Find derivative of the following function 1 + 1 x x in their natural domain.

Let the given function be y = 1 + 1 x x Applying logarithm on both sides, we get log y = x log 1 + 1 x Differentiating on both sides with respect to x . 1 y d y d x = x d d x log 1 + 1 x + log 1 + 1 x d d x x We know that, d d x log n x = 1 x and d d x x n = n x n - 1 . ⇒ 1 y d y d x = x 1 1 + 1 x - 1 x 2 + log 1 + 1 x ⇒ d y d x = 1 + 1 x x log 1 + 1 x - 1 x + 1 Thus the derivative of the given function is 1 + 1 x x log 1 + 1 x - 1 x + 1

Find derivative of the following function ( 1 + x ) x 2 in their natural domain.

Let the given function be y = ( 1 + x ) x 2 Applying logarithm on both sides, we get log y = x 2 log ( 1 + x ) Differentiating on both sides We know that, d d x u · v = u d v d x + v d u d x . ⇒ x 2 d d x log ( 1 + x ) + log ( 1 + x ) d d x x 2 We know that, d d x log n x = 1 x and d d x x n = n x n - 1 . ⇒ 1 y d y d x = x 2 1 ( 1 + x ) 1 2 x + log ( 1 + x ) 2 x ⇒ d y d x = ( 1 + x ) x 2 x 3 2 2 ( 1 + x ) + log ( 1 + x ) 2 x Thus the derivative of the given function is ( 1 + x ) x 2 x 3 2 2 ( 1 + x ) + log ( 1 + x ) 2 x