Differentiate respect to $x$ $(x^2+2x+1)^3$ 
let u=$x^2+2x+1$   $\frac{du}{dx} = 2x$ or $2x+2$ $\frac{dy}{dx}=3u^2 $ 
if $\frac{du}{dx} = 2x$ then 
$3(x^2+2x+1)^2 (2x)$ 
answer is $6x(x^2+2x+1) $  
Or if $\frac{du}{dx} = 2x+2$ then 
$3(x^2+2x+1)^2 (2x+2)$ 
$6x(x^2+2x+1)+2$
however the right answer is $6(x+1)^5$
can please help me out? 
thanks in advance!

by Chain rule :
$f'(x)=3(x^2+2x+1)^2 \cdot(x^2+2x+1)'=3\cdot(x+1)^4\cdot(2x+2)=$
$=6\cdot(x+1)^4\cdot(x+1)=6(x+1)^5$
