Derivatives — where the secant slope from Q1 becomes the tangent slope. The power rule.
Find $f'(-3)$ if $f(x) = -2x^2$.
Remember Q1? We found the slope of the secant line for $f(x) = x^2$ between $x = -1$ and $x = 3$ — the average rate of change between two points. Answer: $2$.
What if, instead of stopping at $x = 3$, we let the second point squeeze in toward the first? The secant line rotates and becomes a tangent line at a single point. Its slope is the instantaneous rate of change — the derivative.
That's what $f'(-3)$ is asking: the slope of the line tangent to $f$ at the single point $x = -3$.
The derivative of $f$ at $x = a$ is the limit of the secant slope as the second point $x = a + h$ approaches $a$ (i.e., $h \to 0$):
$$f'(a) \;=\; \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$This is the limit definition of the derivative, also called the definition from first principles. It is the rigorous version of "squeeze the secant into a tangent."
For Q6 with $a = -3$ and $f(x) = -2x^2$:
$$f'(-3) \;=\; \lim_{h \to 0} \frac{f(-3 + h) - f(-3)}{h}$$For any real exponent $n$ and any constant coefficient $c$:
$$\frac{d}{dx}\left[c \cdot x^n\right] \;=\; c \cdot n \cdot x^{n-1}$$"Multiply by the exponent, drop the exponent by 1."
Applied to $f(x) = -2x^2$ (here $c = -2$, $n = 2$):
$$f'(x) \;=\; (-2)(2)\, x^{2-1} \;=\; -4x$$Now evaluate at $x = -3$:
$$f'(-3) \;=\; -4 \cdot (-3) \;=\; \boxed{12}$$For practice (and because profs love to put "use the limit definition" on exams), here's the long form for $f(x) = -2x^2$ at $a = -3$:
Step 1 — Write the limit.
$$f'(-3) \;=\; \lim_{h \to 0} \frac{f(-3 + h) - f(-3)}{h}$$Step 2 — Compute $f(-3 + h)$.
$$f(-3 + h) \;=\; -2(-3 + h)^2 \;=\; -2(9 - 6h + h^2) \;=\; -18 + 12h - 2h^2$$Step 3 — Compute $f(-3)$.
$$f(-3) \;=\; -2(-3)^2 \;=\; -2(9) \;=\; -18$$Step 4 — Build the numerator.
$$f(-3 + h) - f(-3) \;=\; (-18 + 12h - 2h^2) - (-18) \;=\; 12h - 2h^2$$Step 5 — Divide by $h$ and simplify.
$$\frac{f(-3 + h) - f(-3)}{h} \;=\; \frac{12h - 2h^2}{h} \;=\; \frac{h(12 - 2h)}{h} \;=\; 12 - 2h$$Step 6 — Take the limit as $h \to 0$.
$$\lim_{h \to 0} (12 - 2h) \;=\; 12 - 0 \;=\; 12$$Same answer: 12. The shortcut just skips steps 2–5 with a one-line formula.
Here's the full arc in one table. Q1 was the warm-up for Q6.
| Question | What it computes | How | Geometric meaning |
|---|---|---|---|
| Q1 | Slope of secant line through two fixed points | $\frac{f(b) - f(a)}{b - a}$ | Average rate of change between $a$ and $b$ |
| (intermediate) | Slope of secant with one point fixed and the other at $a + h$ | $\frac{f(a+h) - f(a)}{h}$ | Average rate of change over a shrinking interval |
| Q6 | Slope of tangent at one point (= derivative) | $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ | Instantaneous rate of change at $a$ |
The derivative is the limit of the secant slope. Q1 is the seed of Q6; the entire limit unit has been building toward this.
Everything in calc II onward is built from derivatives:
The power rule is the smallest derivative rule; the bigger ones (chain, product, quotient) are all variations on "differentiate this combination of things." The pattern of limit of a ratio of small changes never goes away.
Use the power rule to find $f'(a)$ for each function and value:
You've now got the full bridge: secant slope (Q1) → limit of secant slope → derivative (Q6). The next derivatives you'll meet in the pretest are probably:
Send the next pretest problem, or tell me which derivative rule to drill next.