# Thread: Calculus - Chain Rule Derivative

1. ## Calculus - 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?

2. Hello, Happy!

Find $\frac{dy}{dx}$ of: . $\frac{1}{4}\sin^2(2x)$.

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

then, using the chain rule, $\frac{1}{2}\sin2x\cos2x$ . . . . You dropped a "2"
$\frac{dy}{dx}\:=\;\frac{1}{4}\cdot2\sin2x\cos2x\cd ot2 \:=\:\sin2x\cos2x$

Then they used the identity: . $\sin2\theta \:=\:2\sin\theta\cos\theta$

3. Oh yes! Thank you, I forgot to "chain" the derivative of the 2x to the rest of the chained derivative.