Rational function edition — $3/x$ is the first time the difference quotient has $h$ in two places at once. The 5-step dance still works, you just have to combine fractions before you can subtract.
The derivative of $3/x$ should be negative: as $x$ grows, $3/x$ shrinks, so the slope is negative everywhere (except at $x = 0$, where the function isn't defined). That immediately rules out C and D — the answer has to be a negative sign over a positive $x^2$.
The interesting question is which negative — and that's what the 5-step dance tells you. The constant $3$ in the numerator has to survive the algebra and land in the answer; if you lose it, you end up with $-1/x^2$ instead of $-3/x^2$.
For any function $f$, the derivative is the limit of the difference quotient:
$$f'(x) \;=\; \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$Substituting $f(x) = 3/x$ gives:
$$f'(x) \;=\; \lim_{h \to 0} \frac{\dfrac{3}{x+h} - \dfrac{3}{x}}{h}$$Now the numerator is a difference of two fractions. That's the new wrinkle in this problem — you have to combine them before you can do anything else.
Given $f(x) = \dfrac{3}{x}$. We compute $\displaystyle f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$ in five steps.
Step 1 — Write the difference quotient.
$$f'(x) \;=\; \lim_{h \to 0} \frac{\dfrac{3}{x+h} - \dfrac{3}{x}}{h}$$Step 2 — Compute $f(x+h) - f(x)$: combine the fractions over a common denominator.
$$\frac{3}{x+h} - \frac{3}{x} \;=\; \frac{3 \cdot x}{(x+h) \cdot x} - \frac{3 \cdot (x+h)}{x \cdot (x+h)} \;=\; \frac{3x - 3(x+h)}{x(x+h)}$$ $$=\; \frac{3x - 3x - 3h}{x(x+h)} \;=\; \frac{-3h}{x(x+h)}$$Common denominator is $x(x+h)$. After expanding, the $3x$ terms cancel and we're left with $-3h$ on top.
Step 3 — Divide by $h$ (i.e. put $h$ in the denominator).
$$f'(x) \;=\; \lim_{h \to 0} \frac{\dfrac{-3h}{x(x+h)}}{h} \;=\; \lim_{h \to 0} \frac{-3h}{h \cdot x(x+h)}$$Just rewrite: a fraction divided by $h$ is the same fraction with $h$ multiplied into the denominator.
Step 4 — Cancel $h$.
$$=\; \lim_{h \to 0} \frac{-3\cancel{h}}{\cancel{h} \cdot x(x+h)} \;=\; \lim_{h \to 0} \frac{-3}{x(x+h)}$$The $h$ in the numerator was the only $h$ on top — that's the whole reason combining fractions in step 2 was worth doing.
Step 5 — Take the limit as $h \to 0$.
$$f'(x) \;=\; \lim_{h \to 0} \frac{-3}{x(x+h)} \;=\; \frac{-3}{x(x+0)} \;=\; \frac{-3}{x \cdot x} \;=\; -\frac{3}{x^2}$$Final answer:
Each wrong answer maps to a specific step where the algebra got cut short or a sign was lost.
Every derivative-via-the-definition problem in this course is these five steps, in this order:
If step 4 doesn't go cleanly, the most common reasons for rational functions are: (a) you didn't expand the common denominator in step 2 carefully, or (b) you lost a sign when distributing the minus sign in $3(x+h)$.
Use the definition (the 5 steps above) to find $f'(x)$ for $f(x) = \dfrac{1}{x}$.
Hint: this is the same shape as Q1, just with a different constant. Watch whether the constant survives into the answer.
Step 1: $\displaystyle f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$
Step 2: $\dfrac{1}{x+h} - \dfrac{1}{x} = \dfrac{x - (x+h)}{x(x+h)} = \dfrac{-h}{x(x+h)}$
Step 3: $\dfrac{-h}{h \cdot x(x+h)}$
Step 4: $\dfrac{-\cancel{h}}{\cancel{h} \cdot x(x+h)} = -\dfrac{1}{x(x+h)}$
Step 5: $f'(x) = \lim_{h \to 0} \left(-\dfrac{1}{x(x+h)}\right) = -\dfrac{1}{x \cdot x} = -\dfrac{1}{x^2}$
Sanity check: $1/x$ is decreasing, so the derivative is negative ✓. And the leading coefficient of $1$ (instead of $3$) survived into the answer — that's where the difference between Q1 and this practice problem lives.
Three directions once you finish Q1: