My answer:
Suppose that f: ℝ $\rightarrow$ ℝ is an even function, that is differentiable everywhere. If f is an even function, then we have that f(x)= f(-x). We now take the derivative using the chain rule $$f'(x)= \frac{df}{dx}$$ $$=\frac{d}{dx}[f(x)]$$ $$=\frac{d}{dx}[f(-x)]$$ $$= f'(-x)\cdot (-x)'$$ $$= f'(-x)(-1)$$ $$=-f'(-x)$$ Since that f '(x) = -f '(-x) we have shown that the derivative of an even function that is differentiable everywhere, is odd.
My question: Why did my TA not accept my solution? I was asked to use the chain rule and did just that. I don't understand what I did wrong.
The answer is correct, if you want to use the chain rule explicitly you should define $\phi(x)=-x$ then you know $\phi '(x)=-1$. Now we can apply the chain rule within the proof:
$$f'(-x)=f'(\phi(x))=-f'(\phi(x))\phi'(x)=-(f\circ\phi)'(x)=-(f\circ\phi)'(-x)=-f'(x)$$
When the last equation is due to the fact that $f(\phi (x))=f(x)$ since $f$ is even.