Answer
Prove that:
(a) The derivative of an even function is an odd function.
If $f(x)$ is even, then $f(-x)=f(x)$,
using the Chain Rule to differentiate this equation, we get
$$
\begin{aligned}
f^{\prime}(x)&=f^{\prime}(-x) \frac{d}{d x}(-x)\\
&=-f^{\prime}(-x)
\end{aligned}
$$
Thus,
$$ f^{\prime}(-x)=-f^{\prime}(x),$$
so $ f^{\prime}$ is odd
(b)The derivative of an odd function is an even function.
If $f(x)$ is odd, then $f(x)=-f(-x)$,
using the Chain Rule to differentiate this equation, we get
$$
\begin{aligned}
f^{\prime}(x)&=-f^{\prime}(-x) \frac{d}{d x}(-x)\\
&=-f^{\prime}(-x)(-1)\\
&=f^{\prime}(-x)
\end{aligned}
$$
Thus,
$$ f^{\prime}(-x)=f^{\prime}(x),$$
so $ f^{\prime}$ is even.
Work Step by Step
Prove that:
(a) The derivative of an even function is an odd function.
If $f(x)$ is even, then $f(-x)=f(x)$,
using the Chain Rule to differentiate this equation, we get
$$
\begin{aligned}
f^{\prime}(x)&=f^{\prime}(-x) \frac{d}{d x}(-x)\\
&=-f^{\prime}(-x)
\end{aligned}
$$
Thus,
$$ f^{\prime}(-x)=-f^{\prime}(x),$$
so $ f^{\prime}$ is odd
(b)The derivative of an odd function is an even function.
If $f(x)$ is odd, then $f(x)=-f(-x)$,
using the Chain Rule to differentiate this equation, we get
$$
\begin{aligned}
f^{\prime}(x)&=-f^{\prime}(-x) \frac{d}{d x}(-x)\\
&=-f^{\prime}(-x)(-1)\\
&=f^{\prime}(-x)
\end{aligned}
$$
Thus,
$$ f^{\prime}(-x)=f^{\prime}(x),$$
so $ f^{\prime}$ is even.
