Differentiate the following w . r . t . x tan - 1 x 3 - x 1 - 3 x

Let y = tan - 1 x 3 - x 1 - 3 x Put x = tan θ y = tan - 1 tan θ 3 - tan 2 θ 1 - 3 tan 2 θ ⇒ y = tan - 1 3 tan θ - tan 3 θ 1 - 3 tan 2 θ ⇒ y = tan - 1 tan 3 θ ⇒ y = 3 θ ⇒ y = 3 tan - 1 x Differentiate both sides with respect to x , we get ⇒ d y d x = 3 1 + x 2 . 1 2 x ⇒ d y d x = 3 2 1 + x x Thus, the derivative of given function w.r.t. x is ⇒ d y d x = 3 2 1 + x x

Differentiate the following w . r . t . x cos - 1 1 + x - 1 - x 2

Let cos - 1 1 + x - 1 - x 2 Put x = sin 2 A ⇒ 1 + x = 1 + sin 2 A = 1 + 2 sin A cos A = sin 2 A + cos 2 A + 2 sin A cos A = sin A + cos A 2 ⇒ 1 + x = sin A + cos A similarly , 1 - x = sin A - cos A ⇒ cos - 1 1 + x - 1 - x 2 = cos - 1 sin A + cos A - sin A + cos A 2 = cos - 1 cos A = A ⇒ cos - 1 1 + x - 1 - x 2 = sin - 1 x 2 Differentiate both sides with respect to x , we get ⇒ d y d x = 1 2 1 - x 2 Thus, the derivative of given function w.r.t. x is d y d x = 1 2 1 - x 2

Differentiate the following w . r . t . x tan - 1 x 1 + 6 x 2 + cot - 1 1 - 10 x 2 7 x

Let y = tan - 1 x 1 + 6 x 2 + cot - 1 1 - 10 x 2 7 x ⇒ y = tan - 1 x 1 + 6 x 2 + tan - 1 7 x 1 - 10 x 2 ⇒ y = tan - 1 3 x - 2 x 1 + 3 x × 2 x + tan - 1 5 x + 2 x 1 - 5 x × 2 x ⇒ y = tan - 1 3 x - tan - 1 2 x + tan - 1 5 x + tan - 1 2 x ⇒ y = tan - 1 3 x + tan - 1 5 x Differentiate both sides with respect to x , we get ⇒ d y d x = 3 1 + 9 x 2 + 5 1 + 25 x 2

Differentiate the following w . r . t . x tan - 1 1 + x 2 + x 1 + x 2 - x

Let y = tan - 1 1 + x 2 + x 1 + x 2 - x Put x = tan θ y = tan - 1 1 + tan 2 θ + tan θ 1 + tan 2 θ - tan θ ⇒ y = tan - 1 sec 2 θ + tan θ sec 2 θ - tan θ ⇒ y = tan - 1 sec θ + tan θ sec θ - tan θ ⇒ y = tan - 1 1 cos θ + sin θ cos θ 1 cos θ - 1 sin θ ⇒ y = tan - 1 1 + sin θ 1 - sin θ ⇒ y = tan - 1 sin 2 θ 2 + cos 2 θ 2 + 2 sin θ 2 cos θ 2 sin 2 θ 2 + cos 2 θ 2 - 2 sin θ 2 cos θ 2 ⇒ y = tan - 1 cos θ 2 + sin θ 2 cos θ 2 - sin θ 2 2 ⇒ y = tan - 1 cos θ 2 + sin θ 2 cos θ 2 - sin θ 2 ⇒ y = tan - 1 1 + tan θ 2 1 - tan θ 2 ⇒ y = tan - 1 tan π 4 + tan θ 2 1 - tan π 4 tan θ 2 ⇒ y = tan - 1 tan π 4 + θ 2 ⇒ y = π 4 + θ 2 ⇒ y = π 4 + tan - 1 x 2 Differentiate both sides with respect to x , we get ⇒ d y d x = 1 2 1 + x 2

Maharashtra Board Solutions for Chapter: Differentiation, Exercise 6: MISCELLANEOUS EXERCISE 1

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Differentiation, Exercise 6: MISCELLANEOUS EXERCISE 1

Attempt the practice questions on Chapter 1: Differentiation, Exercise 6: MISCELLANEOUS EXERCISE 1 with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 2 Standard 12 solutions are prepared by Experienced Embibe Experts.

Questions from Maharashtra Board Solutions for Chapter: Differentiation, Exercise 6: MISCELLANEOUS EXERCISE 1 with Hints & Solutions

EASY

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 3

Suppose that the functions $f$ and $g$ and their derivatives with respect to $= f' (g (0)) . g' (0)$ have the following values at $x = 0$ and $x = 1$ .

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$0$	$1$	$1$	$5$	$\frac{1}{3}$
$1$	$3$	$- 4$	$- \frac{1}{3}$	$- \frac{8}{3}$

The derivative of $f [g (x)] w . r . t . x$ at $x = 0$ is....

EASY

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 3

Suppose that the functions $f$ and $g$ and their derivatives with respect to $x$ have the following values at $x = 0$ and $x = 1$ .

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$0$	$1$	$1$	$5$	$\frac{1}{3}$
$1$	$3$	$- 4$	$- \frac{1}{3}$	$- \frac{8}{3}$

The derivative of $g [f (x)] w . r . t . x$ at $x = 0$ is...

EASY

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 3

Suppose that the functions $f$ and $g$ and their derivatives with respect to $x$ have the following values at $x = 0$ and $x = 1$ .

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$0$	$1$	$1$	$5$	$\frac{1}{3}$
$1$	$3$	$- 4$	$- \frac{1}{3}$	$- \frac{8}{3}$

The value of ${[\frac{d}{d x} {[x^{10} + f (x)]}^{- 2}]}_{x = 1}$ is....

EASY

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 3

Suppose that the functions $f$ and $g$ and their derivatives with respect to $x$ have the following values at $x = 0$ and $x = 1$ .

$x$	$f (x)$	$g (x)$	$f^{'} (x)$	$g^{'} (x)$
$0$	$1$	$1$	$5$	$\frac{1}{3}$
$1$	$3$	$- 4$	$- \frac{1}{3}$	$- \frac{8}{3}$

The derivative of $f [(x + g (x))] w . r . t . x$ at $x = 0$ is

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 4

Differentiate the following $w . r . t . x$

$\tan^{- 1} [\frac{\sqrt{x} (3 - x)}{1 - 3 x}]$

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 4

Differentiate the following $w . r . t . x$

$\cos^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{2})$

EASY

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 4

Differentiate the following $w . r . t . x$

$\tan^{- 1} (\frac{x}{1 + 6 x^{2}}) + \cot^{- 1} (\frac{1 - 10 x^{2}}{7 x})$

MEDIUM

12th Maharashtra Board

IMPORTANT

Mathematics & Statistics (Arts & Science) Part 2 Standard 12>Chapter 1 - Differentiation>MISCELLANEOUS EXERCISE 1>Q 4

Differentiate the following $w . r . t . x$

$\tan^{- 1} [\sqrt{\frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x}}]$

Maharashtra Board Solutions for Chapter: Differentiation, Exercise 6: MISCELLANEOUS EXERCISE 1

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Differentiation, Exercise 6: MISCELLANEOUS EXERCISE 1

Questions from Maharashtra Board Solutions for Chapter: Differentiation, Exercise 6: MISCELLANEOUS EXERCISE 1 with Hints & Solutions

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