Question

Question #7fdfd

Answer 1

(1) `0`

Explanation:

Rewriting the original function,

`f(x)=g(x)/2+g(-x)/2+2h(x)+2h(-x)`

Differentiating, and using the chain rule on `g(-x)` and `h(-x)`:

`f'(x)=(g'(x))/2-(g'(-x))/2+2h'(x)-2h'(-x)`

So:

`f'(0)=(g'(0))/2-(g'(0))/2+2h'(0)-2h'(0)`

`f'(0)=0`