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

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!

• 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. – rschwieb Sep 28 '12 at 12:41
• 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. – S4M Sep 28 '12 at 12:43
• 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? – Yellow Skies Sep 28 '12 at 12:48

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.

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

• 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 – Yellow Skies Sep 28 '12 at 12:50
• 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. – rschwieb Sep 28 '12 at 12:53
• I see. That was helpful! – Yellow Skies Sep 28 '12 at 12:56

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 $f(x)$

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