Chain Rule Derivative

**Happy** · October 10th 2007, 04:10 PM

The question is find dy/dx of $\frac{1}{4}\sin^2(2x)$ .

which I rewrote as $\frac{1}{4}(\sin2x)^2$

then, using the chain rule, $\frac{1}{2}\sin2x\cos2x$ .

But the book gives the answer $\frac{1}{2}\sin4x$ .

Is my answer wrong? Or do I just need to simplify it further?

**Happy** · October 10th 2007, 04:46 PM

Never mind, I've got a new answer that is right, and I checked it by graphing both my new equation and the equation the books gives w/ my TI-89 and the graphs are identical.

Now, the real derivative I got is $\sin2x\cos2x$ .

Now I just need to convert this into the equation the book wants, which is $\frac{1}{2}\sin4x$ .

I'm guessing its a random trig identity...but can't find which one

**Jhevon** · October 10th 2007, 05:11 PM

Originally Posted by Happy

Never mind, I've got a new answer that is right, and I checked it by graphing both my new equation and the equation the books gives w/ my TI-89 and the graphs are identical.

Now, the real derivative I got is $\sin2x\cos2x$ .

Now I just need to convert this into the equation the book wants, which is $\frac{1}{2}\sin4x$ .

I'm guessing its a random trig identity...but can't find which one