$2/x^2$ — a rational function with a square in the denominator. The 5-step dance is the same as Q1, but the common denominator is $x^2(x+h)^2$ and the numerator factors as a difference of squares, which is a different (and very clean) way to surface the $h$.
$2/x^2$ is positive everywhere (except at $x = 0$ where it's undefined) and decreasing — as $x$ grows, $1/x^2$ shrinks very fast. So the derivative should be negative, and the answer rounded to the nearest tenth should be small in magnitude (close to zero).
The "new" algebraic move in this problem is that the numerator, after combining fractions, is a difference of squares: $\dfrac{2}{x^2} - \dfrac{2}{(x+h)^2}$ becomes $\dfrac{2(x+h)^2 - 2x^2}{x^2(x+h)^2} = \dfrac{2[(x+h)^2 - x^2]}{x^2(x+h)^2}$. The thing in brackets is $(A^2 - B^2) = (A+B)(A-B)$, and one of those factors is the $h$ we need.
For a difference of squares, we use:
$$A^2 - B^2 \;=\; (A - B)(A + B)$$With $A = x + h$ and $B = x$:
$$(x+h)^2 - x^2 \;=\; [(x+h) - x] \cdot [(x+h) + x] \;=\; h \cdot (2x + h)$$That's the step that surfaces the $h$ for cancellation — same role as the conjugate in Q3, but for a different shape of problem.
Given $f(x) = \dfrac{2}{x^2}$ and $a = 2$. We compute $\displaystyle f'(2) = \lim_{h \to 0} \dfrac{f(2+h) - f(2)}{h}$ in five steps.
Step 1 — Write the difference quotient.
$$f'(2) \;=\; \lim_{h \to 0} \frac{\dfrac{2}{(2+h)^2} - \dfrac{2}{2^2}}{h} \;=\; \lim_{h \to 0} \frac{\dfrac{2}{(2+h)^2} - \dfrac{1}{2}}{h}$$Step 2 — Combine the two fractions in the numerator over a common denominator.
$$f'(2) \;=\; \lim_{h \to 0} \frac{\dfrac{2 \cdot 2 - 1 \cdot (2+h)^2}{2(2+h)^2}}{h} \;=\; \lim_{h \to 0} \frac{4 - (2+h)^2}{2(2+h)^2 \cdot h}$$The common denominator is $2(2+h)^2$. The first fraction's numerator gets multiplied by $2$ (the other piece's denominator), the second by $(2+h)^2$ (the other piece's denominator).
Step 3 — Expand $(2+h)^2$ and subtract.
$$(2+h)^2 \;=\; 4 + 4h + h^2$$ $$4 - (2+h)^2 \;=\; 4 - (4 + 4h + h^2) \;=\; -4h - h^2$$So far:
$$f'(2) \;=\; \lim_{h \to 0} \frac{-4h - h^2}{2(2+h)^2 \cdot h}$$Step 4 — Factor $h$ out of the numerator and cancel.
$$\frac{-4h - h^2}{2(2+h)^2 \cdot h} \;=\; \frac{h(-4 - h)}{2(2+h)^2 \cdot h} \;=\; \frac{-4 - h}{2(2+h)^2}$$(We could also have done this via the difference-of-squares shortcut: $-4h - h^2 = -h(4 + h)$, which is the same factorization.)
Step 5 — Take the limit as $h \to 0$.
$$f'(2) \;=\; \lim_{h \to 0} \frac{-4 - h}{2(2+h)^2} \;=\; \frac{-4 - 0}{2 \cdot (2+0)^2} \;=\; \frac{-4}{2 \cdot 4} \;=\; \frac{-4}{8} \;=\; -\frac{1}{2}$$Rounded to the nearest tenth:
The function $2/x^2$ is positive and decreasing for $x > 0$, so the derivative should be negative — and $-0.5$ is negative ✓.
Quick check via shortcut: $f(x) = 2x^{-2}$, so the shortcut derivative is $f'(x) = -4x^{-3} = -4/x^3$. At $x = 2$: $f'(2) = -4/8 = -0.5$ ✓. (The shortcut exists; we just did the dance first to prove the 5-step method generalizes.)
As $x$ grows, $-4/x^3 \to 0$ — the slope flattens. At $x = 2$ it's still pretty steep ($-0.5$), but at $x = 10$ it would be $-4/1000 = -0.004$, basically flat. That matches the intuition that $1/x^2$ decays fast.
For rational functions with $x^2$ in the denominator, the 5 steps become:
The "difference of squares" move $(A^2 - B^2) = (A-B)(A+B)$ is a nice alternative way to do step 3–4: it shows that the $h$ comes from the $(A - B)$ factor, with $(A + B)$ surviving into the denominator. Either approach works — pick whichever feels cleaner.
Use the definition (the 5 steps + common denominator) to find $f'(1)$ for $f(x) = \dfrac{3}{x^2}$.
Hint: same shape as Q4, just with $3$ instead of $2$ and $a = 1$ instead of $a = 2$. The leading constant $3$ has to survive into the answer.
Step 1: $\displaystyle f'(1) = \lim_{h \to 0} \dfrac{\dfrac{3}{(1+h)^2} - 3}{h}$
Step 2: Common denominator $(1+h)^2$: $\dfrac{3 - 3(1+h)^2}{(1+h)^2 \cdot h}$
Step 3: $(1+h)^2 = 1 + 2h + h^2$, so $3 - 3(1 + 2h + h^2) = 3 - 3 - 6h - 3h^2 = -6h - 3h^2$
Step 4: $\dfrac{-6h - 3h^2}{(1+h)^2 \cdot h} = \dfrac{h(-6 - 3h)}{(1+h)^2 \cdot h} = \dfrac{-6 - 3h}{(1+h)^2}$
Step 5: $f'(1) = \lim_{h \to 0} \dfrac{-6 - 3h}{(1+h)^2} = \dfrac{-6 - 0}{(1+0)^2} = -6$
Sanity check via shortcut: $f(x) = 3x^{-2}$, so $f'(x) = -6x^{-3} = -6/x^3$. At $x = 1$: $f'(1) = -6$ ✓. And the $3$ from $f(x)$ survived into the answer as a $6$ in the derivative (the constant gets multiplied by $-2$ — that's the pattern for $x^{-2} \to -2x^{-3}$).
Three directions once you finish Q4: