This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even. (Write the equation that says $f$ is even, and differentiate both sides, using the chain rule.)"

I already read numerous solutions online. The following block shows the official solution, but I didn't quite understand it. (Particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$.)

(1F-6). Following the hint, let $z=-x$. If $f$ is even, then $f(x)=f(z).$ Differentiating and using the chain rule: $$f'(x)=f'(z) \frac{dz}{dx}=-f'(z),$$ because $\frac{dz}{dx}=-1.$ But this means that $f'$ is odd.
Similarly, if $g$ is odd, then $g$($x=-g(z)$. Differentiating and using the chain rule: $$g'(x)=-g'(z) \frac{dz}{dx}=g'(z),$$ because $\frac{dz}{dx}=-1.$

Thanks in advance =]

  • 9
    $\begingroup$ Well, what what you say that the derivative of $-x$ is, if not $-1$? $\endgroup$ Sep 11, 2012 at 9:44
  • $\begingroup$ Wow, can't believe I missed that. Thank you =] (I looked at it in a different way altogether.) $\endgroup$
    – Py42
    Sep 11, 2012 at 9:47

4 Answers 4


Official, shmofficial: I think the following might prove to be easier to grasp for some: suppose $\,f\,$ is odd, then

$$f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{-f(x)+f(x_0)}{-x+x_0}=$$

$$=\lim_{x\to x_0}\frac{f(-x)-f(-x_0)}{(-x)-(-x_0)}\stackrel{-x\to y}=\lim_{y\to -x_0}\frac{f(y)-f(-x_0)}{y-(-x_0)}=:f'(-x_0)$$

The above remains, mutatis mutandis, in case $\,f\,$ is even.


Following the official solution, we have $f(-x) = -f(x)$ by assumption. Thus, by considering the function $g(x) = -x = (-1) \cdot x$, we have $f(g(x)) = (-1)\cdot f(x)$. Differentiating on both sides gives $$\frac{d}{dx} f(g(x)) - \frac{d}{dx} (-1)\cdot f(x) = -1 \cdot \frac{d}{dx} f(x).$$

Now, applying the chain rule, we get $$\frac{d}{dx} f(g(x)) = \frac{df}{dx} g(x) \cdot \frac{d}{dx} g(x) = \frac{df}{dx} (-x) \cdot \frac{d}{dx} (-1 \cdot x) = \frac{df}{dx} (-x) \cdot (-1).$$

Equating both sides and simplifying gives $$\frac{d}{dx} f(-x) = \frac{d}{dx} f(x).$$


If $f(x)$ is an even function,

Then: $f(−x)=f(x)$

Now, differentiate above equation both side: $f'(−x)(−1)=f'(x)$

$⇒f'(−x)=−f' (x)$

Hence $f'(x)$ is an odd function.

Rest part can be proved similarly.

  • 4
    $\begingroup$ Funnily, $-x$ is odd but $-1$ is even :) $\endgroup$ Dec 21, 2020 at 8:39
  • 2
    $\begingroup$ beautiful proof $\endgroup$ Nov 6, 2021 at 11:49

Well, geometrically, even function means reflection along y axis, so any direction will reflect, that mean, the derivative on the right is the same as the derivative on the left, but the direction change. It means the value is the same, but with different sign.

Odd function means rotational symmetric, if you rotate an arrow, I.e. direction, you will change by 180 degree, so it is the same slope, hence the derivative of odd function is even.


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