# Thread: Simple Chain Rule Question

1. ## Simple Chain Rule Question

Hello!

I have a quick question about a question I'm not too sure about...

y = 2sin^2(x)

What's the derivative?

I'm thinking 4sin(x), am I right?

2. Originally Posted by lionpants
Hello!

I have a quick question about a question I'm not too sure about...

y = 2sin^2(x)

What's the derivative?

I'm thinking 4sin(x), am I right?
dont forget to chain it....

$\displaystyle y=2sin^2x=2(sin\:x)^2......y'=4(sin\:x)(cos\:x)=2s in\:2x$

3. Well you have a chain rule question. So just use the chain rule, don't guess your answers.

Firstly, the chain rule $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

we have $\displaystyle y = 2 \sin^2 x$

so let $\displaystyle u = \sin x$
$\displaystyle \frac{du}{dx} = \cos x$

$\displaystyle y = 2u^2$
$\displaystyle \frac{dy}{du} = 4u$
$\displaystyle \Rightarrow \frac{dy}{du} = 4 \sin x$

then $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
$\displaystyle \Rightarrow \frac{dy}{dx} = 4 \sin x \cdot \cos x$

4. Thanks for the help, it helped a lot!

I have one more question though...

I have this... y = sin(3x)cos(4x)

I've worked it out using product rule and chain rule and come out with...

(3cos(3x) * cos(4x)) + (sin(3x) * -4sin(4x))

5. It does exist a formula to split the original product into a sum. Have you seen it before?