Differentiate with respect to x : 2 x + 3

Given, 2 x + 3 Let y = 2 x + 3 . . . i Now, let u = 2 x + 3 . . . . i i ⇒ y = u . . . . i i i [From i ] From equation i i , we have d u d x = d d x 2 x + 3 = d d x 2 x + d d x 3 x 0 = 2 d d x x + 3 d d x x 0 = 2 × 1 x 1 - 1 + 3 × 0 x 0 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 2 . . . . i v From equation i i i , d y d u = d d u u 1 2 = 1 2 u 1 2 - 1 = 1 2 u - 1 2 = 1 2 u . . . . v We know that d y d x = d y d u × d u d x = 1 2 u × 2 [using i v , and v ] = 1 u Hence, d d x 2 x + 3 = 1 2 x + 3

Differentiate with respect to x : 2 x + 3

Given, 2 x + 3 Let y = 2 x + 3 . . . i Now, let u = 2 x + 3 . . . . i i ⇒ y = u . . . . i i i [From i ] From equation i i , we have d u d x = d d x 2 x + 3 = d d x 2 x + d d x 3 x 0 = 2 d d x x + 3 d d x x 0 = 2 × 1 x 1 - 1 + 3 × 0 x 0 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 2 . . . . i v From equation i i i , d y d u = d d u u 1 2 = 1 2 u 1 2 - 1 = 1 2 u - 1 2 = 1 2 u . . . . v We know that d y d x = d y d u × d u d x = 1 2 u × 2 [using i v , and v ] = 1 u Hence, d d x 2 x + 3 = 1 2 x + 3

Differentiate with respect to x : 2 x 2 - 1

Given, 2 x 2 - 1 Let y = 2 x 2 - 1 . . . i Now, let u = 2 x 2 - 1 . . . . i i ⇒ y = u . . . . i i i [From i ] From equation i i , we have d u d x = d d x 2 x 2 - 1 = d d x 2 x 2 - d d x 1 x 0 = 2 d d x x 2 - d d x x 0 = 2 × 2 x 2 - 1 - 0 × x 0 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 4 x . . . . i v From equation i i i , d y d u = d d u u 1 2 = 1 2 u 1 2 - 1 = 1 2 u - 1 2 = 1 2 u . . . . v We know that d y d x = d y d u × d u d x = 1 2 u × 4 x [using i v , and v ] = 2 x u Hence, d d x 2 x 2 - 1 = 2 x 2 x 2 - 1

Differentiate with respect to x : x 3 - 5 x

Given, x 3 - 5 x Let y = x 3 - 5 x . . . i Now, let u = x 3 - 5 x . . . . i i ⇒ y = u . . . . i i i [From i ] From equation i i , we have d u d x = d d x x 3 - 5 x = d d x x 3 - d d x 5 x = d d x x 3 - 5 d d x x = 3 x 3 - 1 - 5 × 1 x 1 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 3 x 2 - 5 . . . . i v From equation i i i , d y d u = d d u u 1 2 = 1 2 u 1 2 - 1 = 1 2 u - 1 2 = 1 2 u . . . . v We know that d y d x = d y d u × d u d x = 1 2 u × 3 x 2 - 5 [using i v , and v ] = 3 x 2 - 5 2 u Hence, d d x x 3 - 5 x = 3 x 2 - 5 2 x 3 - 5 x

Differentiate with respect to x : 5 - 2 x 3

Given, 5 - 2 x 3 Let y = 5 - 2 x 3 . . . i Now, let u = 5 - 2 x . . . . i i ⇒ y = u 3 . . . . i i i [From i ] From equation i i , we have d u d x = d d x 5 - 2 x = d d x 5 x 0 - d d x 2 x = 5 d d x x 0 - 2 d d x x = 5 × 0 x 0 - 1 - 2 × 1 x 1 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = - 2 . . . . i v From equation i i i , d y d u = d d u u 3 = d d u u 1 3 = 1 3 u 1 3 - 1 = 1 3 u - 2 3 = 1 3 u 2 3 = 1 3 u 2 3 . . . . v We know that d y d x = d y d u × d u d x = 1 3 u 2 3 × - 2 [using i v , and v ] = - 2 3 u 2 3 Hence, d d x 5 - 2 x 3 = - 2 3 5 - 2 x 2 3

Differentiate with respect to x : 2 3 x + 1

Given, 2 3 x + 1 Let y = 2 3 x + 1 . . . i Now, let u = 3 x + 1 . . . . i i ⇒ y = 2 u . . . . i i i [From i ] From equation i i , we have d u d x = d d x 3 x + 1 = d d x 3 x + d d x 1 x 0 = 3 d d x x + d d x x 0 = 3 × 1 x 1 - 1 + 0 x 0 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 3 . . . . i v From equation i i i , d y d u = d d u 2 u = 2 d d u u 1 2 = 2 × 1 2 u 1 2 - 1 = u - 1 2 = 1 u 1 2 = 1 u . . . . v We know that d y d x = d y d u × d u d x = 1 u × 3 [using i v , and v ] = 3 u Hence, d d x 2 3 x + 1 = 3 3 x + 1

Differentiate with respect to x : 1 2 x - 5

Given, 1 2 x - 5 Let y = 2 x - 5 - 1 2 . . . i Now, let u = 2 x - 5 . . . . i i ⇒ y = u - 1 2 . . . . i i i [From i ] From equation i i , we have d u d x = d d x 2 x - 5 = d d x 2 x - d d x 5 x 0 = 2 d d x x - 5 d d x x 0 = 2 × 1 x 1 - 1 - 5 × 0 x 0 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = 2 . . . . i v From equation i i i , d y d u = d d u u - 1 2 = - 1 2 u - 1 2 - 1 = - 1 2 u - 3 2 = - 1 2 u 3 . . . . v We know that d y d x = d y d u × d u d x = - 1 2 u 3 × 2 [using i v , and v ] = - 1 u 3 Hence, d d x 1 2 x - 5 = - 1 2 x - 5 3

Differentiate with respect to x : 6 2 - 3 x 3

Given, 6 2 - 3 x 3 Let y = 6 2 - 3 x 3 . . . i Now, let u = 2 - 3 x . . . . i i ⇒ y = 6 u 3 . . . . i i i [From i ] From equation i i , we have d u d x = d d x 2 - 3 x = d d x 2 x 0 - d d x 3 x = 2 d d x x 0 - 3 d d x x = 2 × 0 x 0 - 1 - 3 × 1 x 1 - 1 ∵ d d x x n = n x n - 1 , n ∈ R = - 3 . . . . i v From equation i i i , d y d u = d d u 6 u 3 = 6 d d u 1 u 3 = 6 d d u u - 1 3 = 6 × - 1 3 u - 1 3 - 1 = - 2 u - 4 3 . . . . v We know that d y d x = d y d u × d u d x = - 2 u - 4 3 × - 3 [using i v , and v ] = 6 u - 4 3 = 6 × 1 u 4 3 = 6 u 4 3 Hence, d d x 6 2 - 3 x 3 = 6 2 - 3 x 4 3