Question

If $f(x)=\frac{|x|}{x^{2}}$ then $f^{\prime}(2)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $-\frac{1}{2}$
• D. $-\frac{1}{4}$
• E. $-\frac{1}{6}$

Hint:

Logic Tip: Never attempt to differentiate an absolute value function directly using product/quotient rules without resolving the absolute value first. Since we evaluate at $x=2$ (a positive number), $|x|$ acts exactly like a normal $x$, turning this into a trivial power rule problem.

Answer 1

The Correct Option is D

Concept:
The absolute value function $|x|$ behaves differently depending on whether $x$ is positive or negative. Before differentiating, evaluate the behavior of the absolute value function in the local neighborhood of the specific point you are analyzing.
Step 1: Simplify the function for the domain of interest.
We need to find the derivative at $x = 2$. In the neighborhood of $x = 2$, $x$ is strictly positive ($x>0$). By definition, for $x>0$, $|x| = x$. Substitute this into the original function: $$f(x) = \frac{x}{x^2}$$ $$f(x) = \frac{1}{x} = x^{-1} \quad \text{for } x>0$$
Step 2: Differentiate the simplified function.
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$: $$f'(x) = -1 \cdot x^{-2}$$ $$f'(x) = -\frac{1}{x^2}$$
Step 3: Evaluate the derivative at the specific point.
Substitute $x = 2$ into the derivative: $$f'(2) = -\frac{1}{(2)^2}$$ $$f'(2) = -\frac{1}{4}$$