Thread: even/odd function's derivative is odd/even

using differentiation, i want to show

f is even on (-r,r) if and only if f' is odd on (-r,r) and also odd-even relation...

in proving f=>f', differentiating the eqation seems sufficient, but

i dont know how to do the converse, proving f'=>f using differentiation...

Hi

For all x in (-r,r)
-f'(-x) = f'(x)
(f(-x))' = (f(x))'

Integrating gives
f(-x) = f(x) + K

To find K we take x=0
f(0) = f(0) + K
K = 0

And finally
f(-x) = f(x)

is there any way to prove without using integration?

i think mean value theorum might work...

You could use the contrapositive of your original statement:

You're trying to show that: If $\displaystyle f'$ is odd on $\displaystyle (-r,r)$, then $\displaystyle f$ is even on $\displaystyle (-r,r)$.

The contrapositive would be: If $\displaystyle f$ is not even on $\displaystyle (-r,r)$, then $\displaystyle f'$ is not odd on $\displaystyle (-r,r)$.

Both of these statements are logically equivalent.

Use the chain rule: for any function f(x), $\displaystyle \frac{df(-x)}{dx}= \frac{df(u)}{du}\frac{du}{dx}$ with u= -x. That is, $\displaystyle \frac{df(-x)}{dx}= f'(-x)(-1)= -f'(-x)$.

If f is an even function, then f(-x)= f(x) so $\displaystyle \frac{df(-x)}{dx}= \frac{df(x)}{dx}= f'(x)$ and that previous equation becomes f'(x)= -f'(-x) which is the same as -f'(x)= f'(-x) showing that f'(x) is an odd function.

If f is an odd function, then f(-x)= -f(x) so $\displaystyle \frac{df(-x)}{dx}= -\frac{df(x)}{dx}= -f'(x)$ and that previous equation becomes -f'(x)= -f'(-x) which is the same as f'(x)= f'(-x) showing that f'(x) is an even function.