Derivative of cos of sin

1. Apr 28, 2005

monet A

Hi I'm new here, great forum!
On another thread that I can't find I found a message from jamesrc about differentiating cos of sinx that says:

df(g(x))/dx = df(g)/dg * dg(x)/dx by the chain rule, this part I get and I thank james for the help the thing that confuses me is why df(g)/dg = cos(cosx).
Aren't we differentiating df(g) wrt (g) which would give -sin(sinx). What am I missing?

2. Apr 28, 2005

Staff: Mentor

You're not missing anything; you are correct: $d/dx [\cos (\sin x)] = - \sin (\sin x) \cos x$

3. Apr 28, 2005

monet A

Oh thanks Al! :rofl:
I just found the thread I read and realised that james was differentiating sin of cos not cos of sin, lol.

4. Apr 28, 2005

Staff: Mentor

I'm glad you cleared jamesrc's good name!

And welcome to PF, by the way.