Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 88

a.) Show that $\displaystyle \frac{d}{dx} |x| = \frac{x}{|x|}$ where $|x| = \sqrt{x^2}$ using Chain Rule

By using Chain Rule we have,

$ \displaystyle \frac{d}{dx} |x| = \frac{d}{dx} \sqrt{x^2} = \frac{1}{\cancel{2} \sqrt{x^2}} \cdot 2x = \frac{x}{\sqrt{x^2}}; \text{ but } |x| = \sqrt{x^2}$

Therefore,

$\displaystyle \frac{d}{dx} |x| = \frac{x}{|x|}$



b.) Find $f'(x)$ and sketch the graphs of $f$ and $f'$. Also, state on what value(s) is $f$ not differentiable? Suppose that $f(x) = | \sin x| $.







By referring to the graph and by using the property of absolute value, we can deduce $f(x) = | \sin x|$ as

$f(x) = \left\{
\begin{array}{ccc}
\sin x & \text{ for } & x \varepsilon (2n \pi, (2n + 1) \pi) \\
- \sin x & \text{ for } & x \varepsilon ((2n - 1)\pi, 2n \pi)
\end{array}
\right.$ where $n$ is any integer.

So we can form $f'(x)$ as...

$f(x) = \left\{
\begin{array}{ccc}
\cos x & \text{ for } & x \varepsilon (2n \pi, (2n + 1) \pi) \\
- \cos x & \text{ for } & x \varepsilon ((2n - 1)\pi, 2n \pi)
\end{array}
\right.$ where $n$ is any integer.

Referring to the graph, $f(x)$ is not differentiable where $f'(x)$ is not continuous, that is at $x = n \pi$; where $n$ is any integer.


c.) Find $g'(x)$ and sketch the graphs of $g$ and $g'$. Also, state on what value(s) is $g$ not differentiable? Suppose that $f(x) = \sin |x|$.








By referring to the graph and by using the property of absolute value, we can deduce


$
\begin{equation}
\begin{aligned}

f(x) =& \sin |x| \text{ as }
\\
\\
f(x) =& \left\{
\begin{array}{ccc}
\sin (x) & \text{ for } & x \geq 0 \\
\sin (-x) & \text{ for } & x < 0
\end{array}
\right.

\longrightarrow

f(x) =& \left\{
\begin{array}{ccc}
\sin x & \text{ for } & x \geq 0 \\
- \sin x & \text{ for } & x < 0
\end{array}
\right.

\end{aligned}
\end{equation}
$


So we can form $f'(x)$ as..

$f'(x) = \left\{
\begin{array}{ccc}
\cos x & \text{ for } & x \geq 0 \\
- \cos x & \text{ for } & x < 0
\end{array}
\right.$

Referring to the graph, $f(x)$ is not differentiable where $f'(x)$ is not continuous, that is $x = 0$.