If x = 2 cos θ - cos 2 θ and y = 2 sin θ - sin 2 θ , prove that d y d x = tan 3 θ 2 .

Given: x = 2 cos θ - cos 2 θ .....(i) y = 2 sin θ - sin 2 θ .....(ii) Differentiating equation (i) with respect to θ , d x d θ = d d θ 2 cos θ - cos 2 θ d x d θ = - 2 sin θ + 2 sin 2 θ Similarly differentiating equation (ii) with respect to θ , d y d θ = d d θ 2 sin θ - sin 2 θ d y d θ = 2 cos θ - 2 cos 2 θ By differentiation of parametric functions, d y d x = d y d θ d x d θ d y d x = 2 cos θ - 2 cos 2 θ - 2 sin θ + 2 sin 2 θ d y d x = cos θ - cos 2 θ sin 2 θ - sin θ d y d x = - 2 sin 3 θ 2 sin - θ 2 2 cos 3 θ 2 sin θ 2 d y d x = sin 3 θ 2 c os 3 θ 2 d y d x = tan 3 θ 2 Hence proved.

If x = e cos 2 t and y = e sin 2 t , then prove that d y d x = - y log e x x log e y .

Given: x = e cos 2 t .....(i) and y = e sin 2 t .....(ii) Using differentiation of parametric functions, d y d x = d y d t d x d t Differentiating equation (i) with respect to t , d x d t = d d θ e cos 2 t d x d t = 2 e cos 2 t - sin 2 t Differentiating equation (ii) with respect to t , d y d t = d d t e sin 2 t d y d t = 2 e sin 2 t cos 2 t ∴ d y d x = 2 e sin 2 t cos 2 t - 2 e cos 2 t sin 2 t ∴ d y d x = e sin 2 t cos 2 t - e cos 2 t sin 2 t Taking logarithm of equations (i) and (ii), log e x = log e e cos 2 t ⇒ log e x = cos 2 t and log e y = log e e sin 2 t ⇒ log e y = sin 2 t ∴ d y d x = - y log e x x log e y Hence proved.

If x = cos t and y = sin t , then prove that d y d x = 1 3 , at t = 2 π 3 .

Given: x = cos t .....(i) and y = sin t .....(ii) Differentiating equation (i) with respect to t , d x d t = - sin t Similarly differentiating equation (ii) with respect to t , d y d t = cos t Using differentiation of parametric functions, d y d x = d y d t d x d t ⇒ d y d x = cos t - sin t ∴ d y d x = - cot t At t = 2 π 3 , d y d x = - cot 2 π 3 ∴ d y d x = - - 1 3 ⇒ d y d x = 1 3 Hence proved.

If x = a t + 1 t and y = a t - 1 t , then prove that d y d x = x y .

Given: x = a t + 1 t . . . i x = a t 2 + 1 t and y = a t - 1 t . . . i i y = a t 2 - 1 t Differentiating equation (i) with respect to t , d x d t = a 1 - 1 t 2 d x d t = a t 2 - 1 t 2 Similarly differentiating equation (ii) with respect to t , d y d t = a 1 + 1 t 2 d y d t = a t 2 + 1 t 2 Using differentiation of parametric functions, d y d x = d y d t d x d t ⇒ d y d x = a t 2 + 1 t 2 a t 2 - 1 t 2 ⇒ d y d x = a t 2 + 1 t a t 2 - 1 t ∴ d y d x = x y Hence proved.

If x = sin - 1 2 t 1 + t 2 and y = tan - 1 2 t 1 - t 2 , then prove that d y d x = 1 .

Given: x = sin - 1 2 t 1 + t 2 and y = tan - 1 2 t 1 - t 2 Let us assume t = tan a , ⇒ a = tan - 1 t x = sin - 1 2 tan a 1 + tan 2 a ⇒ x = sin - 1 sin 2 a ⇒ x = 2 a ⇒ x = 2 tan - 1 t Similarly, y = tan - 1 2 t 1 - t 2 ⇒ y = tan - 1 2 tan a 1 - tan 2 a ⇒ y = tan - 1 tan 2 a ⇒ y = 2 a ⇒ y = 2 tan - 1 t By differentiation of parametric functions, d y d x = d y d t d x d t d y d t = d d t 2 tan - 1 t d y d t = 2 1 1 + t 2 Similarly, d x d t = d d t 2 tan - 1 t d x d t = 2 1 1 + t 2 ⇒ d y d x = 1 Hence proved.

Differentiate sin - 1 2 a x 1 - a 2 x 2 w.r.t. 1 - a 2 x 2 .

Let u = sin - 1 2 a x 1 - a 2 x 2 and v = 1 - a 2 x 2 We have to find d u d v . Let us substitute x = 1 a sin θ in u and v , u = sin - 1 2 a sin θ a 1 - a 2 sin 2 θ a 2 ⇒ u = sin - 1 2 sin θ 1 - sin 2 θ ⇒ u = sin - 1 2 sin θ cos θ ⇒ u = sin - 1 sin 2 θ ⇒ u = 2 θ ..... ( i ) Similarly v = 1 - a 2 sin 2 θ a 2 v = 1 - sin 2 θ ⇒ v = cos θ ..... ( ii ) Differentiating equation ( i ) with respect to θ , d u d θ = d d θ 2 θ ⇒ d u d θ = 2 Similarly differentiating equation ( ii ) with respect to θ , d v d θ = d d θ cos θ ⇒ d v d θ = - sin θ By differentiation of parametric functions, d u d v = d u d θ d v d θ ∴ d u d v = 2 - sin θ ∵ x = 1 a sin θ ⇒ a x = sin θ ⇒ d u d v = - 2 a x

If y = tan - 1 a 1 x - α a 1 α + x + tan - 1 a 2 - a 1 1 + a 1 a 2 + tan - 1 a 3 - a 2 1 + a 2 a 3 + . . . . + tan - 1 a n - a n - 1 1 + a n a n - 1 - tan - 1 a n , then find d y d x .

Given, y = tan - 1 a 1 x - α a 1 α + x + tan - 1 a 2 - a 1 1 + a 1 a 2 + tan - 1 a 3 - a 2 1 + a 2 a 3 + . . . . + tan - 1 a n - a n - 1 1 + a n a n - 1 - tan - 1 a n ⇒ y = tan - 1 a 1 - α x 1 + a 1 α x + tan - 1 a 2 - a 1 1 + a 1 a 2 + tan - 1 a 3 - a 2 1 + a 2 a 3 + . . . . + tan - 1 a n - a n - 1 1 + a n a n - 1 - tan - 1 a n {taking x common in numerator and denominator of the first term of above series} As we know that, tan - 1 x - tan - 1 y = tan - 1 x - y 1 + x y , x ≥ 0 , y ≥ 0 So, y = tan - 1 a 1 - tan - 1 α x + tan - 1 a 2 - tan - 1 a 1 + tan - 1 a 3 - tan - 1 a 2 + . . . . + tan - 1 a n - tan - 1 a n - 1 - tan - 1 a n ⇒ y = - tan - 1 α x Differentiate the above equation with respect to x , we get ⇒ d y d x = d d x - tan - 1 α x ⇒ d y d x = - 1 1 + α x 2 d d x α x {Using chain rule of differentiation} ⇒ d y d x = - x 2 x 2 + α 2 · - 1 x 2 α ⇒ d y d x = α x 2 + α 2

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5

Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 5: Target Exercise 2.5

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 5: Target Exercise 2.5

Attempt the practice questions on Chapter 2: Differentiation, Exercise 5: Target Exercise 2.5 with hints and solutions to strengthen your understanding. Skills in Mathematics Differential Calculus for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 5: Target Exercise 2.5 with Hints & Solutions

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 19

If $x = 2 \cos (θ) - \cos (2 θ)$ and $y = 2 \sin (θ) - \sin (2 θ)$ , prove that $\frac{d y}{d x} = \tan (\frac{3 θ}{2})$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 20

If $x = e^{\cos (2 t)}$ and $y = e^{\sin (2 t)}$ , then prove that $\frac{d y}{d x} = \frac{- y \log_{e} (x)}{x \log_{e} (y)}$ .

EASY

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 21

If $x = \cos (t)$ and $y = \sin (t)$ , then prove that $\frac{d y}{d x} = \frac{1}{\sqrt{3}}$ , at $t = \frac{2 π}{3}$ .

EASY

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 22

If $x = a (t + \frac{1}{t})$ and $y = a (t - \frac{1}{t})$ , then prove that $\frac{d y}{d x} = \frac{x}{y}$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 23

If $x = \sin^{- 1} (\frac{2 t}{1 + t^{2}})$ and $y = \tan^{- 1} (\frac{2 t}{1 - t^{2}})$ , then prove that $\frac{d y}{d x} = 1$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 24

Differentiate $x^{\sin^{- 1} (x)}$ with respect to $\sin^{- 1} (x)$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 25

Differentiate $\sin^{- 1} (2 a x \sqrt{1 - a^{2} x^{2}})$ w.r.t. $\sqrt{1 - a^{2} x^{2}}$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.5>Q 26

If $y = \tan^{- 1} (\frac{a_{1} x - α}{a_{1} α + x}) + \tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{2} a_{3}}) + . . . . + \tan^{- 1} (\frac{a_{n} - a_{n - 1}}{1 + a_{n} a_{n - 1}}) - \tan^{- 1} (a_{n})$ , then find $\frac{d y}{d x}$ .

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