Answer
a)
If $f$ is even, than $f(-x) = f(x)$
Taking the derivative of both sides and applying the chain rule, we find:
$f'(-x) * (\frac{d}{dx} [-x]) = f'(x)$
$-f'(-x) = f'(x)$
This is then the condition for an odd function ($-f(-x) = f(x)$); thus $f'(x)$ is odd if $f$ is even.
b.
If $f$ is odd, than $-f(-x) = f(x)$
Taking the derivative of both sides and applying the chain rule, we find:
$-f'(-x) * (\frac{d}{dx} [-x]) = f'(x)$
$f'(-x) = f'(x)$
This is then the condition for an even function ($f(-x) = f(x)$); thus $f'(x)$ is even if $f$ is odd.
Work Step by Step
a)
If $f$ is even, than $f(-x) = f(x)$
Taking the derivative of both sides and applying the chain rule, we find:
$f'(-x) * (\frac{d}{dx} [-x]) = f'(x)$
$-f'(-x) = f'(x)$
This is then the condition for an odd function ($-f(-x) = f(x)$); thus $f'(x)$ is odd if $f$ is even.
b.
If $f$ is odd, than $-f(-x) = f(x)$
Taking the derivative of both sides and applying the chain rule, we find:
$-f'(-x) * (\frac{d}{dx} [-x]) = f'(x)$
$f'(-x) = f'(x)$
This is then the condition for an even function ($f(-x) = f(x)$); thus $f'(x)$ is even if $f$ is odd.
