Differentiate with respect to x . tan - 1 x 3

Let y = tan - 1 x 3 Differentiate both sides, with respect to x , we get d y d x = d d x tan - 1 x 3 ⇒ d y d x = 1 1 + x 3 2 × d d x x 3 ⇒ d y d x = 1 3 1 + x 2 9 ⇒ d y d x = 3 9 + x 2

Differentiate with respect to x . tan - 1 2 x x + 1

Let y = tan - 1 2 x x + 1 Differentiate both sides, with respect to x , we get d y d x = d d x tan - 1 2 x x + 1 ⇒ d y d x = 1 1 + 2 x x + 1 2 × d d x 2 x x + 1 ⇒ d y d x = ( x + 1 ) 2 ( x + 1 ) 2 + 4 x 2 × ( x + 1 ) · d d x 2 x - ( 2 x ) · d d x ( x + 1 ) x + 1 2 [Applying quotient rule] ⇒ d y d x = ( x + 1 ) 2 x 2 + 2 x + 1 + 4 x 2 × 2 x + 1 - 2 x ( x + 1 ) 2 ⇒ d y d x = 2 5 x 2 + 2 x + 1

Differentiate with respect to x . tan - 1 2 x x

Let y = tan - 1 2 x x Differentiate both sides, with respect to x , we get d y d x = d d x tan - 1 2 x x ⇒ d y d x = x · d d x tan - 1 2 x - tan - 1 2 x · d d x ( x ) x 2 [Applying quotient rule] ⇒ d y d x = x 1 + 2 x 2 × d d x ( 2 x ) - tan - 1 2 x x 2 ⇒ d y d x = 2 x 1 + 4 x 2 - tan - 1 2 x x 2 ⇒ d y d x = 2 x - ( 1 + 4 x 2 ) tan - 1 2 x x 2 ( 1 + 4 x 2 )