Answer
Prove that:
(a) The derivative of an even function is an odd function.
If $f$ is even, then
$$
\begin{aligned}
f^{\prime}(-x) &=\lim _{h \rightarrow 0} \frac{f(-x+h)-f(-x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{f[-(x-h)]-f(-x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{f(x-h)-f(x)}{h}\\
&=-\lim _{h \rightarrow 0} \frac{f(x-h)-f(x)}{h} \quad[\text { let } \Delta x=-h] \\
&=-\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\\
&=-f^{\prime}(x)
\end{aligned}$$
Therefore, $f^{\prime} $ is odd.
(b)The derivative of an odd function is an even function.
If $f$ is odd, then
$$
\begin{aligned}
f^{\prime}(-x) &=\lim _{h \rightarrow 0} \frac{f(-x+h)-f(-x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{f[-(x-h)]-f(-x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{-f(x-h)+f(x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{f(x-h)-f(x)}{-h} \quad[\text { let } \Delta x=-h] \\
&=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\\
&=f^{\prime}(x)
\end{aligned}
$$
Therefore, $f^{\prime} $ is even.
Work Step by Step
Prove that:
(a) The derivative of an even function is an odd function.
If $f$ is even, then
$$
\begin{aligned}
f^{\prime}(-x) &=\lim _{h \rightarrow 0} \frac{f(-x+h)-f(-x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{f[-(x-h)]-f(-x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{f(x-h)-f(x)}{h}\\
&=-\lim _{h \rightarrow 0} \frac{f(x-h)-f(x)}{h} \quad[\text { let } \Delta x=-h] \\
&=-\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\\
&=-f^{\prime}(x)
\end{aligned}$$
Therefore, $f^{\prime} $ is odd.
(b)The derivative of an odd function is an even function.
If $f$ is odd, then
$$
\begin{aligned}
f^{\prime}(-x) &=\lim _{h \rightarrow 0} \frac{f(-x+h)-f(-x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{f[-(x-h)]-f(-x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{-f(x-h)+f(x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{f(x-h)-f(x)}{-h} \quad[\text { let } \Delta x=-h] \\
&=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\\
&=f^{\prime}(x)
\end{aligned}
$$
Therefore, $f^{\prime} $ is even.
