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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 $\lim_{x\to a^-} f(x)=\lim_{x\to a^+}f(x)=\lim_{x\to a}f(x)$

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

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

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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)

$$

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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