Find the derivative of y = x 2 + 3 x 2 - 5 w.r.t x .

Given, y = x 2 + 3 x 2 - 5 Differentiate y w.r.t x on each side, we get: d y d x = d d x x 2 + 3 x 2 - 5 ⇒ d y d x = x 2 - 5 d d x x 2 + 3 - x 2 + 3 d d x x 2 - 5 x 2 - 5 2 ∵ d d x u v = v · d u d x - u · d v d x v 2 ⇒ d y d x = x 2 - 5 2 x - x 2 + 3 2 x x 2 - 5 2 ⇒ d y d x = 2 x 3 - 10 x - 2 x 3 - 6 x x 2 - 5 2 ⇒ d y d x = - 16 x x 2 - 5 2 Hence, the derivative of y = x 2 + 3 x 2 - 5 is - 16 x x 2 - 5 2 .

Find the derivative of y = x + 5 x - 5 w.r.t x .

Given, y = x + 5 x - 5 Differentiate y w.r.t x on each side, we get: d y d x = d d x x + 5 x - 5 ⇒ d y d x = x - 5 d d x x + 5 - x + 5 d d x x - 5 x - 5 2 ∵ d d x u v = v · d u d x - u · d v d x v 2 ⇒ d y d x = x - 5 1 2 x - x + 5 1 2 x x - 5 2 ⇒ d y d x = x · 1 2 x - 5 · 1 2 x - x · 1 2 x + 5 · 1 2 x x - 5 2 ⇒ d y d x = - 5 · 1 2 x - 5 · 1 2 x x - 5 2 ⇒ d y d x = 1 x - 5 x - 5 2 ⇒ d y d x = - 5 x x - 5 2 Hence, the derivative of y = x + 5 x - 5 is - 5 x x - 5 2 .

Find the derivative of y = x e x x + e x w.r.t x .

Given, y = x e x x + e x Differentiate y w.r.t x on each side, we get: d y d x = d d x x e x x + e x ⇒ d y d x = x + e x d d x x e x - x e x d d x x + e x x + e x 2 ∵ d d x u v = v · d u d x - u · d v d x v 2 ⇒ d y d x = x + e x x d d x e x + e x d d x x - x e x d d x x + d d x e x x + e x 2 ∵ d d x u v = u · d v d x + v · d u d x ⇒ d y d x = x + e x x e x + e x - x e x 1 + e x x + e x 2 ∵ d e x d x = e x ⇒ d y d x = x 2 e x + x e 2 x + x e x + e 2 x - x e 2 x - x e x x + e x 2 ⇒ d y d x = x 2 e x + e 2 x x + e x 2 ⇒ d y d x = e x x 2 + e x x + e x 2 Hence, the derivative of y = x e x x + e x is e x x 2 + e x x + e x 2 .

Differentiate y = 5 e x - 4 3 e x - 2 w.r.t x .

Given y = 5 e x - 4 3 e x - 2 ⇒ d y d x = 3 e x - 2 d d x 5 e x - 4 - 5 e x - 4 d d x 3 e x - 2 3 e x - 2 2 ⇒ d y d x = 3 e x - 2 5 e x - 5 e x - 4 3 e x 3 e x - 2 2 ⇒ d y d x = 15 e 2 x - 10 e x - 15 e 2 x + 12 e x 3 e x - 2 2 ∴ d y d x = 2 e x 3 e x - 2 2

If f x = a sin x - b cos x , f ' π 4 = 2 and f ' π 6 = 2 then find f x .

Given f x = a sin x - b cos x , f ' π 4 = 2 and f ' π 6 = 2 Given f ( x ) = asin x - b cos x - - - ( 1 ) f ' ( x ) = a cos x - b ( - sin x ) ⟹ f ' x = a cos x + b sin x - - - ( 2 ) given f ' π 4 = 2 ⟹ acos π 4 + b sin π 4 = 2 ⟹ a × 1 2 + b × 1 2 = 2 ⟹ 1 2 ( a + b ) = 2 ⟹ a + b = 2 - - - ( 3 ) given f ' π 6 = 2 ⟹ acos π 6 + b sin π 6 = 2 ⟹ a × 1 2 + b × 3 2 = 2 ⟹ 1 2 ( a + 3 b ) = 2 ⟹ a + 3 b = 4 - - - ( 4 ) Subtracting equation (3) from equation (4) , we get a ( 3 - 1 ) = 2 ⟹ a = 2 ( 3 - 1 ) ⟹ a = 2 ( 3 + 1 ) ( 3 - 1 ) ( 3 + 1 ) ⟹ a = 2 ( 3 + 1 ) ( 3 ) 2 - 1 2 ⟹ a = 2 ( 3 + 1 ) ( 3 - 1 ) ⟹ a = 2 ( 3 + 1 ) 2 ⟹ a = 3 + 1 - - - ( 5 ) Substituting value of a in equation 3 , we get 3 + 1 + b = 2 ⟹ b = 1 - 3 - - - ( 6 ) Therefore, f ( x ) = ( 3 + 1 ) sin x + ( 1 - 3 ) cos x

Differentiate y = sin x x 2 + 2 with respect to x.

Given y = sin x x 2 + 2 d y d x = x 2 + 2 d d x sin x - sin x d d x x 2 + 2 x 2 + 2 2 d y d x = x 2 + 2 ( - cos x ) - sin x ( 2 x ) x 2 + 2 2 d y d x = - x 2 cos x - 2 cos x - 2 x s in x x 2 + 2 2

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Mathematics and Statistics (Arts & Science) Part 2>Chapter 9 - Diffrentiation>EXERCISE

Maharashtra Board Solutions for Chapter: Diffrentiation, Exercise 2: EXERCISE

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Diffrentiation, Exercise 2: EXERCISE

Attempt the practice questions on Chapter 9: Diffrentiation, Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. Mathematics and Statistics (Arts & Science) Part 2 solutions are prepared by Experienced Embibe Experts.