Prove that if $f$ is odd and differentiable on $I$ ,then $f^\prime$ is even on $I$

Question

Suppose that $I=(-a,a)$ for some $a>0$. A function $f:I \to \mathbb{R}$ is said to be even if and only if $f(-x)=f(x)$ for all $x \in I$, and said to be of odd if and only if $f(-x)=-f(x)$ for all $x \in I$

Prove that if $f$ is odd and differentiable on $I$ ,then $f^\prime$ is even on $I$

since $f$ is differentiable on $I$, by definition for all $b \in I$

$$f^{\prime}(b) :=\lim_{x\to b} \frac{f(x)-f(b)}{x-b}=\lim_{x\to b} \frac{-f(x)+f(b)}{-x+b}=\lim_{-x\to -b} \frac{f(-x)-f(-b)}{(-x)-(-b)}=f^{\prime}(-b)$$
for all $b \in I$ thus $f^{\prime}(x)=f^{\prime}(-x)$

Is that correct? I know taking derivative of $f(-x)=-f(x)$ is also correct but I want to check this