1. ## f(x)=cos^2(sinx)

I think I know how to differentiate it, this is what I did:

f(x)=cos^2(sinx)
f'(x)=[(2)(cos)(-sin)(sinx)](cosx)
=-2cos(sinx)[sin(sinx)](cosx)

Why are there so many brackets being used in the answer? Couldn't I have went from my step 2 to:

=-2cos(sinx)sin(sinx)(cosx)?

2. Originally Posted by kmjt
I think I know how to differentiate it, this is what I did:

f(x)=cos^2(sinx)
f'(x)=[(2)(cos)(-sin)(sinx)](cosx)
=-2cos(sinx)[sin(sinx)](cosx)

Why are there so many brackets being used in the answer? Couldn't I have went from my step 2 to:

=-2cos(sinx)sin(sinx)(cosx)?
You need to use the chain rule twice.

$y = [\cos{(\sin{x})}]^2$.

To find $\frac{dy}{dx}$ let $u = \cos{(\sin{x})}$ so that $y = u^2$.

$\frac{dy}{du} = 2u = 2\cos{(\sin{x})}$.

To find $\frac{du}{dx}$ let $v = \sin{x}$ so that $u = \cos{v}$.

$\frac{dv}{dx} = \cos{x}$

$\frac{du}{dv} = -\sin{v} = -\sin{(\sin{x})}$.

Therefore $\frac{du}{dx} = -\cos{x}\sin{(\sin{x})}$.

Thus $\frac{dy}{dx} = -2\cos{(\sin{x})}\cos{x}\sin{(\sin{x})}$.

3. So my answer was actually right too?

4. Wouldn't that be the same as my answer of:

=-2cos(sinx)sin(sinx)(cosx)?