1. Differentiating sin(2x)^2

I'm trying to differentiate $\displaystyle \sin^2(2x)$ 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.

2. 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)

3. Re: Differentiating sin(2x)^2

Argh. Trig identities getting me again. I really gotta learn those better. Thank you.

4. Re: Differentiating sin(2x)^2

Originally Posted by Paze
I'm trying to differentiate $\displaystyle \sin^2(2x)$ 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.
You should know that \displaystyle \displaystyle \begin{align*} \cos{(2X)} \equiv 1 - 2\sin^2{(X)} \end{align*}, which means \displaystyle \displaystyle \begin{align*} \sin^2{(X)} \equiv \frac{1}{2} - \frac{1}{2}\cos{(2X)} \end{align*}. So that means

\displaystyle \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*}

5. Re: Differentiating sin(2x)^2

Thanks. I see now that my trig course in school was sub-par. Does anyone know of a trig-course online that nails all identities?

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(sin2x)^2

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