Equivalence Relation between Derivative Being Odd and Function Being Even

Question

In the exercise, I am required to prove that

$f'$ is odd $\iff$ $f$ is even

Moving from right to left was pretty trivial, however, I couldn't move from left to right. Note that we can only use very basic things like the definition of the derivative and rules of differentiation (No mean value theorem or any other theorems.)

I cannot even see the left implying the right, but I hope I will see it after you guys explain how. Thanks in advance.

Answer 1

Use contraposition. Suppose $f(-x)\ne f(x)$, then $-f'(-x)\ne f'(x)$.

Here I used the fact that $f'=g'$ implies that $f$ and $g$ differ only by a constant. If $f(x)-g(x)=h(x)$, then $f'(x)-g'(x)=h'(x)$. This implies $h'(x)=0$. $h(x)$ is then a constant. You might use k_g's suggestion to prove this basic theorem to proceed.

So we also need to note the special case where $f$ and $g$ only differ by a constant, let $f(-x)=f(x)+C$, in this case $-f'(-x)=f'(x)$. But it also implies $f(0)=f(0)+C$, i.e., $C=0$.

Answer 2

Hint: $f$ even $\iff$ $f(x)=f(-x)$ for all $x$. Then $$f'(x)=\frac{d}{dx}f(-x)=\cdots$$ (apply chain rule)

Answer 3

Consider the function $g(x)=f(x)-f(-x)$. If $f'(-x)=-f'(x)$, then $$ \begin{align} g'(x) &=f'(x)+f'(-x)\tag{1}\\ &=f'(x)-f'(x)\tag{2}\\ &=0\tag{3} \end{align} $$ Explanation:
$(1)$: chain rule and the definition of $g$
$(2)$: $f'$ is odd
$(3)$: $a-a=0$

Thus, $g(x)$ is constant. Since $g(0)=f(0)-f(0)=0$, $g(x)=0$ for all $x$. Thus, $$ f(x)=f(-x) $$ Therefore, $f$ is even.