The Even and Odd Nature of the Derivatives of Functions

This problem was inspired from this question .

Given a function $f(x)$ call its derivative $g(x)$ .

By definition, $g(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Then, $g(-x)=\lim_{h\to 0}\frac{f(-x+h)-f(-x)}{h}$

For even function $f(x)$ , $f(x)=f(-x)$ . Thus,

$g(-x)=\lim_{h\to 0}\frac{f(x-h)-f(x)}{h}$

Define $k=-h$ . As $h\to 0$ , $k\to 0$ . Then,

$g(-x)=\lim_{k\to 0}-\frac{f(x+k)-f(x)}{k}=-g(x)$

Thus, for even $f(x)$ , its derivative $g(x)$ will be odd.

A similar proof can be used to show that for odd $f(x)$ , its derivative $g(x)$ will be even.

However, it is not always true that $g(x)$ will be neither even nor odd when $f(x)$ is neither even nor odd. Consider the function $f(x)=x^{3}+1$ . Its derivative $3x^{2}$ is even.

We can easily construct more counterexamples. If $g(x)$ is an even function, its integral will be $f(x)+C$ for some arbitrary constant $C$ . Given a suitable choice of $C$ , $f(x)+C$ will be neither even nor odd.

Bonus

The above shows that the derivative of a function that is neither even nor odd can be even. Does a function exist such that, even though it is continuous and differentiable everywhere and is neither even nor odd, its derivative is odd?

We can simply apply the Chain Rule to prove the first two cases. Let $f(x)$ be an even function. Then

$\begin{aligned} f(x) &= f(-x) \\ f'(x) &= -f'(-x) \end{aligned}$

so $f'(x)$ is odd. Similarly, let $g(x)$ be an odd function. Then

$\begin{aligned} g(x) &= -g(-x) \\ g'(x) &= g'(-x) \end{aligned}$

so $g'(x)$ is even.

To disprove the third case, we take any odd function $g(x)$ . Let $G(x) = g(x) + C$ , for $C$ any non-zero constant. Then $G(-x) = g(-x) + C = -g(x) + C \ne -G(x)$ , so $G(x)$ is not odd. But clearly $G'(x) = g'(x)$ so as shown above $G'(x)$ will be even.