Polynomial with both a linear and a quadratic term. The difference quotient is a sum of two parts, and they each contribute to the answer in different ways. The constants vanish; the variable terms survive.
The derivative of any polynomial is also a polynomial — the only rule at play is the same 5-step dance. The reason to do this problem by the limit definition is to internalize the dance; once it's muscle memory, the shortcut (power rule) lands naturally.
For a polynomial, the $h$ terms always collect into a single factor of $h$ that cancels cleanly — no conjugate tricks, no common denominators, just expand-and-collect. That's why polynomial problems are the cleanest place to see the 5-step shape.
The derivative at a specific point $a$ is:
$$f'(a) \;=\; \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$Same shape as the general formula — just plug $a = -1$ for $x$ throughout. (Some textbooks write the definition with a different letter instead of $a$; they're all the same definition.)
Given $f(x) = -3x + 4x^2$ and $a = -1$. We compute $\displaystyle f'(-1) = \lim_{h \to 0} \dfrac{f(-1+h) - f(-1)}{h}$ in five steps.
Step 1 — Write the difference quotient (with $a = -1$).
$$f'(-1) \;=\; \lim_{h \to 0} \frac{f(-1+h) - f(-1)}{h}$$Plug in $x = -1$ in both spots. Don't substitute $h$ for $x$ — we're using $a = -1$ as a fixed number and $h$ is still our "step size" that goes to zero.
Step 2 — Compute $f(-1+h)$ and $f(-1)$ separately.
$f(-1) = -3(-1) + 4(-1)^2 = 3 + 4 = 7$.
$$f(-1+h) \;=\; -3(-1+h) + 4(-1+h)^2$$Expand each piece carefully:
$$-3(-1+h) \;=\; 3 - 3h$$ $$4(-1+h)^2 \;=\; 4(1 - 2h + h^2) \;=\; 4 - 8h + 4h^2$$Adding them together:
$$f(-1+h) \;=\; (3 - 3h) + (4 - 8h + 4h^2) \;=\; 7 - 11h + 4h^2$$Step 3 — Subtract $f(-1) = 7$.
$$f(-1+h) - f(-1) \;=\; (7 - 11h + 4h^2) - 7 \;=\; -11h + 4h^2$$The $7$'s cancel — same trick as Q1, just with numbers instead of fractions. The constant terms disappear because adding a constant to a function only shifts it up; shifting doesn't change the slope anywhere.
Step 4 — Divide by $h$ and factor.
$$\frac{-11h + 4h^2}{h} \;=\; \frac{h(-11 + 4h)}{h} \;=\; -11 + 4h$$Both terms in the numerator had an $h$ in them, so both contribute. This is the moment to notice: the $4$ in $4x^2$ and the $-3$ in $-3x$ are both doing something — neither is decorative.
Step 5 — Take the limit as $h \to 0$.
$$f'(-1) \;=\; \lim_{h \to 0} (-11 + 4h) \;=\; -11 + 0 \;=\; -11$$Final answer:
The two constants $3$ and $4$ in $f(x) = -3x + 4x^2$ are doing very different jobs. Watch how each one travels through the dance:
A quick sanity check: at $x = -1$, the derivative of the linear part $-3x$ alone would be $-3$ (constant slope). The derivative of the quadratic part $4x^2$ alone at $x = -1$ would be $8 \cdot (-1) = -8$. So the total is $-3 + (-8) = -11$. ✓
Every derivative-via-the-definition problem in this course is these five steps, in this order:
The trick with polynomials is to expand all the way before subtracting — don't try to "see" which terms will cancel in advance, just expand, then let the algebra do its work.
Use the definition (the 5 steps above) to find $f'(2)$ for $f(x) = 5x - 2x^2$.
Hint: same shape, different numbers. Pay attention to the signs — a negative leading coefficient flips the sign of the quadratic's contribution to the derivative.
Step 1: $\displaystyle f'(2) = \lim_{h \to 0} \dfrac{f(2+h) - f(2)}{h}$
Step 2: $f(2) = 5(2) - 2(2)^2 = 10 - 8 = 2$.
$f(2+h) = 5(2+h) - 2(2+h)^2 = (10 + 5h) - 2(4 + 4h + h^2) = 10 + 5h - 8 - 8h - 2h^2 = 2 - 3h - 2h^2$
Step 3: $f(2+h) - f(2) = (2 - 3h - 2h^2) - 2 = -3h - 2h^2$
Step 4: $\dfrac{h(-3 - 2h)}{h} = -3 - 2h$
Step 5: $f'(2) = \lim_{h \to 0}(-3 - 2h) = -3$
Sanity check: derivative of the linear part $5x$ is $5$ (constant), derivative of $-2x^2$ is $-4x$, evaluated at $x = 2$ gives $-8$. Total: $5 + (-8) = -3$. ✓
Three directions once you finish Q2: