Find the differential coefficient of: x log x w.r.t sin x

Let u = x log x and v = sin x Differentiate w.r.t x , we get: d u d x = d x log x d x ⇒ d u d x = log x × d x d x - x × d log x d x log x 2 [Using Quotient rule] ⇒ d u d x = log x × 1 - x × 1 x log x 2 = log ⁡ x − 1 ( log ⁡ x ) 2 Consider, v = sin x Differentiate w.r.t x , we get: d v d x = d sin x d x ⇒ d v d x = cos x So, d u d v = d u d x d v d x = log ⁡ x − 1 ( log ⁡ x ) 2 cos x = log ⁡ x − 1 ( log ⁡ x ) 2 × cos x

Find the differential coefficients of tan - 1 2 x 1 - x 2 w.r.t tan - 1 x [ 0 < x < 1 ] .

Let u = tan - 1 ⁡ 2 x 1 − x 2 and v = tan − 1 ⁡ x Put x = tan θ ⇒ θ = tan - 1 x . Then, u = tan - 1 ⁡ 2 tan θ 1 − tan 2 θ = tan − 1 ⁡ tan 2 θ ∵ tan 2 A = 2 tan ⁡ A 1 − tan 2 ⁡ A = 2 θ ∵ tan - 1 tan A = A = 2 tan - 1 x Differentiate w.r.t x , we get: d u d x = 2 1 + x 2 Now consider, v = tan − 1 ⁡ x Differentiate w.r.t x , we get: d v d x = 1 1 + x 2 Hence, d u d v = d u d x d v d x = 2 1 + x 2 1 1 + x 2 = 2

Find the differential coefficient of cos - 1 1 - x 2 1 + x 2 w.r.t tan - 1 3 x - x 3 1 - 3 x 2 0 < x < 1 3

Let u = cos − 1 ⁡ 1 − x 2 1 + x 2 Put x = tan θ ⇒ θ = tan - 1 x ⇒ u = cos − 1 ⁡ 1 − tan 2 θ 1 + tan 2 θ ⇒ u = cos − 1 ⁡ cos 2 θ ∵ cos 2 A = 1 − tan 2 A 1 + tan 2 A ⇒ u = 2 θ ∵ cos - 1 cos A = A ⇒ u = 2 tan - 1 x Differentiate w.r.t x , we get: d u d x = 2 1 + x 2 Let v = tan - 1 3 x - x 3 1 - 3 x 2 Put x = tan θ ⇒ θ = tan - 1 x ⇒ v = tan − 1 ⁡ 3 tan θ − tan 3 θ 1 − 3 tan 2 θ ⇒ v = tan − 1 ⁡ tan 3 θ ∵ tan 3 A = 3 tan A − tan 3 A 1 − 3 tan 2 A ⇒ v = 3 θ ∵ tan - 1 tan A = A ⇒ v = 3 tan - 1 x Differentiate w.r.t x , we get: d v d x = 3 1 + x 2 Hence, d u d v = d u d x d v d x = 2 1 + x 2 3 1 + x 2 = 2 3

Find the differential coefficients of tan - 1 1 + x 2 - 1 x w.r.t tan - 1 x

Let u = tan − 1 ⁡ 1 + x 2 − 1 x Put x = tan θ ⇒ θ = tan - 1 x ⇒ u = tan − 1 ⁡ 1 + tan 2 θ − 1 tan θ ⇒ u = tan − 1 ⁡ sec θ − 1 tan θ ∵ 1 + tan 2 A = sec 2 A ⇒ u = tan − 1 ⁡ 1 - cos θ sin θ ⇒ u = tan − 1 ⁡ 2 sin 2 θ 2 2 sin θ 2 cos θ 2 ∵ sin 2 A = 2 sin A cos A , cos 2 A = 1 - 2 sin 2 A ⇒ u = tan − 1 ⁡ sin θ 2 cos θ 2 ⇒ u = tan − 1 ⁡ tan θ 2 ⇒ u = θ 2 ∵ tan − 1 ⁡ tan A = A ⇒ u = tan - 1 x 2 Differentiate w.r.t x , we get: d u d x = 1 2 1 + x 2 Now, let v = tan - 1 x Put x = tan ⁡ θ ⇒ θ = tan − 1 ⁡ x ⇒ v = tan - 1 tan θ ⇒ v = θ ∵ tan - 1 tan A = A ⇒ v = tan - 1 x Differentiate w.r.t x , we get: d v d x = 1 1 + x 2 Hence, d u d v = d u d x d v d x = 1 2 1 + x 2 1 1 + x 2 = 1 2

Find the differential coefficient of sec - 1 1 2 x 2 - 1 w.r.t sin - 1 3 x - 4 x 3 0 < x < 1 2

Let u = sec − 1 ⁡ 1 2 x 2 − 1 Put x = cos θ ⇒ θ = cos - 1 x ⇒ u = sec − 1 ⁡ 1 2 cos 2 θ − 1 ⇒ u = sec − 1 ⁡ 1 cos 2 θ ∵ 2 cos 2 A − 1 = cos 2 A ⇒ u = sec - 1 sec 2 θ ⇒ u = 2 θ ∵ sec - 1 sec A = A ⇒ u = 2 cos - 1 x Differentiate w.r.t x , we get: d u d x = - 2 1 - x 2 Now, let v = sin − 1 ⁡ 3 x − 4 x 3 Put x = sin θ ⇒ θ = sin - 1 x ⇒ v = sin - 1 3 sin θ - 4 sin 3 θ ⇒ v = sin - 1 sin 3 θ ∵ sin 3 A = 3 sin A - 4 sin 3 A ⇒ v = 3 θ ∵ sin - 1 sin A = A ⇒ v = 3 sin - 1 x Differentiate w.r.t x , we get: d v d x = 3 1 - x 2 Hence, d u d v = d u d x d v d x = − 2 1 − x 2 3 1 − x 2 = - 2 3

E M B I B E

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - A. Continuity and Differentiability>Chapter 8 - Differentiation [Part-II]>EXERCISE 3.5

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 5: EXERCISE 3.5

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 5: EXERCISE 3.5

Attempt the practice questions from Exercise 5: EXERCISE 3.5 with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 5: EXERCISE 3.5 with Hints & Solutions

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE 3.5>Q 1

Find the differential coefficient of $\log (\sec x)$ w.r.t $\tan x$

EASY

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE 3.5>Q 1

Find the differential coefficient of $(\sin x + \cos x)$ w.r.t $\frac{x}{2}$

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE 3.5>Q 1

Find the differential coefficient of $e^{\sin x}$ w.r.t $\tan^{- 1} x$

EASY

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE 3.5>Q 1

Find the differential coefficient of: $\frac{x}{\log x}$ w.r.t $\sin x$

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE 3.5>Q 1

Find the differential coefficients of $\tan^{- 1} (\frac{2 x}{1 - x^{2}})$ w.r.t $\tan^{- 1} x [0 < x < 1]$ .

MEDIUM

12th West Bengal Board

IMPORTANT

RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Chapter 8 - Differentiation [Part-II]>EXERCISE 3.5>Q 1

Find the differential coefficient of $\cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$ w.r.t $\tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) [0 < x < \frac{1}{\sqrt{3}}]$