Suppose $f(x)=\pm f(-x)$ . We need to show that $f'(x)=\mp f'(-x)$ . To do this, let us define the auxiliary function $m:\mathbb{R}\to \mathbb{R}$ , $m(x)=-x$ . The condition on $f$ is then $f(x) = \pm (f\circ m)(x)$ . Using the chain rule, we have that \begin{eqnarray*} f'(x) &=& \pm(f\circ m)'(x) \\ &=& \pm f'\big(m(x)\big) m'(x)\\ &=& \mp f'(-x), \end{eqnarray*}and the claim follows.