# Differentiating sin(2x)^2

• May 9th 2013, 08:11 PM
Paze
Differentiating sin(2x)^2
I'm trying to differentiate $\sin^2(2x)$ but it's not going very well.

To me it seems natural to use the chain rule and get 2sin(2x) and then multiplied with the derivative of sin(2x) which is 2cos(2x). So my gut tells me the answer is 2sin(2x)*2cos(2x) but that's obviously wrong.

Any insight is well appreciated. Thanks.
• May 9th 2013, 08:27 PM
ibdutt
Re: Differentiating sin(2x)^2
You are right
d/dx ( sin^2 ( 2x)) = 2 sin(2x) * d/dx ( sin (2x) ) = 2 sin (2x) * cos (2x) * d/dx ( 2x) = 2 sin (2x) * cos (2x) * 2 = 2* 2 sin(2x) cos (2x) = 2 sin (4x)
• May 9th 2013, 08:52 PM
Paze
Re: Differentiating sin(2x)^2
Argh. Trig identities getting me again. I really gotta learn those better. Thank you.
• May 9th 2013, 09:19 PM
Prove It
Re: Differentiating sin(2x)^2
Quote:

Originally Posted by Paze
I'm trying to differentiate $\sin^2(2x)$ but it's not going very well.

You should know that \displaystyle \begin{align*} \cos{(2X)} \equiv 1 - 2\sin^2{(X)} \end{align*}, which means \displaystyle \begin{align*} \sin^2{(X)} \equiv \frac{1}{2} - \frac{1}{2}\cos{(2X)} \end{align*}. So that means
\displaystyle \begin{align*} y &= \sin^2{(2x)} \\ &= \frac{1}{2} - \frac{1}{2}\cos{(4x)} \\ \\ \frac{dy}{dx} &= 0 - \frac{1}{2} \left( 4 \right) \left[ -\sin{(4x)} \right] \\ &= 2\sin{(4x)} \end{align*}