a.) Prove that the derivative of an even function is an odd function using Chain Rule.
Suppose that for an even function, $f(x) = f(-x)$ thus,
$
\begin{equation}
\begin{aligned}
f'(x) =& \frac{d}{dx} f(x) = \frac{d}{dx} f(-x)
\\
\\
\frac{d}{dx} f(-x) =& \frac{d}{d(-x)} = \frac{d}{dx}(-x) = -f'(-x)
\end{aligned}
\end{equation}
$
which proves that the derivative of an even function is an odd function.
b.) Prove that the derivative of an odd function is an even function using Chain Rule.
Suppose that for an odd function, $f(x) = - f(-x)$ thus,
$
\begin{equation}
\begin{aligned}
f'(x) = \frac{d}{dx} f(x) =& \frac{d}{dx} [- f(-x)]
\\
\\
\frac{d}{dx} [-f(-x)] =& \frac{d}{d(-x)} [-f(-x)] \cdot \frac{d}{dx} (-x) = f'(-x)
\end{aligned}
\end{equation}
$
which proves that the derivative of an odd function is an even function.
Wednesday, August 6, 2014
Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 83
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