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Prove that the derivative of an even differentiable function is odd, and the derivative of an odd differentiable function is even.

Here are my workings so far.

Lets prove the derivative of an odd differentiable function is even first. Let the odd function be $f(x)$. We have $f(-x)=-f(x)$ and $\lim_{x\to a^-} f(x)=\lim_{x\to a^+}f(x)=\lim_{x\to a}f(x)$

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

And so, I am stuck. Thanks in advance. Hints are appreciated. Solutions are even more welcome!

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    $\begingroup$ I noticed you put $x\to h$ where you needed $h\to 0$, so I fixed that. You must be confounding two different definitions for the derivative. $\endgroup$
    – rschwieb
    Sep 28, 2012 at 12:41
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    $\begingroup$ Why don't you differenciate the equality $f(-x) = -f(x)$ using the chain rule? No need to come back to the definition of the derivative with the limits. $\endgroup$
    – S4M
    Sep 28, 2012 at 12:43
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    $\begingroup$ That's true! Thanks. Now I know a lot of ways to do this. But I was wondering how do I approach it from this definition angle? $\endgroup$ Sep 28, 2012 at 12:48

2 Answers 2

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Continuing from your last line, $$ \lim_{h\to 0} -\frac {f(x-h)-f(x)}{h}=\lim_{h\to 0} \frac {f(x-h)-f(x)}{-h}=f'(x) $$

That completes the proof for $f$ an odd function.

The analogous approach will probably work for $f(x)$ an even function.

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

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    $\begingroup$ I don't understand how $\lim_{h\to 0} \frac {f(x-h)-f(x)}{-h}=f'(x)$ Isn't the definition of $ f'(x)=\lim_{h\to 0} \frac {f(x+h)-f(x)}{-h}$ The difference between the -h and the +h $\endgroup$ Sep 28, 2012 at 12:50
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    $\begingroup$ They are equivalent: you can just write $-h=h'$ and note that $h'$ goes to zero as $-h$ goes to zero. That's a formal trick, but you should think a little about the picture that goes along with the derivative to see why the negative sign doesn't matter. $\endgroup$
    – rschwieb
    Sep 28, 2012 at 12:53
  • $\begingroup$ I see. That was helpful! $\endgroup$ Sep 28, 2012 at 12:56
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If $f(x)$ is odd then,$$f'(x)=\frac{d(f(x))}{dx}=\frac{d(-f(-x))}{dx}=-\frac{d(f(-x))}{dx}=-(-f'(-x))=f'(-x)$$

Here, have used $\frac{d(f(-x))}{dx}=-f'(-x)$ (using chain rule)

You can follow similar approach for even $f(x)$

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  • $\begingroup$ Awesome! Is there a way to do this using $\lim_{h\to 0} \frac {f(-x+h)-f(-x)}{h}$? $\endgroup$ Sep 28, 2012 at 12:46
  • $\begingroup$ @SingaporeanDude. If you want to see that, it's in my solution. I like this solution better though! $\endgroup$
    – rschwieb
    Sep 28, 2012 at 12:49
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    $\begingroup$ For odd $f(x)$, you can do like this : $$f'(-x)= \lim_{h\to 0} \frac {f(-x+h)-f(-x)}{h}= \lim_{h\to 0} \frac {-f(x-h)+f(x)}{h} \lim_{h\to 0} -\frac {f(x-h)-f(x)}{h}=\lim_{h\to 0} \frac {f(x)-f(x-h)}{h}=f'(x) $$ $\endgroup$
    – Aang
    Sep 28, 2012 at 12:51

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