RD Sharma Class 12 Exercise 10.7 Differentiation Solutions Maths

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

Updated on Jan 20, 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 (\frac{θ}{2})

Hint:

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

Given:
$x=a(\theta+\sin x) $ ,

y = a (1 - \cos θ)

Solution:
$x=a(\theta+\sin x) $

\frac{d x}{d θ} = a (1 + \cos θ)

y = a (1 - \cos θ)

\frac{d y}{d θ} = a (- \sin θ) = a \sin θ

\begin{aligned} \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{a \sin θ}{a (1 + \cos θ)} \end{aligned}

\frac{d y}{d x} = \frac{\sin θ}{1 + \cos θ}

\begin{aligned} = \frac{2 \sin \frac{θ}{2} \cdot \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} \end{aligned}

[\begin{array}{l} \sin 2 θ = 2 \sin θ \cos θ \\ 1 + \cos 2 θ = 2 \cos^{2} θ \end{array}]

= \frac{\sin (\frac{θ}{2})}{\cos (\frac{θ}{2})}

[∵ \tan θ = \frac{\sin θ}{\cos θ}]

\frac{d y}{d x} = \tan (\frac{θ}{2})

Differentiation Exercise 10.7 Question 3

Answer:

\frac{d y}{d x} = \frac{- b}{a} \cot θ

Hint:

\frac{d (\sin x)}{d x} = \cos x, \frac{d (\cos x)}{d x} = - \sin x

Given:

x = a \cos θ

y = b \sin θ

Solution:

x = a \cos θ

\frac{d x}{d θ} = - a \sin θ

y = b \sin θ

\begin{aligned} \frac{d y}{d θ} = b \cos θ \end{aligned}

\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{b \cos θ}{- a \sin θ} = \frac{- b}{a} \cot θ

[∵ \cot θ = \frac{\cos θ}{\sin θ}]

Differentiation Exercise 10.7 Question 4

Answer:

\frac{d y}{d x} = \cot θ

Hint:
Use product rule
Given:

x = a e^{θ} (\sin θ - \cos θ)

\begin{aligned} y = a e^{θ} (\sin θ + \cos θ) \end{aligned}

Solution:

x = a e^{θ} (\sin θ - \cos θ)

\begin{aligned} \frac{d x}{d θ} = a e^{θ} \frac{d (\sin θ - \cos θ)}{d θ} + (\sin θ - \cos θ) \frac{d (a e^{θ})}{d θ} \end{aligned}

[Use product rule]

\begin{aligned} \frac{d x}{d θ} = a e^{θ} (\cos θ + \sin θ) + (\sin θ - \cos θ) a \cdot \frac{d e^{θ}}{d θ} \\ \frac{d x}{d θ} = a e^{θ} (\cos θ + \sin θ) + (\sin θ - \cos θ) \cdot a \cdot e^{θ} \\ \frac{d x}{d θ} = a e^{θ} (\cos θ + \sin θ + \sin θ - \cos θ) \end{aligned}

\begin{aligned} \frac{d x}{d θ} = 2 a e^{θ} \sin θ \\ y = a e^{θ} (\sin θ + \cos θ) \end{aligned}

\frac{d y}{d θ} = a e^{θ} \cdot \frac{d (\sin θ + \cos θ)}{d θ} + (\sin θ + \cos θ) \cdot \frac{d (a e^{θ})}{d θ}

[Using product rule]

\begin{aligned} = a e^{θ} (\cos θ - \sin θ) + (\sin θ + \cos θ) a \cdot \frac{d e^{θ}}{d θ} \\ = a e^{θ} (\cos θ - \sin θ) + (\sin θ + \cos θ) a e^{θ} \\ = a e^{θ} (\cos θ - \sin θ + \sin θ + \cos θ) \\ \frac{d y}{d θ} = 2 a e^{θ} \cos θ \end{aligned}

So,

\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{2 a e^{θ} \cos θ}{2 a e^{θ} \sin θ} = \frac{\cos θ}{\sin θ} = \cot θ

Differentiation Excercise 10.7 Question 5

Answer:

\frac{d y}{d x} = - \frac{a}{b}

Hint:
Use chain rule
Given:

x = b \sin^{2} θ

y = a \cos^{2} θ

Solution:

x = b \sin^{2} θ

\frac{d x}{d θ} = b \frac{d (\sin^{2} θ)}{d θ}

$\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 θ \times \cos θ]

\begin{aligned} \frac{d x}{d θ} = 2 b \sin θ \cos θ \end{aligned}

y = a \cos^{2} θ

\frac{d y}{d θ} = a \cdot \frac{d (\cos^{2} θ)}{d θ}

\begin{aligned} = a \cdot \frac{d \cos^{2} θ}{d \cos θ} \times \frac{d \cos θ}{d θ} \end{aligned}

[Using chain rule]

= a (2 \cos θ) (- \sin θ)

\frac{d y}{d θ} = - 2 a \sin θ \cos θ

\begin{aligned} \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{- 2 a \sin θ \cos θ}{2 b \sin θ \cos θ} = - \frac{a}{b} \end{aligned}

Differentiation Excercise 10.7 Question 6

Answer:

\frac{d y}{d x} = 1

θ = \frac{π}{2}

Hint:
Use differentiation formula
Given:

x = a (1 - \cos θ)

\begin{aligned} y = a (θ + \sin θ) \end{aligned}

Solution:

x = a (1 - \cos θ)

\frac{d x}{d θ} = a \frac{d (1 - \cos θ)}{d θ}

\begin{aligned} = a [0 - \frac{d \cos θ}{d θ}] \end{aligned}

(\frac{d (CONSTANT)}{d θ} = 0)

= a (- (- \sin θ))

\frac{d x}{d θ} = a \sin θ

\begin{aligned} y = a (θ + \sin θ) \end{aligned}

\frac{d y}{d θ} = a (\frac{d θ}{d θ} + \frac{d \sin θ}{d θ})

\begin{aligned} = a (1 + \cos θ) \end{aligned}

\frac{d x}{d θ}

θ = \frac{π}{2}

= a (1 + \cos \frac{π}{2})

= a (1 + 0)

\begin{aligned} = a \end{aligned}

\frac{d x}{d θ}

θ = \frac{π}{2}

= a \sin \frac{π}{2}

\begin{aligned} = a \end{aligned}

\frac{d y}{d θ}

At $\quad \theta=\frac{\pi}{2} $

\begin{aligned} \frac{(\frac{d y}{d θ}) a t θ = \frac{π}{2}}{(\frac{d x}{d θ}) at θ = \frac{π}{2}} = \frac{a}{a} = 1 \end{aligned}

Differentiation Excercise 10.7 Question 7

Answer:

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

Hint:
Use

\frac{d (e^{x})}{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 (e^{t} + e^{- t})}{d t} \end{aligned}

= \frac{1}{2} (e^{t} - e^{- t})

\begin{aligned} \frac{d y}{d t} = \frac{1}{2} \frac{d (e^{t} - e^{- t})}{d t} \end{aligned}

= \frac{1}{2} (e^{t} + e^{- t})

\begin{aligned} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{1}{2} (e^{t} + e^{- t})}{\frac{1}{2} (e^{t} - e^{- t})} = \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

(e^{t} + e^{- t}) and (e^{t} - e^{- t})

\frac{d y}{d x}

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} (\frac{3 a t}{1 + t^{2}})

\begin{aligned} = \frac{(1 + t^{2}) \frac{d (3 a t)}{d t} - 3 a t \frac{d (1 + t^{2})}{d t}}{{(1 + t^{2})}^{2}} \end{aligned}

[Using quotient rule]

= \frac{(1 + t^{2}) (3 a) - (3 a t) (2 t)}{{(1 + t^{2})}^{2}}

\begin{aligned} = \frac{3 a (1 + t^{2}) - 6 a t^{2}}{{(1 + t^{2})}^{2}} \end{aligned}

= \frac{3 a + 3 a t^{2} - 6 a t^{2}}{{(1 + t^{2})}^{2}}

= \frac{3 a - 3 a t^{2}}{{(1 + t^{2})}^{2}}

\begin{aligned} \frac{d x}{d t} = \frac{3 a (1 - t^{2})}{{(1 + t^{2})}^{2}} \end{aligned}

(1)

y = \frac{3 a t^{2}}{1 + t^{2}}

\frac{d y}{d t} = \frac{d}{d t} (\frac{3 a t^{2}}{1 + t^{2}})

\begin{aligned} = \frac{(1 + t) \frac{d (3 a t^{2})}{d t} - 3 a t^{2} \frac{d (1 + t^{2})}{d t}}{{(1 + t^{2})}^{2}} \end{aligned}

[Using quotient rule]

= \frac{(1 + t^{2}) (3 a \times \frac{d t^{2}}{d t}) - 3 a t^{2} (\frac{d (1)}{d t} + \frac{d (t^{2})}{d t})}{{(1 + t^{2})}^{2}}

= \frac{(1 + t) \cdot 3 a \cdot (2 t) - 3 a t^{2} (0 + 2 t)}{(1 + t^{2})}

[∵ \frac{d (x^{n})}{d x} = n x^{n - 1}]

= \frac{6 a t (1 + t^{2}) - 6 a t^{3}}{{(1 + t^{2})}^{2}}

= \frac{6 a t + 6 a t^{3} - 6 a t^{3}}{{(1 + t^{2})}^{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{d y}{d t} a n d \frac{d x}{d t}

from the equation (2) and (1)
In

\frac{d y}{d x}

\frac{d y}{d x} = \frac{\frac{6 a t}{{(1 + t^{2})}^{2}}}{\frac{3 a (1 - t^{2})}{{(1 + t^{2})}^{2}}}

= \frac{6 a t}{3 a (1 - t^{2})}

\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 θ

Hint:
Use product rule
Given:

\begin{aligned} x = a (\cos θ + θ \sin θ) \\ y = a (\sin θ - θ \cos θ) \end{aligned}

Solution:

x = a (\cos θ + θ \sin θ)

\frac{d x}{d θ} = a \frac{d}{d θ} (\cos θ + θ \sin θ)

\begin{aligned} = a [\frac{d \cos θ}{d θ} + \frac{d}{d θ} (θ \sin θ)] \end{aligned}

= a [- \sin θ + (θ \frac{d \sin θ}{d θ} + \sin θ \cdot \frac{d θ}{d θ})]

[Using product rule]

= a [- \sin θ + (θ \cdot \cos θ + \sin θ)]

[∵ \frac{d \sin θ}{d θ} = \cos θ]

= a (θ \cos θ)

\begin{aligned} \frac{d x}{d θ} = a θ \cos θ \end{aligned}

(1)

y = a (\sin θ - θ \cos θ)

\frac{d y}{d θ} = a \frac{d}{d θ} (\sin θ - θ \cos θ)

= a [\frac{d \sin θ}{d θ} - \frac{d (θ \cos θ)}{d θ}]

\begin{aligned} = a [\cos θ - (θ \cdot \frac{d \cos θ}{d θ} + \cos θ \cdot \frac{d θ}{d θ})] \end{aligned}

[Using product rule]

= a [\cos θ - (θ (- \sin θ) + \cos θ)]

[\begin{array}{l} ∵ \frac{d \cos θ}{d θ} = - \sin θ \\ \frac{d x}{d x} = 1 \end{array}]

= a [\cos θ + θ \sin θ - \cos θ]

\begin{aligned} \frac{d y}{d θ} = a θ \sin θ \end{aligned}

(2)

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

Put the value of

\frac{d y}{d θ} and \frac{d x}{d θ}

from equations (2) and (1) respectively

\frac{d y}{d x} = \frac{a θ \sin θ}{a θ \cos θ} = \tan θ

\begin{aligned} \frac{d y}{d x} = \tan θ \end{aligned}

Differentiation Exercise 10.7 question 10

Answer:

\frac{d y}{d x} = e^{2 θ} [\frac{θ^{2} - θ^{3} + θ + 1}{θ^{3} + θ^{2} - 1}]

Hint:
Use product rule
Given:

\begin{aligned} x = e^{θ} (θ + \frac{1}{θ}) \\ y = e^{- θ} (θ - \frac{1}{θ}) \end{aligned}

Solution:

x = e^{θ} (θ + \frac{1}{θ})

\frac{d x}{d θ} = \frac{d}{d θ} [e^{θ} (θ + \frac{1}{θ})]

\begin{aligned} = e^{θ} \frac{d (θ + \frac{1}{θ})}{d θ} + (θ + \frac{1}{θ}) \cdot \frac{d (e^{θ})}{d θ} \end{aligned}

[Using product rule]

= e^{θ} [\frac{d θ}{d θ} + \frac{d}{d θ} (\frac{1}{θ})] + (θ + \frac{1}{θ}) \cdot e^{θ}

[∵ \frac{d (e^{θ})}{d θ} = e^{θ}]

= e^{θ} [1 + (- \frac{1}{θ^{2}})] + (θ + \frac{1}{θ}) \cdot e^{θ}

[∵ \frac{d (\frac{1}{x})}{d x} = \frac{- 1}{x^{2}}]

= e^{θ} [1 - \frac{1}{θ^{2}}] + (θ + \frac{1}{θ}) \cdot e^{θ}

= e^{θ} [(1 - \frac{1}{θ^{2}}) + (θ + \frac{1}{θ})]

= e^{θ} [\frac{θ^{2} - 1 - θ^{3} + θ}{θ^{2}}]

\begin{aligned} \frac{d y}{d θ} = e^{θ} (\frac{θ^{3} + θ^{2} + θ - 1}{θ^{2}}) \end{aligned}

(1)

y = e^{- θ} (θ - \frac{1}{θ})

\frac{d y}{d θ} = \frac{d}{d θ} (e^{- θ} \cdot (θ - \frac{1}{θ}))

\begin{aligned} = e^{- θ} (\frac{d θ}{d θ} - \frac{d (\frac{1}{θ})}{d θ}] + (θ - \frac{1}{θ}) (- e^{- θ}) \end{aligned}

= e^{- θ} [1 - (- \frac{1}{θ^{2}})] - e^{- θ} (θ - \frac{1}{θ})

= e^{- θ} [1 + \frac{1}{θ^{2}}] - e^{- θ} [θ - \frac{1}{θ}]

\begin{aligned} = e^{- θ} [1 + \frac{1}{θ^{2}} - θ + \frac{1}{θ}] \end{aligned}

= e^{- θ} [\frac{θ^{2} + 1 - θ^{3} + θ}{θ^{2}}]

[∵ L C M (θ, θ^{2}) = θ^{2}]

= e^{- θ} [\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{2}}]

\frac{d y}{d θ} = e^{- θ} [\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{2}}]

(2)
Put the values of

\frac{d y}{d θ} and \frac{d x}{d θ}

from equation (2) and (1)

\frac{d y}{d x} = \frac{e^{θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{2}})}{e^{- θ} (\frac{θ^{3} + θ^{2} + θ - 1}{θ^{2}})}

\begin{aligned} = e^{θ} \cdot e^{θ} \frac{(- θ^{3} + θ^{2} + θ + 1)}{(θ^{3} + θ^{2} + θ - 1)} \end{aligned}

= e^{2 θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{3} + θ^{2} + θ - 1})

\begin{aligned} \frac{d y}{d x} = e^{2 θ} [\frac{θ^{2} - θ^{3} + θ + 1}{θ^{3} + θ^{2} + θ - 1}] \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 (\frac{2 t}{1 + t^{2}})}{d t} = \frac{(1 + t^{2}) \frac{d (2 t)}{d t} - 2 t \frac{d (1 + t^{2})}{d t}}{{(1 + t^{2})}^{2}} \end{aligned}

[Using quotient rule]

\frac{d x}{d t} = \frac{{(1 + t^{2})}^{2} \times 2 - 2 t (2 t)}{{(1 + t^{2})}^{2}}

= \frac{2 (1 + t^{2}) - 4 t^{2}}{{(1 + t^{2})}^{2}}

= \frac{2 + 2 t^{2} - 4 t^{2}}{{(1 + t^{2})}^{2}}

\begin{aligned} \frac{d x}{d t} = \frac{2 - 2 t^{2}}{{(1 + t^{2})}^{2}} \end{aligned}

(1)

y = \frac{1 - t^{2}}{1 + t^{2}}

\begin{aligned} \frac{d y}{d x} = \frac{d}{d x} (\frac{1 - t^{2}}{1 + t^{2}}) \end{aligned}

= \frac{(1 + t^{2}) \cdot \frac{d (1 - t^{2})}{d t} - (1 - t^{2}) \cdot \frac{d (1 + t^{2})}{1 + t}}{(1 + t^{2})}

[Use quotient rule]

= \frac{(1 + t^{2}) (- 2 t) - (1 - t^{2}) (2 t)}{{(1 + t^{2})}^{2}}

\frac{d y}{d x} = \frac{- 2 t - 2 t^{3} - 2 t + 2 t^{3}}{{(1 + t^{2})}^{2}}

\begin{aligned} \frac{d y}{d t} = \frac{- 4 t}{{(1 + t^{2})}^{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} and \frac{d y}{d t}

from the equation (2) and (1) respectively

\frac{d y}{d x} = \frac{\frac{- 4 t}{{(1 + t^{2})}^{2}}}{\frac{(2 - 2 t^{2})}{{(1 + t^{2})}^{2}}}

= \frac{- 4 t}{2 (1 - t^{2})}

\begin{aligned} = \frac{- 2 t}{1 - t^{2}} \end{aligned}

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

[\frac{x}{y} = \frac{\frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}}} = \frac{2 t}{1 - t^{2}}]

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} (\frac{1}{\sqrt{1 + t^{2}}})

\begin{aligned} \frac{d x}{d t} = \frac{d \cos^{- 1} (\frac{1}{\sqrt{1 + t^{2}}})}{d t} \end{aligned}

= \frac{d \cos^{- 1} (\frac{1}{\sqrt{1 + t^{2}}})}{d (\frac{1}{\sqrt{1 + t^{2}}})} \times \frac{d (\frac{1}{\sqrt{1 + t^{2}}})}{d (1 + t^{2})} \times \frac{d (1 + t^{2})}{d t}

[Using chain rule]

= \frac{- 1}{\sqrt{1 - {(\frac{1}{\sqrt{1 + t^{2}}})}^{2}}} \times \frac{- 1}{2 {(1 + t^{2})}^{\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 (1 + t^{2})} \times 2 t \end{aligned}

= \frac{\sqrt{1 + t^{2}}}{t} \times \frac{1}{\sqrt{1 + t^{2}} (1 + t^{2})} \times (t)

\begin{aligned} \frac{d x}{d t} = \frac{1}{1 + t^{2}} \end{aligned}

(1)
Now

y = \sin^{- 1} (\frac{1}{\sqrt{1 + t^{2}}})

\begin{aligned} \frac{d y}{d t} = \frac{d \sin^{- 1} (\frac{1}{\sqrt{1 + t^{2}}})}{d t} \end{aligned}

= \frac{d \sin^{- 1} (\frac{1}{\sqrt{1 + t^{2}}})}{d (\frac{1}{\sqrt{1 + t^{2}}})} \times \frac{d (\frac{1}{\sqrt{1 + t^{2}}})}{d (1 + t^{2})} \times \frac{d (1 + t^{2})}{d t}

[Using chain rule]

\frac{d y}{d x} = \frac{1}{\sqrt{1 - {(\frac{1}{\sqrt{1 + t^{2}}})}^{2}}} \times \frac{- 1}{2 {(1 + t^{2})}^{\frac{3}{2}}} \times 2 t

= \frac{1}{\sqrt{1 - \frac{1}{1 + t^{2}}}} \times \frac{- 1}{2 \sqrt{1 + t^{2}} \cdot (1 + t^{2})} \times 2 t

\begin{aligned} = \frac{\sqrt{1 + t^{2}}}{\sqrt{1 + t^{2} - 1}} \times \frac{- 1}{2 \sqrt{1 + t^{2}} \cdot (1 + t^{2})} \times 2 t \end{aligned}

$=\frac{-1}{t\left(1+t^{2}\right)} \times t $

\begin{aligned} \frac{d y}{d t} = \frac{- 1}{(1 + t^{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} and \frac{d y}{d t}

from the equation (1) and (2) respectively

\frac{d y}{d x} = \frac{\frac{- 1}{(1 + t^{2})}}{\frac{1}{(1 + t^{2})}} = - 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} (\frac{1 - t^{2}}{1 + t^{2}})

\begin{aligned} = \frac{(1 + t^{2}) \frac{d (1 - t^{2})}{d t} - (1 - t^{2}) \frac{d (1 + t^{2})}{d t}}{{(1 + t^{2})}^{2}} \end{aligned}

[Using quotient rule]

\frac{d x}{d t} = \frac{(1 + t^{2}) (- 2 t) - (1 - t^{2}) (2 t)}{{(1 + t^{2})}^{2}}

\begin{aligned} = \frac{- 2 t - 2 t^{3} - 2 t + 2 t^{3}}{{(1 + t^{2})}^{2}} \end{aligned}

\frac{d x}{d t} = \frac{- 4 t}{{(1 + t^{2})}^{2}}

(1)

y = \frac{2 t}{1 + t^{2}}

\begin{aligned} \frac{d y}{d t} = \frac{(1 + t^{2}) \frac{d (2 t)}{d t} - 2 t \frac{d (1 + t^{2})}{d t}}{{(1 + t^{2})}^{2}} \end{aligned}

[Using quotient rule]

= \frac{(1 + t^{2}) (2 t) - 2 t (2 t)}{{(1 + t^{2})}^{2}}

\frac{d y}{d t} = \frac{2 + 2 t^{2} - 4 t^{2}}{{(1 + t^{2})}^{2}}

\begin{aligned} \frac{d y}{d t} = \frac{2 - 2 t}{{(1 + t^{2})}^{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} and \frac{d y}{d t}

from equation (1) and (2) respectively

\frac{d y}{d x} = \frac{[\frac{2 - 2 t^{2}}{{(1 + t^{2})}^{2}}]}{[\frac{- 4 t}{{(1 + t^{2})}^{2}}]} = \frac{2 (1 - t^{2})}{- 4 t} = - \frac{(1 - t^{2})}{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 (\frac{3 θ}{2})

Hint:
Use chain rule
Given:

\begin{aligned} x = 2 \cos θ - \cos 2 θ \\ y = 2 \sin θ - \sin 2 θ \end{aligned}

Solution:

= 2 \cos θ - \cos 2 θ

\frac{d x}{d θ} = \frac{d (2 \cos θ)}{d θ} - \frac{d (\cos 2 θ)}{d θ}

\begin{aligned} x & = 2 (- \sin θ) \cdot \frac{d \cos 2 θ}{d (2 θ)} \times \frac{d (2 θ)}{d θ} \end{aligned}

[Using chain rule]

= - \sin θ - (- \sin 2 θ) \times 2

= - 2 \sin θ + 2 \sin 2 θ

\begin{aligned} \frac{d x}{d θ} = 2 \sin 2 θ - 2 \sin θ \end{aligned}

(1)

y = 2 \sin θ - \sin 2 θ

\frac{d y}{d θ} = \frac{d (2 \sin θ)}{d θ} - \frac{d (\sin 2 θ)}{d θ}

\begin{aligned} = 2 \cos θ - [\frac{d (\sin 2 θ)}{d (2 θ)} \times \frac{d (2 θ)}{d θ}] \end{aligned}

[Using chain rule]

= 2 \cos θ - [\cos 2 θ \times 2]

\begin{aligned} \frac{d y}{d θ} = 2 \cos θ - 2 \cos 2 θ \end{aligned}

(2)

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

So, put the values of

\frac{d x}{d θ} and \frac{d y}{d θ}

from the equation (1) and (2)

\frac{d y}{d x} = \frac{2 (\cos θ - \cos 2 θ)}{2 (\sin 2 θ - \sin θ)}

\begin{aligned} \frac{d y}{d x} = \frac{\cos θ - \cos 2 θ}{\sin 2 θ - \sin θ} \end{aligned}

= \frac{- 2 \sin (\frac{θ + 2 θ}{2}) \cdot \sin (\frac{θ - 2 θ}{2})}{2 \cos (\frac{2 θ + θ}{2}) \cdot \sin (\frac{2 θ - θ}{2})}

[\begin{array}{l} \cos A - \cos B = - 2 \sin (\frac{A + B}{2}) \cdot \sin (\frac{A - B}{2}) \\ \sin A - \sin B = 2 \cos (\frac{A + B}{2}) \cdot \sin (\frac{A - B}{2}) \end{array}]

= \frac{- \sin (\frac{3 θ}{2}) \cdot \sin (\frac{- θ}{2})}{\cos (\frac{3 θ}{2}) \cdot \sin (\frac{θ}{2})}

\begin{aligned} = \frac{- \sin (\frac{3 θ}{2}) \cdot (- \sin \frac{θ}{2})}{\cos (\frac{3 θ}{2}) \cdot \sin \frac{θ}{2}} \end{aligned}

= \frac{\sin (\frac{3 θ}{2})}{\cos (\frac{3 θ}{2})}

\begin{aligned} \frac{d y}{d x} = \tan (\frac{3 θ}{2}) \end{aligned}

[∵ \tan θ = \frac{\sin θ}{\cos θ}]

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 (e^{\cos 2 t})}{d t}

\begin{aligned} = \frac{d (e^{\cos 2 t})}{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

[∵ \frac{d (e^{x})}{d x} = e^{x} \frac{d (\cos θ)}{d θ} = - \sin θ]

\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 (e^{\sin 2 t})}{d t}

\begin{aligned} = \frac{d (e^{\sin 2 t})}{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

[∵ \frac{d e^{x}}{d x} = e^{x}, \frac{d \sin θ}{d θ} = \cos θ]

\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} 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 (e^{\cos 2 t})

\log x = \cos 2 t \cdot \log e

[∵ \log a^{m} = m \log a]

\begin{aligned} \log x = \cos 2 t \end{aligned}

[∵ \log e = 1]

Also

y = e^{\sin 2 t}

Take log on both side

\begin{aligned} \log y = \log (e^{\sin 2 t}) \\ \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}} At t = \frac{2 π}{3}

Hint:
Use

[\tan (π - θ) = - \tan θ]

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} and \frac{d y}{d t}

from equation (1) and (2) respectively

= \frac{\cos t}{- \sin t}

\frac{d y}{d x} A t (x = \frac{2 π}{3})

\begin{aligned} = - \cot (\frac{2 π}{3}) \\ = - \cot (π - \frac{π}{3}) \\ = - (- \cot \frac{π}{3}) \\ = \cot \frac{π}{3} \end{aligned}

\frac{d y}{d x} A t (x = \frac{2 π}{3}) = \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 (t + \frac{1}{t}) \\ y = a (t - \frac{1}{t}) \end{aligned}

Solution:

x = a (t + \frac{1}{t})

\begin{aligned} \frac{d x}{d t} = a \frac{d (t + \frac{1}{t})}{d t} \end{aligned}

\begin{aligned} = a [\frac{d t}{d t} + \frac{d (\frac{1}{t})}{d t}] \\ = a [1 + \frac{d (t)^{- 1}}{d t}] \end{aligned}

= a [1 + {(- 1) t^{- 1 - 1}}]

[∵ \frac{d x^{n}}{d x} = n x^{n - 1}]

\begin{aligned} \frac{d x}{d t} = a (1 - t^{2}) \end{aligned}

(1)

y = a (t - \frac{1}{t})

\frac{d y}{d t} = a \frac{d (t - \frac{1}{t})}{d t}

\begin{aligned} = a [\frac{d t}{d t} - \frac{d (\frac{1}{t})}{d t}] \end{aligned}

= a [1 - \frac{d t^{- 1}}{d t}]

= a [1 - (- 1) t^{- 1 - 1}]

\begin{aligned} \frac{d y}{d t} = a (1 + t^{- 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 x}{d t} 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 (t + \frac{1}{t})}{t (t - \frac{1}{t})} \end{aligned}

= \frac{(t + \frac{1}{t})}{(t - \frac{1}{t})}

\begin{aligned} = \frac{a (t + \frac{1}{t})}{a (t - \frac{1}{t})} \end{aligned}

[Multiply and divide by a]

= \frac{x}{y}

[∵ x = a (t + \frac{1}{t}) & y = a (t - \frac{1}{t})]

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} (\frac{2 t}{1 + t^{2}}), y = \tan^{- 1} (\frac{2 t}{1 - t^{2}}); - 1 < t < 1

Solution:

x = \sin^{- 1} (\frac{2 t}{1 + t^{2}})

\begin{aligned} \frac{d x}{d t} = \frac{d \sin^{- 1} (\frac{2 t}{1 + t^{2}})}{d t} \end{aligned}

= \frac{d \sin^{- 1} (\frac{2 t}{1 + t^{2}})}{d (\frac{2 t}{1 + t^{2}})} \times \frac{d (\frac{2 t}{1 + t^{2}})}{d t}

[Using chain rule]

= \frac{1}{\sqrt{1 - {(\frac{2 t}{1 + t^{2}})}^{2}}} \times \frac{(1 + t^{2}) \frac{d (2 t)}{d t} - 2 t \frac{d (1 + t^{2})}{d t}}{{(1 + t^{2})}^{2}}

[Using quotient rule]

= \frac{1 + t^{2}}{\sqrt{{(1 + t^{2})}^{2} - 4 t^{2}}} \times \frac{(1 + t^{2}) (2) - (2 t) (2 t)}{{(1 + t^{2})}^{2}}

\begin{aligned} = \frac{2 + 2 t^{2} - 4 t^{2}}{\sqrt{1 + t^{4} + 2 t^{2} - 4 t^{2}} \times (1 + t^{2})} \end{aligned}

= \frac{2 - 2 t^{2}}{\sqrt{1 + t^{4} - 2 t^{2}} \times (1 + t^{2})}

\begin{aligned} = \frac{2 (1 - t^{2})}{\sqrt{{(1 - t^{2})}^{2}} (1 + t^{2})} \end{aligned}

\frac{d x}{d t} = \frac{2}{1 + t^{2}}

\begin{aligned} y = \tan^{- 1} (\frac{2 t}{1 - t^{2}}) \end{aligned}

(1)

\frac{d y}{d t} = \frac{d \tan^{- 1} (\frac{2 t}{1 - t^{2}})}{d t}

\begin{aligned} \frac{d y}{d t} = \frac{d \tan^{- 1} (\frac{2 t}{1 - t^{2}})}{d (\frac{2 t}{1 - t^{2}})} \times \frac{d (\frac{2 t}{1 - t^{2}})}{d t} \end{aligned}

= \frac{1}{1 + (\frac{2 t}{1 - t})} \times \frac{(1 - t^{2}) \frac{d (2 t)}{d t} - (2 t) \frac{d (1 - t^{2})}{d t}}{{(1 - t^{2})}^{2}}

[Using chain rule]

= \frac{1}{1 + \frac{4 t^{2}}{{(1 - t^{2})}^{2}}} \times \frac{(1 - t^{2}) \times 2 - (2 t) (- 2 t)}{{(1 - t^{2})}^{2}}

\begin{aligned} = \frac{{(1 - t^{2})}^{2}}{{(1 - t^{2})}^{2} + 4 t^{2}} \times \frac{2 - 2 t^{2} + 4 t^{2}}{{(1 - t^{2})}^{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 (1 + t^{2})}{{(1 + t^{2})}^{2}}

\begin{aligned} \frac{d y}{d t} = \frac{2}{(1 + t^{2})} \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} and \frac{d y}{d t}

from the equation (1) and (2) respectively

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

= \frac{2 (1 + t^{2})}{2 (1 + t^{2})}

\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}}

[∵ \frac{1}{x^{n}} = (x)^{- n}]

\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]

= [(\cos 2 t)^{\frac{- 1}{2}} \times \frac{d \sin^{3} t}{d \sin t} \times \frac{d \sin t}{d t}] + [\sin^{3} t \times \frac{d (\cos 2 t)^{\frac{- 1}{2}}}{d \cos 2 t} \times \frac{d \cos 2 t}{d t}]

[Using chain rule]

\frac{d x}{d t} = [\frac{1}{\sqrt{\cos 2 t}} \times 3 \sin^{2} t] + [\sin^{3} t \times (\frac{- 1}{2} (\cos 2 t)^{\frac{- 1}{2} - 1}) \times (- 2 \sin 2 t)]

\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 (1 - 2 \sin^{2} t) + \sin^{3} t (2 \sin t \cdot \cos t)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}

[\begin{array}{l} ∵ \cos 2 θ = 1 - 2 \sin^{2} θ \\ \sin 2 θ = 2 \sin θ \cos θ \end{array}]

= \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 (3 \sin t - 4 \sin^{3} t)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}

= \frac{\sin t \cos t \cdot \sin 3 t}{(\cos 2 t)^{\frac{3}{2}}}

[∵ \sin 3 θ = 3 \sin θ - 4 \sin^{3} θ]

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}}

[∵ \frac{1}{x^{n}} = (x)^{- n}]

\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]

= [\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}] + [(\cos 2 t)^{\frac{- 1}{2}} \times \frac{\cos^{3} t}{d \cos t} \times \frac{d \cos t}{d t}]

[Using chain rule]

= [\cos^{3} t \times (\frac{- 1}{2} (\cos 2 t)^{\frac{- 3}{2}}) \times (- \sin 2 t) \times 2] + [\frac{1}{\sqrt{\cos 2 t}} \times 3 \cos^{2} t \times (- \sin t)]

= \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 (2 \cos^{2} t - 1) + \cos^{3} t \times 2 \sin t \times \cos t}{(\cos 2 t)^{\frac{3}{2}}}

[\begin{array}{l} ∵ \cos 2 θ = 2 \cos^{2} θ - 1 \\ \sin 2 θ = 2 \sin θ \cos θ \end{array}]

\frac{d y}{d t} = \frac{- 3 \cos^{2} t (2 \sin t \cos^{2} t) + 3 \cos^{2} t \sin t + 2 \cos^{4} t \sin t}{(\cos 2 t)^{\frac{3}{2}}}

= \frac{(- 6 \cos^{4} t \sin t + 2 \cos^{4} t \sin t) + 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 (4 \cos^{3} t - 3 \cos t)}{(\cos 2 t)^{\frac{3}{2}}}

\begin{aligned} = \frac{- \sin t \cos t (\cos 3 t)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}

[∵ \cos 3 θ = 4 \cos^{3} θ - 3 \cos θ]

= \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} 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 {(t + \frac{1}{t})}^{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 = {(t + \frac{1}{t})}^{a} \\ y = (a)^{t + \frac{1}{t}} \end{aligned}

Solution:

x = {(t + \frac{1}{t})}^{a}

\begin{aligned} \frac{d x}{d t} = \frac{d (t + \frac{1}{t})}{d t} \end{aligned}

= \frac{d {(t + \frac{1}{t})}^{a}}{d (t + \frac{1}{t})} \times \frac{d (t + \frac{1}{t})}{d t}

[Using chain rule]

= a {(t + \frac{1}{t})}^{a - 1} \times (\frac{d t}{d t} + \frac{d (\frac{1}{t})}{d t})

[∵ \frac{d x^{n}}{d x} = n x^{n - 1}]

= a {(t + \frac{1}{t})}^{a - 1} \times (1 + (\frac{- 1}{t^{2}}))

\begin{aligned} \frac{d x}{d t} = a {(t + \frac{1}{t})}^{a - 1} \times (1 - \frac{1}{t^{2}}) \end{aligned}

(1)

y = a^{t + \frac{1}{t}}

\frac{d y}{d x} = \frac{d (a^{1 + \frac{1}{t}})}{d t}

\begin{aligned} = \frac{d (a^{t + \frac{1}{t}})}{d (t + \frac{1}{t})} \times \frac{d (t + \frac{1}{t})}{d t} \end{aligned}

= (a^{t + \frac{1}{t}}) \log a \times (\frac{d t}{d t} + \frac{d (\frac{1}{t})}{d t})

[∵ \frac{d (a^{x})}{d x} = a^{x} \log a]

\frac{d y}{d t} = (a)^{t + \frac{1}{t}} \log a \times [1 - \frac{1}{t^{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 x}{d t} 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 (1 - \frac{1}{t^{2}})}{a {(t + \frac{1}{t})}^{a - 1} \times (1 - \frac{1}{t^{2}})}

\begin{aligned} \frac{d y}{d x} = \frac{(a)^{t + \frac{1}{t}} \log a}{a {(t + \frac{1}{t})}^{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 (\frac{1 + t^{2}}{1 - t^{2}}) \\ y = \frac{2 t}{1 - t^{2}} \end{aligned}

Solution:

x = a (\frac{1 + t^{2}}{1 - t^{2}})

$\frac{d x}{d t}=a \frac{d\left(\frac{1+t^{2}}{1-t^{2}}\right)}{d t} $

\begin{aligned} = a \times \frac{(1 - t^{2}) \frac{d (1 + t^{2})}{d t} - (1 + t^{2}) \frac{d (1 - t^{2})}{d t}}{{(1 - t^{2})}^{2}} \end{aligned}

[using quotient rule]

= a [\frac{(1 - t^{2}) \cdot (2 t) - (1 + t^{2}) \cdot (- 2 t)}{{(1 - t^{2})}^{2}}]

\begin{aligned} = a [\frac{2 t - 2 t^{3} + 2 t + 2 t^{3}}{{(1 - t^{2})}^{2}}] \end{aligned}

= a [\frac{4 t}{{(1 - t^{2})}^{2}}]

= \frac{4 a t}{{(1 - t^{2})}^{2}}

(1)

\begin{aligned} y = \frac{2 t}{(1 - t^{2})} \end{aligned}

\frac{d y}{d t} = \frac{(1 - t^{2}) \frac{d (2 t)}{d t} - 2 t \cdot \frac{(1 - t^{2})}{d t}}{{(1 - t^{2})}^{2}}

[Using quotient rule]

= \frac{(1 - t^{2}) (2) - 2 t (- 2 t)}{{(1 - t^{2})}^{2}}

\begin{aligned} = \frac{2 - 2 t^{2} + 4 t^{2}}{{(1 - t^{2})}^{2}} \end{aligned}

\frac{d y}{d x} = \frac{2 + 2 t^{2}}{{(1 - t^{2})}^{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} and \frac{d x}{d t}

from equation (2) and (1) respectively

\frac{d y}{d x} = \frac{\frac{(2 + 2 t^{2})}{{(1 - t^{2})}^{2}}}{\frac{4 a t}{{(1 - t^{2})}^{2}}} = \frac{(2 + 2 t^{2}) \times {(1 - t^{2})}^{2}}{4 a t \times {(1 - t^{2})}^{2}}

= \frac{2 (1 + t^{2})}{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 (\frac{t}{2})

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 (\frac{d t}{d t} - \frac{d \sin t}{d t}) \end{aligned}

[\frac{d \sin t}{d t} = \cos t]

\frac{d x}{d t} = 10 (1 - \cos t)

(1)

y = 12 (1 - \cos t)

\frac{d y}{d t} = 12 (\frac{d (1)}{d t} - \frac{d \cos t}{d t})

\begin{aligned} = 12 (0 - (- \sin t)) \end{aligned}

[∵ \frac{d \cos t}{t} = - \sin t]

\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} 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}

[∵ \cot θ = \frac{\cos θ}{\sin θ}]

Differentiation Exercise 10.7 Question 23

Answer:

\frac{d y}{d x} At (θ = \frac{π}{3}) = - \sqrt{3}

Hint:
Use chain rule and

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

Given:

\begin{aligned} x = a (θ - \sin θ) \\ y = a (1 + \cos θ) \end{aligned}

Solution:

x = a (θ - \sin θ)

\frac{d x}{d θ} = a [\frac{d θ}{d θ} - \frac{d \sin θ}{d θ}]

[\frac{d \sin θ}{d θ} = \cos θ]

\frac{d x}{d θ} = a (1 - \cos θ)

(1)

\begin{aligned} y = a (1 + \cos θ) \end{aligned}

\frac{d y}{d θ} = a (\frac{d 1}{d θ} + \frac{d \cos θ}{d θ})

= a (0 + (- \sin θ))

[∵ \frac{d \cos θ}{d θ} = - \sin θ]

\begin{aligned} \frac{d y}{d θ} = - a \sin θ \end{aligned}

(2)

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

Put the values of

\frac{d y}{d θ} and \frac{d x}{d θ}

from the equation (2) and (1) respectively

\frac{d y}{d x} = \frac{- a \sin θ}{a (1 - \cos θ)}

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

\begin{aligned} At θ = \frac{π}{3} \end{aligned}

\frac{d y}{d x} = - (\frac{\sin \frac{π}{3}}{1 - \cos \frac{π}{3}})

= - (\frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}})

[\begin{aligned} ∵ \sin \frac{π}{3} & = \frac{\sqrt{3}}{2} \\ \cos \frac{π}{3} & = \frac{1}{2} \end{aligned}]

= - (\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}})

\begin{aligned} \frac{d y}{d x} a t θ = \frac{π}{3} = - \sqrt{3} \end{aligned}

Differentiation Excercise 10.7 Question 24

Answer:

\frac{d y}{d x} At (t = \frac{π}{4}) = \frac{b}{a}

Hint:
Use product rule and

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

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 [\sin 2 t \times \frac{d (1 + \cos 2 t)}{d t} + (1 + \cos 2 t) \frac{d (\sin 2 t)}{d t}] \end{aligned}

[Using product rule]

= a [\sin 2 t \times (0 + (- 2 \sin 2 t)) + (1 + \cos 2 t) (2 \cos 2 t)]

= a [- 2 \sin^{2} 2 t + 2 \cos 2 t + 2 \cos^{2} 2 t]

= a [2 \cos 2 t + 2 \cos^{2} 2 t - 2 \sin^{2} 2 t]

$=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}

[∵ \cos 2 θ = \cos^{2} θ - \sin^{2} θ]

\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 [\cos 2 t \times \frac{d (1 - \cos 2 t)}{d t} + (1 - \cos 2 t) \frac{d \cos 2 t}{d t}] \end{aligned}

[Using product rule]

= b [\cos 2 t (0 + 2 \sin 2 t)] + (1 - \cos 2 t) (- 2 \sin 2 t)

[∵ \frac{d \cos θ}{d θ} = - \sin θ]

= 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}

[∵ \sin 2 θ = 2 \sin θ \cos θ]

\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} 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 (\frac{\sin 4 t - \sin 2 t}{\cos 2 t + \cos 4 t})

\begin{aligned} At t = \frac{π}{4} \end{aligned}

\frac{d y}{d x} = \frac{b}{a} \times \frac{(\sin (4 \times \frac{π}{4}) - \sin (2 \times \frac{π}{4}))}{\cos (2 \times \frac{π}{4}) + \cos (4 \times \frac{π}{4})}

= \frac{b}{a} (\frac{\sin π - \sin \frac{π}{2}}{\cos \frac{π}{2} + \cos π})

\begin{aligned} ∵ \sin \frac{π}{2} = 1, \cos \frac{π}{2} = 0, \sin π = 0, \cos π = - 1 \end{aligned}

= \frac{b}{a} [\frac{0 - 1}{0 + (- 1)}]

= \frac{b}{a} (\frac{- 1}{- 1})

\begin{aligned} {\frac{d y}{d x}}_{At} (t = \frac{π}{4}) = (\frac{b}{a}) \end{aligned}

Hence proved

Differentiation Excercise 10.7 Question 25

Answer:

\frac{d y}{d x} At (t = \frac{π}{4}) = 1

Hint:
Use product rule and

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

Given:

\begin{aligned} x = \cos t (3 - 2 \cos^{2} t) \\ y = \sin t (3 - 2 \sin^{2} t) \end{aligned}

Solution:

x = \cos t (3 - 2 \cos^{2} t)

\frac{d x}{d t} = \cos t \cdot \frac{d (3 - 2 \cos^{2} t)}{d t} + (3 - 2 \cos^{2} t) \frac{d \cos t}{d t}

[Using product rule]

\frac{d x}{d t} = \cos t [0 - 4 \cos t (- \sin t)] + (3 - 2 \cos^{2} t) \times (- \sin t)

[∵ \frac{d (\cos θ)}{d θ} = - \sin θ]

\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 (2 \cos^{2} t - 1) \end{aligned}

(1)

y = \sin t (3 - 2 \sin^{2} t)

\begin{aligned} \frac{d y}{d t} = \sin t \frac{(3 - 2 \sin^{2} t)}{d t} + (3 - 2 \sin^{2} t) \frac{d \sin t}{d t} \end{aligned}

[Using product rule]

= \sin t \times [\frac{d 3}{d t} - \frac{d (\sin^{2} t)}{d t}] + (3 - 2 \sin^{2} t) \cos t

[∵ \frac{d \sin θ}{d θ} = \cos θ]

= \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} ∴ \frac{d y}{d t} = 3 \cos t (- 2 \sin^{2} t + 1) \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} and \frac{d x}{d t}

from the equation (2) and (1) respectively

\frac{d y}{d x} = \frac{3 \cos t (1 - \sin^{2} t)}{3 \sin t (2 \cos^{2} t - 1)}

\begin{aligned} \frac{d y}{d x} = \cot t \frac{(1 - 2 \sin^{2} t)}{(2 \cos^{2} t - 1)} \end{aligned}

[∵ \tan θ = \frac{\sin θ}{\cos θ}]

\frac{d y}{d x} = \cot t \frac{(1 - 2 (1 - \cos^{2} t))}{(2 \cos^{2} t - 1)}

[∵ \sin^{2} θ + \cos^{2} θ = 1]

\frac{d y}{d x} = \cot t \times \frac{(1 - 2 + 2 \cos^{2} t)}{(2 \cos^{2} t - 1)}

= \cot t \times \frac{(2 \cos^{2} t - 1)}{(2 \cos^{2} t - 1)}

= \cot t

\begin{aligned} \frac{d y}{d x} = \cot t \end{aligned}

At t = \frac{π}{4}, \frac{d y}{d x} = \cot \frac{π}{4} = 1

\begin{aligned} \frac{d y}{d x} At (t = \frac{π}{4}) = 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 θ}}{\frac{d x}{d θ}}

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 (\frac{1 + \log t}{t^{2}})}{d t}

\begin{aligned} = \frac{t^{2} \frac{d (1 + \log t)}{d t} - (1 + \log t) \frac{d t^{2}}{d t}}{{(t^{2})}^{2}} \end{aligned}

[Using quotient rule]

\frac{d x}{d t} = \frac{t [\frac{d (1)}{d t} + \frac{d \log t}{d t}] - (1 + \log t) \cdot 2 t}{t^{4}}

= \frac{t^{2} (0 + \frac{1}{t}) - 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 (\frac{3 + 2 \log t}{t})}{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 (\frac{2}{t}) - (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} 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} At (t = \frac{π}{3}) = \frac{- 1}{\sqrt{3}}

Hint:
Use

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

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

[∵ \frac{d \sin x}{d x} = \cos x]

\begin{aligned} \frac{d x}{d t} = 3 (\cos t - \cos 3 t) \end{aligned}

(1)

∴ 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}

[∵ \frac{d \cos x}{d x} = - \sin x]

= - 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} 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}

At t = \frac{π}{3}

\begin{aligned} \frac{d y}{d x} = \frac{\sin (3 \times \frac{π}{3}) - \sin \frac{π}{3}}{\cos \frac{π}{3} - \cos (3 \times \frac{π}{3})} \end{aligned}

= \frac{\sin π - \sin \frac{π}{3}}{\cos \frac{π}{3} - \cos π}

\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} At (t = \frac{π}{3}) = \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 θ}}{\frac{d x}{d θ}}

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} (\frac{2 t}{1 + t^{2}}) \end{aligned}

Differentiate w.r.t t

\frac{d x}{d t} = \frac{d \sin^{- 1} (\frac{2 t}{1 + t^{2}})}{d t}

\begin{aligned} = \frac{d \sin^{- 1} (\frac{2 t}{1 + t^{2}})}{d (\frac{2 t}{1 + t^{2}})} \times \frac{d (\frac{2 t}{1 + t^{2}})}{d t} \end{aligned}

[Using chain rule]

= \frac{1}{\sqrt{1 - {(\frac{2 t}{1 + t^{2}})}^{2}}} \times \frac{(1 + t^{2}) \frac{d (2 t)}{d t} - 2 t \frac{d (1 + t^{2})}{d t}}{{(1 + t^{2})}^{2}}

[Using quotient rule]

\frac{d x}{d t} = \frac{(1 + t^{2})}{\sqrt{{(1 + t^{2})}^{2} - (2 t)^{2}}} \times \frac{(1 + t^{2}) (2) - 2 t (2 t)}{{(1 + t^{2})}^{2}}

= \frac{2 + 2 t^{2} - 4 t^{2}}{\sqrt{1 + t^{4} + 2 t^{2} - 4 t^{2}} \cdot (1 + t^{2})}

= \frac{2 - 2 t^{2}}{\sqrt{1 + t^{4} - 2 t^{2}} \cdot (1 + t^{2})}

\begin{aligned} = \frac{2 (1 - t^{2})}{(1 - t^{2}) (1 + t^{2})} \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} (\frac{2 t}{1 - t^{2}}) \end{aligned}

Differentiate w.r.t

t

\frac{d y}{d t} = \frac{d (\tan^{- 1} (\frac{2 t}{1 - t^{2}}))}{d t}

\begin{aligned} = \frac{d (\tan^{- 1} (\frac{2 t}{1 - t^{2}}))}{d (\frac{2 t}{1 - t^{2}})} \times \frac{d (\frac{2 t}{1 - t^{2}})}{d t} \end{aligned}

[Using chain rule]

= \frac{1}{1 + {(\frac{2 t}{1 - t^{2}})}^{2}} \times \frac{(1 - t^{2}) \frac{d (2 t)}{d t} - 2 t \frac{d (1 - t^{2})}{d t}}{{(1 - t^{2})}^{2}}

\begin{aligned} = \frac{1}{1 + \frac{4 t^{2}}{{(1 - t^{2})}^{2}}} \times \frac{(1 - t^{2}) \cdot 2 - 2 t (- 2 t)}{{(1 - t^{2})}^{2}} \end{aligned}

= \frac{{(1 - t^{2})}^{2}}{{(1 - t^{2})}^{2} + 4 t^{2}} \times \frac{2 - 2 t^{2} + 4 t^{2}}{{(1 - t^{2})}^{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 (1 + t^{2})}{(1 + t^{4} + 2 t^{2})}

= \frac{2 (1 + t^{2})}{{(1 + t^{2})}^{2}}

\begin{aligned} \frac{d y}{d t} = \frac{2}{(1 + t^{2})} \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} and \frac{d y}{d t}

from the equation (1) and (2) respectively

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

\begin{aligned} \frac{d y}{d x} = 1 \end{aligned}

Differentiation Exercise 10.7 Question 29

Answer:
$\frac{d y}{d x} At (t = \frac{π}{3}) = \frac{1}{\sqrt{3}}$
Hint:
Use $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$
Given:
$\begin{aligned} x = a (2 θ - \sin 2 θ) \\ y = a (1 - \cos 2 θ) \end{aligned}$
Solution:
$x = a (2 θ - \sin 2 θ)$
$\frac{d x}{d θ} = a \frac{d (2 θ - \sin 2 θ)}{d θ}$
$\begin{aligned} = a \times [\frac{d (2 θ)}{d θ} - \frac{d (\sin 2 θ)}{d θ}] \end{aligned}$
$= a \times [2 - 2 \cos 2 θ]$ $[∵ \frac{d (\sin θ)}{d θ} = \cos θ]$
$= a \times 2 (1 - \cos 2 θ)$
$\begin{aligned} \frac{d x}{d θ} = 2 a (1 - \cos 2 θ) \end{aligned}$ (1)
$y = a (1 - \cos 2 θ)$
$\frac{d y}{d θ} = a \frac{d (1 - \cos 2 θ)}{d θ}$
$\begin{aligned} = a \times [\frac{d 1}{d θ} - \frac{d \cos 2 θ}{d θ}] \end{aligned}$
$\begin{aligned} = a [0 + 2 \sin 2 θ] \end{aligned}$ $[∵ \frac{d \cos θ}{d θ} = - \sin θ]$
$\frac{d y}{d θ} = 2 a \sin 2 θ$ (2)
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$
So, put the value of $\frac{d y}{d θ} and \frac{d x}{d θ}$ from equation (2) and (1) respectively
$\frac{d y}{d x} = \frac{2 a \sin 2 θ}{2 a (1 - \cos 2 θ)}$
$\begin{aligned} \frac{d y}{d x} = \frac{\sin 2 θ}{1 - \cos 2 θ} \end{aligned}$
$At θ = \frac{π}{3}$
$\begin{aligned} \frac{d y}{d x} = \frac{\sin (2 \times \frac{π}{3})}{1 - \cos (2 \times \frac{π}{3})} \end{aligned}$
$= \frac{\sin \frac{2 π}{3}}{1 - \cos \frac{2 π}{3}}$
$= \frac{\sin (π - \frac{π}{3})}{1 - \cos (π - \frac{π}{3})}$
$\begin{aligned} = & \frac{\sin \frac{π}{3}}{1 - (- \cos \frac{π}{3})} \end{aligned}$ $[\begin{matrix} \sin (π - θ) = \sin θ \\ \cos (π - θ) = - \cos θ \end{matrix}]$
$= \frac{\sin \frac{π}{3}}{1 + \cos \frac{π}{3}}$
$\begin{aligned} = \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} \end{aligned}$ $[∵ \sin \frac{π}{3} = \frac{\sqrt{3}}{2}, \cos \frac{π}{3} = \frac{1}{2}]$
$= \frac{\frac{\sqrt{3}}{2}}{\frac{(2 + 1)}{2}}$
$= \frac{\sqrt{3} \times 2}{3 \times 2}$
$\begin{aligned} \frac{d y}{d x} at (θ = \frac{π}{3}) = \frac{1}{\sqrt{3}} \end{aligned}$

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4. Can I use RD Sharma Solutions for homework?

School teachers often give questions from RD Sharma Solutions to test students. Students can easily find the answers to their homework questions in the RD Sharma solutions.

5. 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.

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

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

Differentiation Excercise: 10.7

RD Sharma Chapter-wise Solutions

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