta^{2}+\theta-1\right)} \end{aligned}$
$=e^{2 \theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right) \\$
$\begin{aligned} & &\frac{d y}{d x}=e^{2 \theta}\left[\frac{\theta^{2}-\theta^{3}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right] \end{aligned}$

Differentiation Exercise 10.7 Question 11

Answer:
$\frac{d y}{d x}=-\frac{x}{y}$
Hint:
Use quotient rule
Given:
$\begin{aligned} x &=\frac{2 t}{1+t^{2}} \\ y &=\frac{1-t^{2}}{1+t^{2}} \end{aligned}$
Solution:
$x=\frac{2 t}{1+t^{2}}$
$\begin{aligned} & \\ &\frac{d x}{d t}=\frac{d\left(\frac{2 t}{1+t^{2}}\right)}{d t}=\frac{\left(1+t^{2}\right) \frac{d(2 t)}{d t}-2 t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}$ [Using quotient rule]
$\frac{d x}{d t}=\frac{\left(1+t^{2}\right)^{2} \times 2-2 t(2 t)}{\left(1+t^{2}\right)^{2}} \\$
$=\frac{2\left(1+t^{2}\right)-4 t^{2}}{\left(1+t^{2}\right)^{2}} \\$
$=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}} \end{aligned}$ (1)
$y=\frac{1-t^{2}}{1+t^{2}} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1-t^{2}}{1+t^{2}}\right) \end{aligned}$
$=\frac{\left(1+t^{2}\right) \cdot \frac{d\left(1-t^{2}\right)}{d t}-\left(1-t^{2}\right) \cdot \frac{d\left(1+t^{2}\right)}{1+t}}{\left(1+t^{2}\right)}$ [Use quotient rule]
$=\frac{\left(1+t^{2}\right)(-2 t)-\left(1-t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}} \\$
$\frac{d y}{d x}=\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} &\frac{d y}{d t}=\frac{-4 t}{\left(1+t^{2}\right)^{2}} \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Put the values of $\frac{d y}{d t} \text { and } \frac{d y}{d t}$ from the equation (2) and (1) respectively
$\frac{d y}{d x}=\frac{\frac{-4 t}{\left(1+t^{2}\right)^{2}}}{\frac{\left(2-2 t^{2}\right)}{\left(1+t^{2}\right)^{2}}}$
$=\frac{-4 t}{2\left(1-t^{2}\right)}$
$\begin{aligned} &\\ &=\frac{-2 t}{1-t^{2}} \end{aligned}$
$\frac{d y}{d x}=-\frac{x}{y}$ $\left[\frac{x}{y}=\frac{\frac{2 t}{1+t^{2}}}{\frac{1-t^{2}}{1+t^{2}}}=\frac{2 t}{1-t^{2}}\right]$

Differentiation Exercise 10.7 Question 12

Answer:
$\frac{d y}{d x}=-1$
Hint:
Use chain rule
Given:
$\begin{aligned} &x=\cos ^{-1} \frac{1}{\sqrt{1+t^{2}}} \\ &y=\sin ^{-1} \frac{1}{\sqrt{1+t^{2}}} \\ &t \in R \end{aligned}$
Solution:
$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right) \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{d \cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)}{d t} \end{aligned}$
$=\frac{d \cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)}{d\left(\frac{1}{\sqrt{1+t^{2}}}\right)} \times \frac{d\left(\frac{1}{\sqrt{1+t^{2}}}\right)}{d\left(1+t^{2}\right)} \times \frac{d\left(1+t^{2}\right)}{d t}$ [Using chain rule]
$=\frac{-1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}} \times \frac{-1}{2\left(1+t^{2}\right)^{\frac{3}{2}}} \times 2 t \\$
$\begin{aligned} & &=\frac{1}{\sqrt{\frac{1+t^{2}-1}{1+t^{2}}}} \times \frac{1}{2 \sqrt{1+t^{2}} \times\left(1+t^{2}\right)} \times 2 t \end{aligned}$
$=\frac{\sqrt{1+t^{2}}}{t} \times \frac{1}{\sqrt{1+t^{2}}\left(1+t^{2}\right)} \times(t) \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{1}{1+t^{2}} \end{aligned}$ (1)
Now
$y=\sin ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right) \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{d \sin ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)}{d t} \end{aligned}$
$=\frac{d \sin ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)}{d\left(\frac{1}{\sqrt{1+t^{2}}}\right)} \times \frac{d\left(\frac{1}{\sqrt{1+t^{2}}}\right)}{d\left(1+t^{2}\right)} \times \frac{d\left(1+t^{2}\right)}{d t}$ [Using chain rule]
$\frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}} \times \frac{-1}{2\left(1+t^{2}\right)^{\frac{3}{2}}} \times 2 t$
$=\frac{1}{\sqrt{1-\frac{1}{1+t^{2}}}} \times \frac{-1}{2 \sqrt{1+t^{2}} \cdot\left(1+t^{2}\right)} \times 2 t \\$
$\begin{aligned} & &=\frac{\sqrt{1+t^{2}}}{\sqrt{1+t^{2}-1}} \times \frac{-1}{2 \sqrt{1+t^{2}} \cdot\left(1+t^{2}\right)} \times 2 t \end{aligned}$
$=\frac{-1}{t\left(1+t^{2}\right)} \times t \$
$\begin{aligned} &\ &\frac{d y}{d t}=\frac{-1}{\left(1+t^{2}\right)} \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
So put the values of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from the equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{\frac{-1}{\left(1+t^{2}\right)}}{\frac{1}{\left(1+t^{2}\right)}}=-1 \\$
$\begin{aligned} & &\frac{d y}{d x}=-1 \end{aligned}$

Differentiation Exercise 10.7 Question 13

Answer:
$\frac{d y}{d x}=\frac{t^{2}-1}{2 t}$
Hint:
Use quotient rule
Given:
$\begin{aligned} x &=\frac{1-t^{2}}{1+t^{2}} \\\\ \end{aligned}$
$y =\frac{2 t}{1+t^{2}}$
Solution:
$x=\frac{1-t^{2}}{1+t^{2}} \\$
$\frac{d x}{d t}=\frac{d}{d t}\left(\frac{1-t^{2}}{1+t^{2}}\right) \\$
$\begin{aligned} & &=\frac{\left(1+t^{2}\right) \frac{d\left(1-t^{2}\right)}{d t}-\left(1-t^{2}\right) \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}$ [Using quotient rule]
$\frac{d x}{d t}=\frac{\left(1+t^{2}\right)(-2 t)-\left(1-t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}}$
$\begin{aligned} &\\ &=\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}} \end{aligned}$
$\frac{d x}{d t}=\frac{-4 t}{\left(1+t^{2}\right)^{2}}$ (1)
$y=\frac{2 t}{1+t^{2}} \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{\left(1+t^{2}\right) \frac{d(2 t)}{d t}-2 t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}$ [Using quotient rule]
$=\frac{\left(1+t^{2}\right)(2 t)-2 t(2 t)}{\left(1+t^{2}\right)^{2}} \\$
$\frac{d y}{d t}=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{2-2 t}{\left(1+t^{2}\right)^{2}} \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
So put the values of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{\left[\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}}\right]}{\left[\frac{-4 t}{\left(1+t^{2}\right)^{2}}\right]}=\frac{2\left(1-t^{2}\right)}{-4 t}=-\frac{\left(1-t^{2}\right)}{2 t}$
$\frac{d y}{d x}=\frac{t^{2}-1}{2 t} \$
$\begin{aligned} &\ &\frac{d y}{d x}=\frac{t^{2}-1}{2 t} \end{aligned}$

Differentiation Exercise 10.7 question 14

Answer:
$\frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right)$
Hint:
Use chain rule
Given:
$\begin{aligned} &x=2 \cos \theta-\cos 2 \theta \\ &y=2 \sin \theta-\sin 2 \theta \end{aligned}$
Solution:
$=2 \cos \theta-\cos 2 \theta \\$
$\frac{d x}{d \theta}=\frac{d(2 \cos \theta)}{d \theta}-\frac{d(\cos 2 \theta)}{d \theta} \\$
$\begin{aligned} &x &=2(-\sin \theta) \cdot \frac{d \cos 2 \theta}{d(2 \theta)} \times \frac{d(2 \theta)}{d \theta} \end{aligned}$ [Using chain rule]
$=-\sin \theta-(-\sin 2 \theta) \times 2 \\$
$=-2 \sin \theta+2 \sin 2 \theta \\$
$\begin{aligned} & &\frac{d x}{d \theta}=2 \sin 2 \theta-2 \sin \theta \end{aligned}$ (1)
$y=2 \sin \theta-\sin 2 \theta \\$
$\frac{d y}{d \theta}=\frac{d(2 \sin \theta)}{d \theta}-\frac{d(\sin 2 \theta)}{d \theta} \\$
$\begin{aligned} & &=2 \cos \theta-\left[\frac{d(\sin 2 \theta)}{d(2 \theta)} \times \frac{d(2 \theta)}{d \theta}\right] \end{aligned}$ [Using chain rule]
$=2 \cos \theta-[\cos 2 \theta \times 2] \\$
$\begin{aligned} & &\frac{d y}{d \theta}=2 \cos \theta-2 \cos 2 \theta \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
So, put the values of $\frac{d x}{d \theta} \text { and } \frac{d y}{d \theta}$ from the equation (1) and (2)
$\frac{d y}{d x}=\frac{2(\cos \theta-\cos 2 \theta)}{2(\sin 2 \theta-\sin \theta)}$
$\begin{aligned} &\\ &\frac{d y}{d x}=\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta} \end{aligned}$
$=\frac{-2 \sin \left(\frac{\theta+2 \theta}{2}\right) \cdot \sin \left(\frac{\theta-2 \theta}{2}\right)}{2 \cos \left(\frac{2 \theta+\theta}{2}\right) \cdot \sin \left(\frac{2 \theta-\theta}{2}\right)}$ $\left[\begin{array}{l} \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right) \\ \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right) \end{array}\right]$
$=\frac{-\sin \left(\frac{3 \theta}{2}\right) \cdot \sin \left(\frac{-\theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right) \cdot \sin \left(\frac{\theta}{2}\right)} \\$
$\begin{aligned} & &=\frac{-\sin \left(\frac{3 \theta}{2}\right) \cdot\left(-\sin \frac{\theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right) \cdot \sin \frac{\theta}{2}} \end{aligned}$
$=\frac{\sin \left(\frac{3 \theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right)} \\$
$\begin{aligned} & &\frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right) \end{aligned}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

Differentiation Exercise 10.7 question 15

Answer:
$\frac{d y}{d x}=-\frac{y \log x}{x \log y}$
Hint:
Use chain rule and properties of logarithm
Given:
$\begin{aligned} &x=e^{\cos 2 t} \\ &y=e^{\sin 2 t} \end{aligned}$
Solution:
$x=e^{\cos 2 t} \\$
$\frac{d x}{d t}=\frac{d\left(e^{\cos 2 t}\right)}{d t} \\$
$\begin{aligned} &=\frac{d\left(e^{\cos 2 t}\right)}{d(\cos 2 t)} \times \frac{d(\cos 2 t)}{d(2 t)} \times \frac{d(2 t)}{d t} \end{aligned}$ [Using chain rule]
$=e^{\cos 2 t} \times(-\sin 2 t) \times 2 \\$ $\left[\because \frac{d\left(e^{x}\right)}{d x}=e^{x} \frac{d(\cos \theta)}{d \theta}=-\sin \theta\right]$
$\begin{aligned} & &\frac{d x}{d t}=-2 \sin 2 t e^{\cos 2 t} \end{aligned}$ (1)
$y=e^{\sin 2 t} \\$
$\frac{d y}{d t}=\frac{d\left(e^{\sin 2 t}\right)}{d t} \\$
$\begin{aligned} & &=\frac{d\left(e^{\sin 2 t}\right)}{d(\sin 2 t)} \times \frac{d(\sin 2 t)}{d(2 t)} \times \frac{d(2 t)}{d t} \end{aligned}$ [Using chain rule]
$=e^{\sin 2 t} \times \cos 2 t \times 2 \\$ $\left[\because \frac{d e^{x}}{d x}=e^{x}, \frac{d \sin \theta}{d \theta}=\cos \theta\right]$
$\frac{d y}{d t}=2 \cos 2 t e^{\sin 2 t} \\$ (2)
$\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}$
Putting the value of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from the equation (1) and (2) respectively
$=\frac{2 \cos 2 t \cdot e^{\sin 2 t}}{-2 \sin 2 t \cdot e^{\sin 2 t}}$
$\begin{aligned} &\frac{d y}{d x}=-\frac{\cos 2 t \cdot e^{\sin 2 t}}{\sin 2 t \cdot e^{\cos 2 t}} \\ & \end{aligned}$ (3)
$x=e^{\cos 2 t}$
Take log on both sides
$\log x=\log \left(e^{\cos 2 t}\right) \\$
$\log x=\cos 2 t \cdot \log e \\$ $\left[\because \log a^{m}=m \log a\right]$
$\begin{aligned} & &\log x=\cos 2 t \end{aligned}$ $[\because \log e=1]$
Also
$y=e^{\sin 2 t}$
Take log on both side
$\begin{aligned} &\log y=\log \left(e^{\sin 2 t}\right) \\ &\log y=\sin 2 t \cdot \log e \\ &\log y=\sin 2 t \end{aligned}$
Put $e^{\cos 2 t}=x, \cos 2 t=\log x, e^{\sin 2 t}=y, \sin 2 t=\log y$ in equation (3)
$\frac{d y}{d x}=\frac{-y \log x}{x \log y}$
$\begin{aligned} & \\ &\frac{d y}{d x}=\frac{-y \log x}{x \log y} \end{aligned}$

Differentiation Exercise 10.7 question 16

Answer:
$\frac{d y}{d x}=\frac{1}{\sqrt{3}} \; \; \text { At }\; \; t=\frac{2 \pi}{3}$
Hint:
Use $[\tan (\pi-\theta)=-\tan \theta]$ and $\frac{d(\sin t)}{d t}=\cos t$
Given:
$\begin{aligned} &x=\cos t \\ &y=\sin t \end{aligned}$
Solution:
$x=\cos t \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{d \cos t}{d t}=-\sin t \end{aligned}$ (1)
$y=\sin t$
$\\ \frac{d y}{d t}=\frac{d \sin t}{d t} \\$
$\begin{aligned} & &\frac{d y}{d t}=\cos t \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Putting the value of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from equation (1) and (2) respectively
$=\frac{\cos t}{-\sin t}$
$\frac{d y}{d x}\; \; At\; \; \left(x=\frac{2 \pi}{3}\right)$
$\begin{aligned} &=-\cot \left(\frac{2 \pi}{3}\right) \\ &=-\cot \left(\pi-\frac{\pi}{3}\right) \\ &=-\left(-\cot \frac{\pi}{3}\right) \\ &=\cot \frac{\pi}{3} \end{aligned}$
$\frac{d y}{d x}\;\; At\; \; \left(x=\frac{2 \pi}{3}\right)=\frac{1}{\sqrt{3}}$
Hence proved

Differentiation Exercise 10.7 Question 17

Answer:
$\frac{d y}{d x}=\frac{x}{y}$
Hint:
Use $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Given:
$\begin{aligned} &x=a\left(t+\frac{1}{t}\right) \\ &y=a\left(t-\frac{1}{t}\right) \end{aligned}$
Solution:
$x=a\left(t+\frac{1}{t}\right) \\$
$\begin{aligned} & &\frac{d x}{d t}=a \frac{d\left(t+\frac{1}{t}\right)}{d t} \end{aligned}$
$\begin{aligned} &=a\left[\frac{d t}{d t}+\frac{d\left(\frac{1}{t}\right)}{d t}\right] \\ &=a\left[1+\frac{d(t)^{-1}}{d t}\right] \end{aligned}$
$=a\left[1+\left\{(-1) t^{-1-1}\right\}\right] \\$ $\left[\because \frac{d x^{n}}{d x}=n x^{n-1}\right]$
$\begin{aligned} & &\frac{d x}{d t}=a\left(1-t^{2}\right) \end{aligned}$ (1)
$y=a\left(t-\frac{1}{t}\right) \\$
$\frac{d y}{d t}=a \frac{d\left(t-\frac{1}{t}\right)}{d t} \\$
$\begin{aligned} & &=a\left[\frac{d t}{d t}-\frac{d\left(\frac{1}{t}\right)}{d t}\right] \end{aligned}$
$=a\left[1-\frac{d t^{-1}}{d t}\right] \\$
$=a\left[1-(-1) t^{-1-1}\right] \\$
$\begin{aligned} & &\frac{d y}{d t}=a\left(1+t^{-2}\right) \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Put the values of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from the equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{t^{2}+1}{t^{2}-1} \\$
$\begin{aligned} & &=\frac{t\left(t+\frac{1}{t}\right)}{t\left(t-\frac{1}{t}\right)} \end{aligned}$
$=\frac{\left(t+\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)} \\$
$\begin{aligned} & &=\frac{a\left(t+\frac{1}{t}\right)}{a\left(t-\frac{1}{t}\right)} \end{aligned}$ [Multiply and divide by a]
$=\frac{x}{y}$ $\left[\because x=a\left(t+\frac{1}{t}\right) \& y=a\left(t-\frac{1}{t}\right)\right]$
Hence proved

Differentiation Exercise 10.7 Question 18

Answer:
$\frac{d y}{d x}=1$
Hint:
Use chain rule, product rule and $\frac{d y}{d x}=\frac{\frac{d x}{d t}}{\frac{d x}{d t}}$
Given:
$x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right), y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right) ;-1<t<1$
Solution:
$x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right) \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{d \sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)}{d t} \end{aligned}$
$=\frac{d \sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)}{d\left(\frac{2 t}{1+t^{2}}\right)} \times \frac{d\left(\frac{2 t}{1+t^{2}}\right)}{d t}$ [Using chain rule]
$=\frac{1}{\sqrt{1-\left(\frac{2 t}{1+t^{2}}\right)^{2}}} \times \frac{\left(1+t^{2}\right) \frac{d(2 t)}{d t}-2 t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}}$ [Using quotient rule]
$=\frac{1+t^{2}}{\sqrt{\left(1+t^{2}\right)^{2}-4 t^{2}}} \times \frac{\left(1+t^{2}\right)(2)-(2 t)(2 t)}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} & &=\frac{2+2 t^{2}-4 t^{2}}{\sqrt{1+t^{4}+2 t^{2}-4 t^{2}} \times\left(1+t^{2}\right)} \end{aligned}$
$=\frac{2-2 t^{2}}{\sqrt{1+t^{4}-2 t^{2}} \times\left(1+t^{2}\right)} \\$
$\begin{aligned} & &=\frac{2\left(1-t^{2}\right)}{\sqrt{\left(1-t^{2}\right)^{2}}\left(1+t^{2}\right)} \end{aligned}$
$\frac{d x}{d t}=\frac{2}{1+t^{2}} \\$
$\begin{aligned} & &y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right) \end{aligned}$ (1)
$\frac{d y}{d t}=\frac{d \tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)}{d t} \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{d \tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)}{d\left(\frac{2 t}{1-t^{2}}\right)} \times \frac{d\left(\frac{2 t}{1-t^{2}}\right)}{d t} \end{aligned}$
$=\frac{1}{1+\left(\frac{2 t}{1-t}\right)} \times \frac{\left(1-t^{2}\right) \frac{d(2 t)}{d t}-(2 t) \frac{d\left(1-t^{2}\right)}{d t}}{\left(1-t^{2}\right)^{2}}$ [Using chain rule]
$=\frac{1}{1+\frac{4 t^{2}}{\left(1-t^{2}\right)^{2}}} \times \frac{\left(1-t^{2}\right) \times 2-(2 t)(-2 t)}{\left(1-t^{2}\right)^{2}} \\$
$\begin{aligned} & &=\frac{\left(1-t^{2}\right)^{2}}{\left(1-t^{2}\right)^{2}+4 t^{2}} \times \frac{2-2 t^{2}+4 t^{2}}{\left(1-t^{2}\right)^{2}} \end{aligned}$
$=\frac{2+2 t^{2}}{1+t^{4}-2 t^{2}+4 t^{2}} \\$
$\begin{aligned} & &=\frac{2(1+(-2))}{1+t^{4}+2 t^{2}} \end{aligned}$
$=\frac{2\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{2}{\left(1+t^{2}\right)} \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
So putting the value of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from the equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{\frac{2}{\left(1+t^{2}\right)}}{\frac{2}{\left(1+t^{2}\right)}} \\$
$=\frac{2\left(1+t^{2}\right)}{2\left(1+t^{2}\right)} \\$
$\begin{aligned} &\ &\frac{d y}{d x}=1 \end{aligned}$

Differentiation Exercise 10.7 Question 19

Answer:
$\frac{d y}{d x}=-\cot 3 t$
Hint:
Use chain rule, product rule and $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Given:
$\begin{aligned} &x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}} \\ &y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}} \end{aligned}$
Solution:
$x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}} \\$
$x=\sin ^{3} t(\cos 2 t)^{\frac{-1}{2}}\\$ $\left[\because \frac{1}{x^{n}}=(x)^{-n}\right]$
$\begin{aligned} & &\frac{d x}{d t}=(\cos 2 t)^{\frac{-1}{2}} \frac{d \sin ^{3} t}{d t}+\sin ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d t} \end{aligned}$ [Using product rule]
$=\left[(\cos 2 t)^{\frac{-1}{2}} \times \frac{d \sin ^{3} t}{d \sin t} \times \frac{d \sin t}{d t}\right]+\left[\sin ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d \cos 2 t} \times \frac{d \cos 2 t}{d t}\right]$ [Using chain rule]
$\frac{d x}{d t}=\left[\frac{1}{\sqrt{\cos 2 t}} \times 3 \sin ^{2} t\right]+\left[\sin ^{3} t \times\left(\frac{-1}{2}(\cos 2 t)^{\frac{-1}{2}-1}\right) \times(-2 \sin 2 t)\right]$
$\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t}{\sqrt{\cos 2 t}}+\frac{\sin ^{3} t \cdot \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t}{\sqrt{\cos 2 t}}+\frac{\sin ^{3} t \cdot \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$ (1)
$=\frac{3 \sin ^{2} t \cos t \cdot \cos 2 t+\sin ^{3} t \cdot \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} &=\frac{3 \sin ^{2} t \cos t\left(1-2 \sin ^{2} t\right)+\sin ^{3} t(2 \sin t \cdot \cos t)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$ $\left[\begin{array}{l} \because \cos 2 \theta=1-2 \sin ^{2} \theta \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]$
$=\frac{3 \sin ^{2} t \cos t-4 \sin ^{4} t \cos t+2 \sin ^{4} t \cdot \cos t}{(\cos 2 t)^{\frac{3}{2}}} \\$
$\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t-4 \sin ^{4} t \cos t}{(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} & &=\frac{\sin t \cos t\left(3 \sin t-4 \sin ^{3} t\right)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$
$=\frac{\sin t \cos t \cdot \sin 3 t}{(\cos 2 t)^{\frac{3}{2}}}$ $\left[\because \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta\right]$
Multiply and divide by 2
$\frac{d x}{d t}=\frac{2 \sin t \cos t \sin 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{\sin 2 t \cdot \sin 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$
$\frac{d x}{d t}=\frac{\sin 2 t \cdot \sin 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \\$ (2)
$\begin{aligned} & &y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}} \end{aligned}$
$y=\cos ^{3} t(\cos 2 t)^{\frac{-1}{2}}$ $\left[\because \frac{1}{x^{n}}=(x)^{-n}\right]$
$\begin{aligned} & \\ &y=\cos ^{3} t(\cos 2 t)^{\frac{-1}{2}} \end{aligned}$
$\frac{d y}{d x}=\cos ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d t}+(\cos 2 t)^{\frac{-1}{2}} \frac{d \cos ^{3} t}{d t}$ [Using product rule]
$=\left[\cos ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d \cos 2 t} \times \frac{d \cos 2 t}{d(2 t)} \times \frac{d(2 t)}{d t}\right]+\left[(\cos 2 t)^{\frac{-1}{2}} \times \frac{\cos ^{3} t}{d \cos t} \times \frac{d \cos t}{d t}\right]$ [Using chain rule]
$=\left[\cos ^{3} t \times\left(\frac{-1}{2}(\cos 2 t)^{\frac{-3}{2}}\right) \times(-\sin 2 t) \times 2\right]+\left[\frac{1}{\sqrt{\cos 2 t}} \times 3 \cos ^{2} t \times(-\sin t)\right]$
$=\frac{-3 \cos ^{2} t \sin t}{\sqrt{\cos 2 t}}+\frac{\cos ^{3} t \times \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} & &=\frac{-3 \cos ^{2} t \sin t(\cos 2 t)+\cos ^{3} t \times \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$
$=\frac{-3 \cos ^{2} t \cdot \sin t\left(2 \cos ^{2} t-1\right)+\cos ^{3} t \times 2 \sin t \times \cos t}{(\cos 2 t)^{\frac{3}{2}}}$ $\left[\begin{array}{l} \because \cos 2 \theta=2 \cos ^{2} \theta-1 \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]$
$\frac{d y}{d t}=\frac{-3 \cos ^{2} t\left(2 \sin t \cos ^{2} t\right)+3 \cos ^{2} t \sin t+2 \cos ^{4} t \sin t}{(\cos 2 t)^{\frac{3}{2}}}$
$=\frac{\left(-6 \cos ^{4} t \sin t+2 \cos ^{4} t \sin t\right)+3 \cos ^{2} t \sin t}{(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} & &=\frac{-4 \cos ^{4} t \sin t+3 \cos ^{2} t \sin t}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$
$=\frac{-\sin t \cos t\left(4 \cos ^{3} t-3 \cos t\right)}{(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} & &=\frac{-\sin t \cos t(\cos 3 t)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$ $\left[\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right]$
$=\frac{-2 \sin (-\cos t \cdot \cos 3 t)}{2(\cos 2 t)^{\frac{3}{2}}} \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{-\sin 2 t \cdot \cos 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \end{aligned}$ (3)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Put the values of $\frac{d y}{d t} \text { and } \frac{d x}{d t}$ from equation (2) and (1) respectively
$\frac{d y}{d x}=\frac{\frac{(-\sin 2 t \times \cos 3 t)}{2(\cos 2 t)^{\frac{3}{2}}}}{\frac{(\sin 2 t \times \sin 3 t)}{2(\cos 2 t)^{\frac{3}{2}}}}$
$=\frac{(-\sin 2 t \times \cos 3 t) \times 2(\cos 2 t)^{\frac{3}{2}}}{(\sin 2 t \times \cos 3 t) \times(2 \cos 2 t)^{\frac{3}{2}}} \\$
$=\frac{-\cos 3 t}{\sin 3 t} \\$
$\begin{aligned} & &\frac{d y}{d x}=-\cot 3 t \end{aligned}$

Differentiation Exercise 10.7 Question 20

Answer:
$\frac{d y}{d x}=\frac{(a)^{t+\frac{1}{t}} \cdot \log a}{a\left(t+\frac{1}{t}\right)^{a-1}}$
Hint:
Use chain rule and $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Given:
$\begin{aligned} &x=\left(t+\frac{1}{t}\right)^{a} \\ &y=(a)^{t+\frac{1}{t}} \end{aligned}$
Solution:
$x=\left(t+\frac{1}{t}\right)^{a} \\$
$\begin{aligned} & &\frac{d x}{d t}=\frac{d\left(t+\frac{1}{t}\right)}{d t} \end{aligned}$
$=\frac{d\left(t+\frac{1}{t}\right)^{a}}{d\left(t+\frac{1}{t}\right)} \times \frac{d\left(t+\frac{1}{t}\right)}{d t}$ [Using chain rule]
$=a\left(t+\frac{1}{t}\right)^{a-1} \times\left(\frac{d t}{d t}+\frac{d\left(\frac{1}{t}\right)}{d t}\right)$ $\left[\because \frac{d x^{n}}{d x}=n x^{n-1}\right]$
$=a\left(t+\frac{1}{t}\right)^{a-1} \times\left(1+\left(\frac{-1}{t^{2}}\right)\right) \\$
$\begin{aligned} & &\frac{d x}{d t}=a\left(t+\frac{1}{t}\right)^{a-1} \times\left(1-\frac{1}{t^{2}}\right) \end{aligned}$ (1)
$y=a^{t+\frac{1}{t}} \\$
$\frac{d y}{d x}=\frac{d\left(a^{1+\frac{1}{t}}\right)}{d t} \\$
$\begin{aligned} & &=\frac{d\left(a^{t+\frac{1}{t}}\right)}{d\left(t+\frac{1}{t}\right)} \times \frac{d\left(t+\frac{1}{t}\right)}{d t} \end{aligned}$
$=\left(a^{t+\frac{1}{t}}\right) \log a \times\left(\frac{d t}{d t}+\frac{d\left(\frac{1}{t}\right)}{d t}\right)$ $\quad\left[\because \frac{d\left(a^{x}\right)}{d x}=a^{x} \log a\right]$
$\frac{d y}{d t}=(a)^{t+\frac{1}{t}} \log a \times\left[1-\frac{1}{t^{2}}\right]$ (2)
$\begin{aligned} & \\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}$
Put the values of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{(a)^{r+\frac{1}{1}} \times \log a \times\left(1-\frac{1}{t^{2}}\right)}{a\left(t+\frac{1}{t}\right)^{a-1} \times\left(1-\frac{1}{t^{2}}\right)} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{(a)^{t+\frac{1}{t}} \log a}{a\left(t+\frac{1}{t}\right)^{a-1}} \end{aligned}$

Differentiation Exercise 10.7 Question 21

Answer:
$\frac{d y}{d x}=\frac{1+t^{2}}{2 a t}$
Hint:
Use quotient rule and $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Given:
$\begin{aligned} &x=a\left(\frac{1+t^{2}}{1-t^{2}}\right) \\ &y=\frac{2 t}{1-t^{2}} \end{aligned}$
Solution:
$x=a\left(\frac{1+t^{2}}{1-t^{2}}\right) \\$
$\frac{d x}{d t}=a \frac{d\left(\frac{1+t^{2}}{1-t^{2}}\right)}{d t} \$
$\begin{aligned} &\ &=a \times \frac{\left(1-t^{2}\right) \frac{d\left(1+t^{2}\right)}{d t}-\left(1+t^{2}\right) \frac{d\left(1-t^{2}\right)}{d t}}{\left(1-t^{2}\right)^{2}} \end{aligned}$ [using quotient rule]
$=a\left[\frac{\left(1-t^{2}\right) \cdot(2 t)-\left(1+t^{2}\right) \cdot(-2 t)}{\left(1-t^{2}\right)^{2}}\right] \\$
$\begin{aligned} & &=a\left[\frac{2 t-2 t^{3}+2 t+2 t^{3}}{\left(1-t^{2}\right)^{2}}\right] \end{aligned}$
$=a\left[\frac{4 t}{\left(1-t^{2}\right)^{2}}\right] \\$
$=\frac{4 a t}{\left(1-t^{2}\right)^{2}} \\$ (1)
$\begin{aligned} &y=\frac{2 t}{\left(1-t^{2}\right)} \end{aligned}$
$\frac{d y}{d t}=\frac{\left(1-t^{2}\right) \frac{d(2 t)}{d t}-2 t \cdot \frac{\left(1-t^{2}\right)}{d t}}{\left(1-t^{2}\right)^{2}}$ [Using quotient rule]
$=\frac{\left(1-t^{2}\right)(2)-2 t(-2 t)}{\left(1-t^{2}\right)^{2}} \\$
$\begin{aligned} &\\ &=\frac{2-2 t^{2}+4 t^{2}}{\left(1-t^{2}\right)^{2}} \end{aligned}$
$\frac{d y}{d x}=\frac{2+2 t^{2}}{\left(1-t^{2}\right)^{2}}$ (2)
$\begin{aligned} &\\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}$
Put the values of $\frac{d y}{d x} \text { and } \frac{d x}{d t}$ from equation (2) and (1) respectively
$\frac{d y}{d x}=\frac{\frac{\left(2+2 t^{2}\right)}{\left(1-t^{2}\right)^{2}}}{\frac{4 a t}{\left(1-t^{2}\right)^{2}}}=\frac{\left(2+2 t^{2}\right) \times\left(1-t^{2}\right)^{2}}{4 a t \times\left(1-t^{2}\right)^{2}} \\$
$=\frac{2\left(1+t^{2}\right)}{4 a t} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{1+t^{2}}{2 a t} \end{aligned}$

Differentiation Exercise 10.7 Question 22

Answer:
$\frac{d y}{d x}=\frac{6}{5} \cot \left(\frac{t}{2}\right)$
Hint:
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Given:
$\begin{aligned} &x=10(t-\sin t) \\ &y=12(1-\cos t) \end{aligned}$
Solution:
$x=10(t-\sin t) \\$
$\begin{aligned} & &\frac{d x}{d t}=10\left(\frac{d t}{d t}-\frac{d \sin t}{d t}\right) \end{aligned}$ $\left[\frac{d \sin t}{d t}=\cos t\right]$
$\frac{d x}{d t}=10(1-\cos t) \\$ (1)
$y=12(1-\cos t) \\$
$\frac{d y}{d t}=12\left(\frac{d(1)}{d t}-\frac{d \cos t}{d t}\right) \\$
$\begin{aligned} & &=12(0-(-\sin t)) \end{aligned}$ $\left[\because \frac{d \cos t}{t}=-\sin t\right]$
$\frac{d y}{d t}=12 \sin t \\$ (2)
$\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}$
Put the values of $\frac{d y}{d t} \text { and } \frac{d x}{d t}$ from equation (2) and (1) respectively
$\frac{d y}{d x}=\frac{12 \sin t}{10(1-\cos t)}=\frac{6 \sin t}{5(1-\cos t)} \\$
$\begin{aligned} & &=\frac{6}{5} \times \frac{2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \sin ^{2} \frac{t}{2}} \end{aligned}$
$=\frac{6}{5} \times \frac{2 \times \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \times \sin \frac{t}{2} \cdot \sin \frac{t}{2}} \\$
$\begin{aligned} & &=\frac{6}{5} \cot \frac{t}{2} \end{aligned}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

Differentiation Exercise 10.7 Question 23

Answer:
$\frac{d y}{d x}\; \: \mathrm{At}\; \; \left(\theta=\frac{\pi}{3}\right)=-\sqrt{3}$
Hint:
Use chain rule and $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Given:
$\begin{aligned} &x=a(\theta-\sin \theta) \\ &y=a(1+\cos \theta) \end{aligned}$
Solution:
$x=a(\theta-\sin \theta) \\$
$\frac{d x}{d \theta}=a\left[\frac{d \theta}{d \theta}-\frac{d \sin \theta}{d \theta}\right] \\$ $\left[\frac{d \sin \theta}{d \theta}=\cos \theta\right]$
$\frac{d x}{d \theta}=a(1-\cos \theta) \\$ (1)
$\begin{aligned} & &y=a(1+\cos \theta) \end{aligned}$
$\frac{d y}{d \theta}=a\left(\frac{d 1}{d \theta}+\frac{d \cos \theta}{d \theta}\right) \\$
$=a(0+(-\sin \theta)) \\$ $\left[\because \frac{d \cos \theta}{d \theta}=-\sin \theta\right]$
$\begin{aligned} & &\frac{d y}{d \theta}=-a \sin \theta \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Put the values of $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$ from the equation (2) and (1) respectively
$\frac{d y}{d x}=\frac{-a \sin \theta}{a(1-\cos \theta)} \\$
$\frac{d y}{d x}=\frac{-\sin \theta}{1-\cos \theta} \\$
$\begin{aligned} &\text { At } \theta=\frac{\pi}{3} \end{aligned}$
$\frac{d y}{d x}=-\left(\frac{\sin \frac{\pi}{3}}{1-\cos \frac{\pi}{3}}\right)$
$=-\left(\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}\right)$ $\left[\begin{array}{rl} \because \sin \frac{\pi}{3} & =\frac{\sqrt{3}}{2} \\\\ \cos \frac{\pi}{3} & =\frac{1}{2} \end{array}\right]$
$=-\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) \\$
$\begin{aligned} & &\frac{d y}{d x} a t \theta=\frac{\pi}{3}=-\sqrt{3} \end{aligned}$

Differentiation Excercise 10.7 Question 24

Answer:
$\frac{d y}{d x}\; {\mathrm{At}}\; \left(t=\frac{\pi}{4}\right)=\frac{b}{a}$
Hint:
Use product rule and $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Given:
$\begin{aligned} &x=a \sin 2 t(1+\cos 2 t) \\ &y=b \cos 2 t(1-\cos 2 t) \end{aligned}$
Solution:
$x=a \sin 2 t(1+\cos 2 t) \\$
$\begin{aligned} & &\frac{d x}{d t}=a\left[\sin 2 t \times \frac{d(1+\cos 2 t)}{d t}+(1+\cos 2 t) \frac{d(\sin 2 t)}{d t}\right] \end{aligned}$ [Using product rule]
$=a[\sin 2 t \times(0+(-2 \sin 2 t))+(1+\cos 2 t)(2 \cos 2 t)] \\$
$=a\left[-2 \sin ^{2} 2 t+2 \cos 2 t+2 \cos ^{2} 2 t\right] \\$
$=a\left[2 \cos 2 t+2 \cos ^{2} 2 t-2 \sin ^{2} 2 t\right] \\$
$=a\left[2 \cos 2 t+2\left(\cos ^{2} 2 t-\sin ^{2} 2 t\right)\right] \$
$\begin{aligned} &\ &=2 a[\cos 2 t+\cos 4 t] \end{aligned}$ $\left[\because \cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\right]$
$\frac{d x}{d t}=2 a(\cos t+\cos 4 t)$ (1)
$y=b \cos 2 t(1-\cos 2 t) \\$
$\begin{aligned} & &\frac{d y}{d t}=b\left[\cos 2 t \times \frac{d(1-\cos 2 t)}{d t}+(1-\cos 2 t) \frac{d \cos 2 t}{d t}\right] \end{aligned}$ [Using product rule]
$=b[\cos 2 t(0+2 \sin 2 t)]+(1-\cos 2 t)(-2 \sin 2 t)$ $\quad\left[\because \frac{d \cos \theta}{d \theta}=-\sin \theta\right]$
$=b[2 \sin 2 t \cos 2 t-2 \sin 2 t+2 \sin 2 t \cos 2 t] \\$
$=b[4 \sin 2 t \cos 2 t-2 \sin t 2 t] \\$
$=b[2(2 \sin 2 t \cos 2 t)-2 \sin 2 t] \\$
$\begin{aligned} & &=2 b[\sin 4 t-\sin 2 t] \end{aligned}$ $[\because \sin 2 \theta=2 \sin \theta \cos \theta]$
$\frac{d y}{d t}=2 b(\sin 4 t-\sin 2 t) \\$ (2)
$\begin{aligned} & &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}$
Put the values of $\frac{d y}{d t} \text { and } \frac{d x}{d t}$ from the equations (2) and (1) respectively
$\frac{d y}{d x}=\frac{2 b(\sin 4 t-\sin 2 t)}{2 a(\cos 2 t+\cos 4 t)} \\$
$\frac{d y}{d x}=\frac{b}{a} \times\left(\frac{\sin 4 t-\sin 2 t}{\cos 2 t+\cos 4 t}\right) \\$
$\begin{aligned} & &\text { At } t=\frac{\pi}{4} \end{aligned}$
$\frac{d y}{d x}=\frac{b}{a} \times \frac{\left(\sin \left(4 \times \frac{\pi}{4}\right)-\sin \left(2 \times \frac{\pi}{4}\right)\right)}{\cos \left(2 \times \frac{\pi}{4}\right)+\cos \left(4 \times \frac{\pi}{4}\right)} \\$
$=\frac{b}{a}\left(\frac{\sin \pi-\sin \frac{\pi}{2}}{\cos \frac{\pi}{2}+\cos \pi}\right) \\$
$\begin{aligned} & &\because \sin \frac{\pi}{2}=1, \cos \frac{\pi}{2}=0, \sin \pi=0, \cos \pi=-1 \end{aligned}$
$=\frac{b}{a}\left[\frac{0-1}{0+(-1)}\right] \\$
$=\frac{b}{a}\left(\frac{-1}{-1}\right) \\$
$\begin{aligned} & &\frac{d y}{d x}_{\mathrm{At}}\left(t=\frac{\pi}{4}\right)=\left(\frac{b}{a}\right) \end{aligned}$
Hence proved

Differentiation Excercise 10.7 Question 25

Answer:
$\frac{d y}{d x}\; {\mathrm{At}}\; \left(t=\frac{\pi}{4}\right)=1$
Hint:
Use product rule and $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Given:
$\begin{aligned} &x=\cos t\left(3-2 \cos ^{2} t\right) \\ &y=\sin t\left(3-2 \sin ^{2} t\right) \end{aligned}$
Solution:
$x=\cos t\left(3-2 \cos ^{2} t\right)$
$\frac{d x}{d t}=\cos t \cdot \frac{d\left(3-2 \cos ^{2} t\right)}{d t}+\left(3-2 \cos ^{2} t\right) \frac{d \cos t}{d t}$ [Using product rule]
$\frac{d x}{d t}=\cos t[0-4 \cos t(-\sin t)]+\left(3-2 \cos ^{2} t\right) \times(-\sin t)$ $\quad\left[\because \frac{d(\cos \theta)}{d \theta}=-\sin \theta\right]$
$\frac{d x}{d t}=4 \cos ^{2} t \sin t-3 \sin t+2 \cos ^{2} t \sin t \\$
$=6 \cos ^{2} t \sin t-3 \sin t \\$
$\begin{aligned} &\frac{d x}{d t}=3 \sin t\left(2 \cos ^{2} t-1\right) \end{aligned}$ (1)
$y=\sin t\left(3-2 \sin ^{2} t\right) \\$
$\begin{aligned} &\frac{d y}{d t}=\sin t \frac{\left(3-2 \sin ^{2} t\right)}{d t}+\left(3-2 \sin ^{2} t\right) \frac{d \sin t}{d t} \end{aligned}$ [Using product rule]
$=\sin t \times\left[\frac{d 3}{d t}-\frac{d\left(\sin ^{2} t\right)}{d t}\right]+\left(3-2 \sin ^{2} t\right) \cos t$ $\left[\because \frac{d \sin \theta}{d \theta}=\cos \theta\right]$
$=\sin t[0-4 \sin t \cos t]+3 \cos t-2 \sin ^{2} t \cos t \\$
$=-4 \sin ^{2} t \cos t+3 \cos t-2 \sin ^{2} t \cos t \\$
$=-6 \sin ^{2} t \cos t+3 \cos t \\$
$\begin{aligned} & &\therefore \frac{d y}{d t}=3 \cos t\left(-2 \sin ^{2} t+1\right) \end{aligned}$ (2)
$\because \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Put the values of $\frac{d y}{d t} \text { and } \frac{d x}{d t}$ from the equation (2) and (1) respectively
$\frac{d y}{d x}=\frac{3 \cos t\left(1-\sin ^{2} t\right)}{3 \sin t\left(2 \cos ^{2} t-1\right)} \\$
$\begin{aligned} &\frac{d y}{d x}=\cot t \frac{\left(1-2 \sin ^{2} t\right)}{\left(2 \cos ^{2} t-1\right)} \end{aligned}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$\frac{d y}{d x}=\cot t \frac{\left(1-2\left(1-\cos ^{2} t\right)\right)}{\left(2 \cos ^{2} t-1\right)}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$\frac{d y}{d x}=\cot t \times \frac{\left(1-2+2 \cos ^{2} t\right)}{\left(2 \cos ^{2} t-1\right)} \\$
$=\cot t \times \frac{\left(2 \cos ^{2} t-1\right)}{\left(2 \cos ^{2} t-1\right)} \\$
$=\cot t \\$
$\begin{aligned} & &\frac{d y}{d x}=\cot t \end{aligned}$
$\text { At } t=\frac{\pi}{4}, \frac{d y}{d x}=\cot \frac{\pi}{4}=1 \\$
$\begin{aligned} & &\frac{d y}{d x}\; {\mathrm{At}}\; \left(t=\frac{\pi}{4}\right)=1 \end{aligned}$

Differentiation Excercise 10.7 Question 26

Answer:
$\frac{d y}{d x}=t$
Hint:
Use quotient rule and $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Given:
$\begin{aligned} &x=\frac{1+\log t}{t^{2}} \\ &y=\frac{3+2 \log t}{t} \end{aligned}$
Solution:
$x=\frac{1+\log t}{t^{2}} \\$
$\frac{d x}{d t}=\frac{d\left(\frac{1+\log t}{t^{2}}\right)}{d t} \\$
$\begin{aligned} &=\frac{t^{2} \frac{d(1+\log t)}{d t}-(1+\log t) \frac{d t^{2}}{d t}}{\left(t^{2}\right)^{2}} \end{aligned}$ [Using quotient rule]
$\frac{d x}{d t}=\frac{t\left[\frac{d(1)}{d t}+\frac{d \log t}{d t}\right]-(1+\log t) \cdot 2 t}{t^{4}} \\$
$=\frac{t^{2}\left(0+\frac{1}{t}\right)-2 t-2 t \log t}{t^{4}} \\$
$=\frac{t-2 t-2 t \log t}{t^{4}} \\$
$\begin{aligned} & &=\frac{-t-2 t \log t}{t^{4}} \end{aligned}$
$\frac{d x}{d t}=\frac{-(1+2 \log t)}{t^{3}} \\$ (1)
$\begin{aligned} & &y=\frac{3+2 \log t}{t} \end{aligned}$
$\frac{d y}{d t}=\frac{d\left(\frac{3+2 \log t}{t}\right)}{d t} \\$
$\begin{aligned} &=\frac{t \cdot \frac{d(3+2 \log t)}{d t}-(3+2 \log t) \frac{d t}{d t}}{t^{2}} \end{aligned}$ [Using quotient rule]
$=\frac{t\left(\frac{2}{t}\right)-(3+2 \log t)}{t^{2}} \\$
$\frac{d y}{d t}=\frac{2-3-2 \log t}{t^{2}} \\$
$=\frac{-1-2 \log t}{t^{2}} \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{-(1+2 \log t)}{t^{2}} \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
So put $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{\frac{-(1+2 \log t)}{t^{2}}}{\frac{-(1+2 \log t)}{t^{3}}}=\frac{-(1+2 \log t) \times t^{3}}{-(1+2 \log t) \times t^{2}}$
$\\ =\frac{t^{3}}{t^{2}}=t \\$
$\begin{aligned} & &\frac{d y}{d x}=t \end{aligned}$

Differentiation Exercise 10.7 question 27

Answer:
$\frac{d y}{d x}\; {\mathrm{At}}\; \left(t=\frac{\pi}{3}\right)=\frac{-1}{\sqrt{3}}$
Hint:
Use $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Given:
$\begin{aligned} &x=3 \sin t-\sin 3 t \\ &y=3 \cos t-\cos 3 t \end{aligned}$
Solution:
$x=3 \sin t-\sin 3 t$
$\begin{aligned} &\\ &\frac{d x}{d t}=3 \frac{d(\sin t)}{d t}-\frac{d(\sin 3 t)}{d t} \end{aligned}$
$\frac{d x}{d t}=3 \cos t-3 \cos 3 t \\$ $\left[\because \frac{d \sin x}{d x}=\cos x\right]$
$\begin{aligned} & &\frac{d x}{d t}=3(\cos t-\cos 3 t) \end{aligned}$ (1)
$\therefore y=3 \cos t-\cos 3 t \\$
$\begin{aligned} & &\frac{d y}{d t}=\frac{d(3 \cos t)}{d t}-\frac{d(\cos 3 t)}{d t} \end{aligned}$ $\left[\because \frac{d \cos x}{d x}=-\sin x\right]$
$=-3 \sin t-(-3 \sin 3 t) \\$
$\frac{d y}{d t}=-3 \sin t+3 \sin 3 t \\$
$\frac{d y}{d t}=3(\sin 3 t-\sin t) \\$ (2)
$\begin{aligned} &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}$
Putting the value of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from the equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{3(\sin 3 t-\sin t)}{3(\cos t-\cos 3 t)} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{\sin 3 t-\sin t}{\cos t-\cos 3 t} \end{aligned}$
$\text { At } t=\frac{\pi}{3} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{\sin \left(3 \times \frac{\pi}{3}\right)-\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}-\cos \left(3 \times \frac{\pi}{3}\right)} \end{aligned}$
$=\frac{\sin \pi-\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}-\cos \pi} \\$
$\begin{aligned} & &=\frac{0-\frac{\sqrt{3}}{2}}{\frac{1}{2}-(-1)} \end{aligned}$

$=\frac{\frac{-\sqrt{3}}{2}}{\frac{3}{2}}=\frac{-\sqrt{3} \times 2}{3 \times 2}=\frac{-1}{\sqrt{3}}$
$\frac{d y}{d x} \text { At }\left(t=\frac{\pi}{3}\right)=\frac{-1}{\sqrt{3}}$

Differentiation Exercise 10.7 question 28

Answer:
$\frac{d y}{d x}=1$
Hint:
Use product rule, quotient rule and $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Given:
$\begin{aligned} &\sin x=\frac{2 t}{1+t^{2}} \\\\ &\tan y=\frac{2 t}{1-t^{2}} \end{aligned}$
Solution:
$\begin{aligned} &\sin x=\frac{2 t}{1+t^{2}} \\\\ &x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right) \end{aligned}$
Differentiate w.r.t t
$\frac{d x}{d t}=\frac{d \sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)}{d t} \\$
$\begin{aligned} & &=\frac{d \sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)}{d\left(\frac{2 t}{1+t^{2}}\right)} \times \frac{d\left(\frac{2 t}{1+t^{2}}\right)}{d t} \end{aligned}$ [Using chain rule]
$=\frac{1}{\sqrt{1-\left(\frac{2 t}{1+t^{2}}\right)^{2}}} \times \frac{\left(1+t^{2}\right) \frac{d(2 t)}{d t}-2 t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}}$ [Using quotient rule]
$\frac{d x}{d t}=\frac{\left(1+t^{2}\right)}{\sqrt{\left(1+t^{2}\right)^{2}-(2 t)^{2}}} \times \frac{\left(1+t^{2}\right)(2)-2 t(2 t)}{\left(1+t^{2}\right)^{2}} \\$
$=\frac{2+2 t^{2}-4 t^{2}}{\sqrt{1+t^{4}+2 t^{2}-4 t^{2}} \cdot\left(1+t^{2}\right)} \\$
$=\frac{2-2 t^{2}}{\sqrt{1+t^{4}-2 t^{2}} \cdot\left(1+t^{2}\right)} \\$
$\begin{aligned} & &=\frac{2\left(1-t^{2}\right)}{\left(1-t^{2}\right)\left(1+t^{2}\right)} \end{aligned}$
$\frac{d x}{d t}=\frac{2}{1+t^{2}} \\$ (1)
$\tan y=\frac{2 t}{1-t^{2}} \\$
$\begin{aligned} & &y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right) \end{aligned}$
Differentiate w.r.t $t$
$\frac{d y}{d t}=\frac{d\left(\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)\right)}{d t} \\$
$\begin{aligned} & &=\frac{d\left(\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)\right)}{d\left(\frac{2 t}{1-t^{2}}\right)} \times \frac{d\left(\frac{2 t}{1-t^{2}}\right)}{d t} \end{aligned}$ [Using chain rule]
$=\frac{1}{1+\left(\frac{2 t}{1-t^{2}}\right)^{2}} \times \frac{\left(1-t^{2}\right) \frac{d(2 t)}{d t}-2 t \frac{d\left(1-t^{2}\right)}{d t}}{\left(1-t^{2}\right)^{2}} \\$
$\begin{aligned} & &=\frac{1}{1+\frac{4 t^{2}}{\left(1-t^{2}\right)^{2}}} \times \frac{\left(1-t^{2}\right) \cdot 2-2 t(-2 t)}{\left(1-t^{2}\right)^{2}} \end{aligned}$
$=\frac{\left(1-t^{2}\right)^{2}}{\left(1-t^{2}\right)^{2}+4 t^{2}} \times \frac{2-2 t^{2}+4 t^{2}}{\left(1-t^{2}\right)^{2}} \\$
$\begin{aligned} & &=\frac{2+2 t^{2}}{1+t^{4}-2 t^{2}+4 t^{2}} \end{aligned}$
$\frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1+t^{4}+2 t^{2}\right)} \\$
$=\frac{2\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}} \\$
$\begin{aligned} &\frac{d y}{d t}=\frac{2}{\left(1+t^{2}\right)} \end{aligned}$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Put $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from the equation (1) and (2) respectively
$\frac{d y}{d x}=\frac{\left(\frac{2}{1+t^{2}}\right)}{\left(\frac{2}{1+t^{2}}\right)}=1 \\$
$\begin{aligned} & &\frac{d y}{d x}=1 \end{aligned}$

Differentiation Exercise 10.7 Question 29

Answer:
$\frac{d y}{d x} \text { At }\left(t=\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}}$
Hint:
Use $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
Given:
$\begin{aligned} &x=a(2 \theta-\sin 2 \theta) \\ &y=a(1-\cos 2 \theta) \end{aligned}$
Solution:
$x=a(2 \theta-\sin 2 \theta) \\$
$\frac{d x}{d \theta}=a \frac{d(2 \theta-\sin 2 \theta)}{d \theta} \\$
$\begin{aligned} & &=a \times\left[\frac{d(2 \theta)}{d \theta}-\frac{d(\sin 2 \theta)}{d \theta}\right] \end{aligned}$
$=a \times[2-2 \cos 2 \theta]$ $\quad\left[ \because \frac{d(\sin \theta)}{d \theta}=\cos \theta\right]$
$=a \times 2(1-\cos 2 \theta) \\$
$\begin{aligned} & &\frac{d x}{d \theta}=2 a(1-\cos 2 \theta) \end{aligned}$ (1)
$y=a(1-\cos 2 \theta) \\$
$\frac{d y}{d \theta}=a \frac{d(1-\cos 2 \theta)}{d \theta} \\$
$\begin{aligned} & &=a \times\left[\frac{d 1}{d \theta}-\frac{d \cos 2 \theta}{d \theta}\right] \end{aligned}$
$\begin{aligned} &=a[0+2 \sin 2 \theta] \\ & \end{aligned}$ $\left[\because \frac{d \cos \theta}{d \theta}=-\sin \theta\right]$
$\frac{d y}{d \theta}=2 a \sin 2 \theta$ (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$
So, put the value of $\frac{d y}{d \theta} \text { and } \frac{d x}{d \theta}$ from equation (2) and (1) respectively
$\frac{d y}{d x}=\frac{2 a \sin 2 \theta}{2 a(1-\cos 2 \theta)} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{\sin 2 \theta}{1-\cos 2 \theta} \end{aligned}$
$\text { At } \theta=\frac{\pi}{3} \\$
$\begin{aligned} & &\frac{d y}{d x}=\frac{\sin \left(2 \times \frac{\pi}{3}\right)}{1-\cos \left(2 \times \frac{\pi}{3}\right)} \end{aligned}$
$=\frac{\sin \frac{2 \pi}{3}}{1-\cos \frac{2 \pi}{3}} \\$
$=\frac{\sin \left(\pi-\frac{\pi}{3}\right)}{1-\cos \left(\pi-\frac{\pi}{3}\right)} \\$
$\begin{aligned} & =& \frac{\sin \frac{\pi}{3}}{1-\left(-\cos \frac{\pi}{3}\right)} \end{aligned}$ $\left[\begin{array}{c} \sin (\pi-\theta)=\sin \theta \\ \cos (\pi-\theta)=-\cos \theta \end{array}\right]$
$=\frac{\sin \frac{\pi}{3}}{1+\cos \frac{\pi}{3}} \\$
$\begin{aligned} & &=\frac{\frac{\sqrt{3}}{2}}{1+\frac{1}{2}} \end{aligned}$ $\left[\because \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}, \cos \frac{\pi}{3}=\frac{1}{2}\right]$
$=\frac{\frac{\sqrt{3}}{2}}{\frac{(2+1)}{2}} \\$
$=\frac{\sqrt{3} \times 2}{3 \times 2} \\$
$\begin{aligned} & &\frac{d y}{d x} \text { at }\left(\theta=\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \end{aligned}$

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