Answer
The question asks you to prove two statements:
(a) $f(-x) = - f(x) \implies f'(-x) = f'(x)$
(b) $f(-x) = f(x) \implies f'(-x) = -f'(x) $
The proofs appear below.
Work Step by Step
In order to prove these statements, we apply the chain rule for differentiation:
(a) Since $f(-x) = - f(x)$, we have that: $\frac{d(-f(x))}{dx}$ = $ \frac{d(f(-x))}{dx} $
On the left side of the equality, we apply linearity of the derivative $\implies$
$\frac{d(-f(x))}{dx}$ = $-\frac{d(f(x))}{dx} = -f'(x)$
On the right side of the equality, we apply the chain rule for differentiation, with $u = (-x) \implies \frac{d(f(-x))}{dx} = \frac{d(u)}{dx} * \frac{d(f(u))}{du} = (-1)*(f'(u)) = -f'(-x) $
Now, by the equality, we have:
$-f'(x) = -f'(-x) \implies$ (divide by (-1)) $ f'(x) = f'(-x)$, proving the desired statement.
(b) Similarly, since $f(-x) = f(x)$, we have that: $\frac{d(f(x))}{dx}$ = $ \frac{d(f(-x))}{dx} $
On the left side of the equality, we just have:
$\frac{d(f(x))}{dx}$ = $f'(x)$
On the right side of the equality, we apply the chain rule for differentiation, with $u = (-x) \implies \frac{d(f(-x))}{dx} = \frac{d(u)}{dx} * \frac{d(f(u))}{du} = (-1)*(f'(u)) = -f'(-x) $
Now, by the equality, we have:
$f'(x) = -f'(-x) \implies$ (divide by (-1)) $ -f'(x) = f'(-x)$, proving the desired statement.
