# Simplifying wrong or completely wrong?

iRaid

## Homework Statement

Find dy/dx.
$y=sin^{2}(3x-2)$

## The Attempt at a Solution

Using the chain rule,
$y=sin^{2}u$
$y'=2sinucosu$
$u=3x-2$
$u'=3$

$\frac{dy}{dx}=6sin(3x-2)cos(3x-2)$

$3sin(6x-4)$

What am I missing here ?

## Answers and Replies

Mentor
Your answer is correct, and so is the other answer you show. They are using the identity sin(2A) = 2sin(A)cos(A). In your problem A = 3x - 2.

Homework Helper
The identity

sin2(u)=(1/2)[1-cos(2x)]
so
[sin2(u)]'=[(1/2)[1-cos(2x)]]'=sin(2u)u'

iRaid
Ah, I remember that now, now I'll remember that identity, thank you.