Let $f$ be a function such that $f(x) = f(1-x)$ for all real numbers $x$. If $f$ is differentiable everywhere, then $f'(0)$

Question

I do not know how to solve this question:

Could anyone help me please?

Answer 1

Differentiating both sides of $f(x)=f(1-x)$ leads to $$ f^{\prime}(x)=-f^{\prime}(1-x)$$ by the chain rule. Now plug in $x=0$.