f ' ( x ) = 1 ( x + 1 ) 2
Explanation:
differentiating from first principles
f ' ( x ) = lim h → 0
f ( x + h ) − f ( x ) h
f ' ( x ) = lim h → 0
x + h x + h + 1 − x x + 1 h
the aim now is to eliminate h from the denominator
f ' ( x ) = lim h =0
( x + h ) ( x + 1 )− x ( x + h + 1) h ( x + 1 ) ( x + h + 1 )
f ' ( x ) = lim h → 0
x 2 + h x + x + h − x 2 − h x − x h ( x + 1 ) ( x+h + 1 )
f ' ( x ) = lim h → 0
h 1 h 1 ( x + 1 ) ( x + h +1 )
f ' ( x ) = 1 ( x + 1 ) 2