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Without using the chain rule, I want to show that the derivative of an even function is an odd function; here's my attempt:

We have

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

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

from using the fact that $f(-x) = f(x)$. This appears to show that the derivative is again even ...

Where's my mistake?

I have also tried subtracting the above two limits to get:

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

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

which doesn't appear to help much. Lastly, I tried multiplying both sides by $h$, but then I'd end up with a difference of function evaluations and no longer have a difference quotient to work with.

Thanks,

D.Hutchinson
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2 Answers2

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Hint: when you replace $x$ with $-x$ in the difference quotient, also replace $h$ with $-h$. That last replacement does not change the value of the limit, since $h\to 0$.

Rory Daulton
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Suppose $f$ is even, so that $f(x)=f(-x)$. We want to show that $f^\prime (-x)=-f^\prime (x)$. We have that

$$ f^\prime(-x) = \lim_{h\to0} \frac{f(-x+h)-f(-x)}{h}. $$

Since $f$ is even, $f(-x+h)=f(x-h)$ and $f(-x)=f(x)$. Therefore,

$$ f^\prime(-x) = \lim_{h\to0} \frac{f(x-h)-f(x)}{h}. $$

If we let $h^\prime =- h$, we have that

$$ \lim_{h\to0} \frac{f(x-h)-f(x)}{h} = -\lim_{h^\prime\to0} \frac{f(x+h^\prime)-f(x)}{h^\prime} = -f^\prime (x). $$

Theoretical Economist
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  • Nice proof. One note you might add, as $h\to 0 $ it is true that $-h \to 0 $ and thus by substitution $h' \to 0 $. To be honest these substitutions make me nervous and seem a little handwavy. It would be nice to know under what conditions can we substitute for $h$ in the limit, though it is a placeholder. Do we need epsilon delta to make this more rigorous. Is this merely a heuristic proof? – john Oct 18 '22 at 02:29
  • The only requirement is sufficient continuity, which we have in this problem. ($h \mapsto -h$ is continuous, and $f$ must be continuous at the points it is differentiable.) I glossed over those details, and I agree that a careful proof would at least mention them. See https://math.stackexchange.com/q/167926. – Theoretical Economist Oct 18 '22 at 08:27