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 
First we show that 
Thus, 
Now let’s differentiate
where the last step follows by the chain rule.
And since
we see that the derivative of
Last line, after 3rd = sign, I’m not seeing it.
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!
Impressive derivation, I enjoyed it. Any other similar proofs you’d recommend?
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.
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.
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.