# Chain Rule Derivative

• Oct 10th 2007, 05:10 PM
Happy
Chain Rule Derivative
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?
• Oct 10th 2007, 05:46 PM
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(Doh)
• Oct 10th 2007, 06:11 PM
Jhevon
Quote:

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(Doh)

recall that $\sin 2u = 2 \sin u \cos u$

$\Rightarrow \sin u \cos u = \frac 12 \sin 2u$

Here you have $u = 2x$
• Oct 10th 2007, 09:04 PM
Happy
Ah, I see. Thank you for your help. (Handshake)