1
$\begingroup$

$(x^2+2x+1)^3$
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!

  • 0
    @copper: it was $y=u^3$ (it should be separated from 2x+2 and written... :-))2012-04-06

2 Answers 2

1

by Chain rule :

f'(x)=3(x^2+2x+1)^2 \cdot(x^2+2x+1)'=3\cdot(x+1)^4\cdot(2x+2)=

$=6\cdot(x+1)^4\cdot(x+1)=6(x+1)^5$

2

Hint

$\begin{align}\dfrac{du}{dx}&=2x+2=2(x+1)\\x^2+2x+1 &=(x+1)^2\\(x^a)^b&=x^{ab}\end{align}$