Solving h'(x) = f'[g(x)] * g'(x)

[Dustobusto]

Homework Statement

h(x) = f[g(x)]

h'(x) = f'[g(x)] * g'(x)

Homework Equations

h(x) = sin(-x)

The Attempt at a Solution

So, this one is pretty simple, except I just want to confirm something. When I do it it, it looks like this:

The derivative of sin = cos,

so you have

h(x) = cos(-x)

then you multiply the outside by g'(x). The derivative of -x is negative one. So it's

h(x) = cos(-x) * -1

h(x) = -cos(-x)

h'(x) = cos(x)

But the computer program bagatrix insists the final answer is

h'(x) = -cos(x)

Am I wrong, and if so, where did I go wrong?

 

[Averki]

Cosine is an even function.

 

[Dustobusto]

Cosine is an even function.

As in the opposite angle identity?

sin(-x) = -sin(x)

and

cos(-x) = cos(x) ?

 

[MarneMath]

Never heard it called that, but yes to the above. A good way to have checked this is to look at the unit circle :]