Prove that (a) the derivative of an even function is odd

Prove that (a) the derivative of an even function is odd, and (b) the derivative of an odd function is even.

Step 1: Part (a): $f$ even $\Longrightarrow$ $f'$ odd

Suppose $f$ is differentiable and even: \[f(-x) = f(x) \qquad\text{for all } x. \tag{1}\]

Differentiate both sides with respect to $x$. On the left the chain rule contributes the inner derivative $-1$: $\frac{d}{dx}f(-x) = -f'(-x), \qquad \frac{d}{dx}f(x) = f'(x)$.

Equating and multiplying by $-1$: $-f'(-x) = f'(x) \;\Longrightarrow\; f'(-x) = -f'(x)$.

which is precisely the statement that $f'$ is odd.

Step 2: Part (b): $f$ odd $\Longrightarrow$ $f'$ even

Now suppose

\[ f(-x) = -f(x) \qquad\text{for all } x. \tag{2} \]

Differentiating both sides in the same way: $-f'(-x) = -f'(x) \;\Longrightarrow\; f'(-x) = f'(x)$.

so $f'$ is even.

Step 3: Verification

\[(\cos x)' = -\sin x:\quad \text{even} \to \text{odd} \;\;\checkmark\]

\[(\sin x)' = \cos x:\quad \text{odd} \to \text{even} \;\;\checkmark\]

\[\bigl(x^{2}\bigr)' = 2x:\ \text{even}\to\text{odd}; \qquad \bigl(x^{3}\bigr)' = 3x^{2}:\ \text{odd}\to\text{even} \;\;\checkmark\]