Question:

Step 1: Determine the derivative of $f(x) = |x|$ for $x \neq 0$. The function $f(x) = |x|$ can be defined piecewise as: $$f(x) = \begin{cases} x & \text{if } x > 0 \\ -x & \text{if } x < 0 \end{cases}$$ For $x > 0$, $f(x) = x$, so its derivative is $f'(x) = \frac{d}{dx}(x) = 1$. For $x < 0$, $f(x) = -x$, so its derivative is $f'(x) = \frac{d}{dx}(-x) = -1$. Thus, the derivative of $f(x) = |x|$ for $x \neq 0$ is: $$f'(x) = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \end{cases}$$ Step 2: Evaluate the expression $\frac{|x|}{x}$ for $x \neq 0$. Consider the expression $\frac{|x|}{x}$: For $x > 0$, $|x| = x$, so $\frac{|x|}{x} = \frac{x}{x} = 1$. For $x < 0$, $|x| = -x$, so $\frac{|x|}{x} = \frac{-x}{x} = -1$. Thus, the expression $\frac{|x|}{x}$ for $x \neq 0$ is: $$\frac{|x|}{x} = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \end{cases}$$ Step 3: Compare the results. From Step 1, $f'(x) = 1$ for $x > 0$ and $f'(x) = -1$ for $x < 0$. From Step 2, $\frac{|x|}{x} = 1$ for $x > 0$ and $\frac{|x|}{x} = -1$ for $x < 0$. Since the expressions for $f'(x)$ and $\frac{|x|}{x}$ are identical for all $x \neq 0$, the statement is true.