Derivatives of Even Functions

Published by MrHonner on September 14, 2012September 14, 2012

A recent tweet from @AnalysisFact noted that the derivative of an even function is an odd function. There are many ways to explore and understand this fact, but here’s a simple algebraic approach that uses a neat little trick in representing even and odd functions.

Claim: The derivative of a [differentiable] even function is an odd function.

Proof: Suppose $f(x)$ is an even, differentiable function. Consider the function

$a(x) = \frac {f(x) + f(-x)}{2}$

First we show that $a(x) = f(x)$ . Since $f(x)$ is even, we know

$f(x) = f(-x)$ .

Thus, $a(x) = \frac{f(x)+f(-x)}{2} = \frac{2f(x)}{2} = f(x)$ .

Now let’s differentiate $f(x)$ . We have

$f'(x) = a'(x) = \frac{(f(x)+f(-x))'}{2} = \frac{f'(x) - f'(-x)}{2}$

where the last step follows by the chain rule.

And since

$f'(-x) = a'(-x) = \frac{f'(-x) -f'(x)}{2} = -\frac{(-f'(-x)+f'(x))}{2} = -a'(x) = -f'(x)$

we see that the derivative of $f(x)$ is an odd function, as desired.

Related Posts

Decomposing Functions Into Even and Odd Parts

8 Comments

Sue VanHattum · September 14, 2012 at 2:31 pm

Last line, after 3rd = sign, I’m not seeing it.

MrHonner · September 14, 2012 at 4:05 pm

Sue-

I factored out a (-1) from the numerator. Now rearrange the terms in the numerator and you’ll have (f'(x) – f'(-x)), which is the numerator of a'(x).

Sorry if I sacrificed a bit of clarity for brevity!

Oliver Prior · September 14, 2012 at 4:46 pm

Impressive derivation, I enjoyed it. Any other similar proofs you’d recommend?

MrHonner · September 14, 2012 at 5:00 pm

I think a more natural proof would show that the tangents to an even function at x = c and x = -c have opposite slopes. This probably follows pretty quickly from the definition of derivative.

Hao Ye · November 23, 2013 at 5:33 am

How about take the Taylor expansion of f(x). Group the terms with even exponents into g(x) and the terms with odd exponents into h(x). Then f(x) = g(x) + h(x). Note that g(x) is an even function and h(x) is an odd function. Also note that g'(x) is an odd function [all the exponents of g'(x) are odd] and h'(x) is an even function. If we posit that there is a *unique* decomposition of f(x) into an even and odd function, then it follows that f(x) is even implies f(x) = g(x) and h(x) = 0, so f'(x) = g'(x) is odd. Similarly, f(x) is odd implies f'(x) = h'(x) is even.

It should be relatively simple to prove a unique decomposition using a proof by contradiction.

MrHonner · November 23, 2013 at 7:07 am

Yes, I think Taylor expansion is a nice way to think about this, as well. The only drawback to this argument is that it only works in the interval of convergence of the Taylor series.

Nat · September 19, 2012 at 6:00 pm

hmm..
What about writing f(x) = f(-x) (by definition)?
Differentiating both sides gives us
f'(x) = f'(-x)*(-1) (by the Chain Rule).
So by definition f’ is odd.

MrHonner · September 19, 2012 at 9:58 pm

Yes, that is probably the quickest proof. I shared the original proof because I really like that trick of splitting up every function an even part and an odd part (here, since the function here is even to begin with, it has only an even part).

I don’t think either of these algebraic proofs are very illuminating, though, which is why I like the tangent line argument.