The Derivative
Slope of the tangent line, instantaneous rate of change, velocity at an instant — one limit defines them all.
01From secant to tangent
Pick two points on a curve, $(a, f(a))$ and $(a+h,\, f(a+h))$. The line through them — a secant line — has slope
$$\frac{f(a+h) - f(a)}{h}$$
which is the average rate of change of $f$ over the interval. Now slide the second point toward the first by letting $h \to 0$. The secant lines pivot into a single limiting line — the tangent line — and their slopes converge to the instantaneous rate of change. That sliding picture is the hero graph above, and it is the entire conceptual content of the derivative. Everything else is bookkeeping.
02The definition
The derivative of $f$ at $a$ is
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
provided the limit exists. There's an equivalent form that's sometimes easier to use and to recognize on exams:
$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$
Notice this is always a $\frac{0}{0}$ indeterminate form — the numerator and denominator both vanish — which is why every Module 02 technique (expand, factor, cancel, conjugate) reappears here. The derivative is not new machinery; it is a specific limit you already know how to compute.
03Two worked computations
Example 1: $f(x) = x^2$.
$$f'(x) = \lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}\frac{2xh + h^2}{h} = \lim_{h\to 0}\,(2x + h) = 2x$$
Example 2: $f(x) = \dfrac{1}{x}$. Combine the fractions in the numerator first:
$$f'(x) = \lim_{h\to 0}\frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h\to 0}\frac{\frac{-h}{x(x+h)}}{h} = \lim_{h\to 0}\frac{-1}{x(x+h)} = -\frac{1}{x^2}$$
The rhythm is always: set up the difference quotient, do algebra until the lone $h$ in the denominator cancels, then let $h \to 0$.
04The derivative as a function, and notation
Computing $f'$ at a general $x$ (as above) produces a new function $f'(x)$ whose value at each point is the slope there. You'll see several notations for the same object, and you need to read all of them fluently:
$$f'(x) \qquad y' \qquad \frac{dy}{dx} \qquad \frac{d}{dx}f(x)$$
Lagrange's prime notation is compact; Leibniz's $\frac{dy}{dx}$ makes the "change in $y$ per change in $x$" meaning visible and will shine in the chain rule and related rates. Evaluation at a point is written $f'(3)$ or $\left.\frac{dy}{dx}\right|_{x=3}$.
05What the derivative means
Interpretation is half the course. If $s(t)$ is position at time $t$, then $s'(t)$ is velocity — the speedometer reading at instant $t$, not an average over a trip. If $C(x)$ is the cost of producing $x$ units, $C'(x)$ is marginal cost, roughly the cost of one more unit. In every case the units of $f'$ are (units of output) per (unit of input): a population in thousands measured over years has a derivative in thousands per year. Exam tip: when asked to "interpret $f'(5) = 12$," answer with the rate, the units, and the instant — "at time 5, the quantity is increasing at 12 units per hour."
06The tangent line equation
The tangent line to $y = f(x)$ at $x = a$ passes through $(a, f(a))$ with slope $f'(a)$, so point–slope form gives
$$y = f(a) + f'(a)\,(x - a)$$
Example: for $f(x) = x^2$ at $a = 3$: $f(3) = 9$, $f'(3) = 6$, so $y = 9 + 6(x - 3) = 6x - 9$. This little formula is also the linear approximation of $f$ near $a$ — the best straight-line stand-in for the curve — an idea that returns throughout calculus.
07Where derivatives fail to exist
The defining limit can fail, and the failures have geometric signatures:
- Corner — one-sided slopes disagree. $f(x) = |x|$ at $0$: slope $-1$ from the left, $+1$ from the right.
- Cusp — slopes blow up to $+\infty$ on one side and $-\infty$ on the other, like $x^{2/3}$ at $0$.
- Vertical tangent — the tangent exists but is vertical (infinite slope), like $x^{1/3}$ at $0$.
- Discontinuity — no continuity, no derivative. Ever.
That last item is a theorem worth stating: differentiable $\Rightarrow$ continuous. The converse is false — $|x|$ is the standard counterexample — so "continuous but not differentiable" is a phrase to keep handy.
08Reading derivatives from graphs
Given only a graph of $f$, you can sketch $f'$: where $f$ rises, $f' > 0$; where $f$ falls, $f' < 0$; at smooth peaks and valleys, $f' = 0$. Steepness translates to magnitude. This graphical fluency — matching a function's graph to its derivative's graph — is a favorite exam question and builds the intuition that Module 08 (curve sketching) will formalize. When you only have a table of values, estimate $f'(a)$ with the symmetric difference quotient $\frac{f(a+h) - f(a-h)}{2h}$ using the nearest available points.
The one-sentence summary. The derivative is a limit of difference quotients: $f'(x) = \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}$ — slope of the tangent, instantaneous rate of change, and velocity are all this same number wearing different outfits. Compute it with Module 02 algebra; interpret it with units.
Using the limit definition, find $f'(x)$ for $f(x) = 3x^2 - 2$.
Solution
$\dfrac{3(x+h)^2 - 2 - (3x^2 - 2)}{h} = \dfrac{6xh + 3h^2}{h} = 6x + 3h \to \mathbf{6x}$.
Using the limit definition, find $f'(x)$ for $f(x) = x^2 - 5x$.
Solution
Numerator: $(x+h)^2 - 5(x+h) - x^2 + 5x = 2xh + h^2 - 5h$.
Divide by $h$: $2x + h - 5 \to \mathbf{2x - 5}$.
Using the limit definition, find $f'(x)$ for $f(x) = \sqrt{x}$.
Solution
$\dfrac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \dfrac{h}{h(\sqrt{x+h}+\sqrt{x})} = \dfrac{1}{\sqrt{x+h}+\sqrt{x}} \to \mathbf{\dfrac{1}{2\sqrt{x}}}$.
Using the limit definition, find $f'(x)$ for $f(x) = \dfrac{1}{x + 1}$.
Solution
Numerator: $\dfrac{1}{x+h+1} - \dfrac{1}{x+1} = \dfrac{-h}{(x+h+1)(x+1)}$.
Divide by $h$ and let $h \to 0$: $\mathbf{-\dfrac{1}{(x+1)^2}}$.
Find the equation of the tangent line to $f(x) = x^2$ at $x = 3$.
Solution
$f(3) = 9$, $f'(x) = 2x$ so $f'(3) = 6$.
$y = 9 + 6(x - 3) = \mathbf{6x - 9}$.
Find the tangent line to $f(x) = \sqrt{x}$ at $x = 4$.
Solution
$f(4) = 2$, $f'(x) = \dfrac{1}{2\sqrt{x}}$ so $f'(4) = \dfrac{1}{4}$.
$y = 2 + \dfrac{1}{4}(x - 4) = \mathbf{\dfrac{x}{4} + 1}$.
The limit $\lim\limits_{h \to 0} \dfrac{(5+h)^3 - 125}{h}$ is a derivative in disguise. Identify $f$ and $a$, then evaluate it.
Solution
It's $f'(5)$ for $f(x) = x^3$ (since $125 = 5^3$).
$f'(x) = 3x^2$ (or expand directly), so the limit is $3 \cdot 25 = \mathbf{75}$.
Identify and evaluate: $\lim\limits_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9}$.
Solution
Alternate form of $f'(9)$ for $f(x) = \sqrt{x}$: value $\dfrac{1}{2\sqrt{9}} = \mathbf{\dfrac{1}{6}}$.
(Or algebraically: $x - 9 = (\sqrt{x}-3)(\sqrt{x}+3)$, cancel, get $\frac{1}{\sqrt{x}+3} \to \frac16$.)
A ball's height is $s(t) = 64t - 16t^2$ feet after $t$ seconds. Find the velocity at $t = 1$ and the time when velocity is zero.
Solution
$v(t) = s'(t) = 64 - 32t$. $v(1) = \mathbf{32 \text{ ft/s}}$.
$v = 0$ at $t = \mathbf{2}$ s — the peak of the flight.
Where does $f(x) = |x - 3|$ fail to be differentiable, and why?
Solution
At $\mathbf{x = 3}$: slope from the left is $-1$, from the right $+1$. Disagreeing one-sided slopes → a corner, so the defining limit DNE. ($f$ is continuous there.)
True or false: if $f$ is continuous at $a$, then $f$ is differentiable at $a$.
Solution
False. $f(x) = |x|$ at $0$ is continuous but not differentiable. (The true direction: differentiable $\Rightarrow$ continuous.)
Sketch-reasoning: $f$ increases on $(-\infty, 2)$, has a smooth peak at $x=2$, and decreases after. What are the signs of $f'$ on each interval and the value of $f'(2)$?
Solution
$f' > 0$ on $(-\infty, 2)$, $f'(2) = 0$, $f' < 0$ on $(2, \infty)$.
Given the table $f(1) = 4.0$, $f(1.1) = 4.63$, $f(0.9) = 3.42$, estimate $f'(1)$ with the symmetric difference quotient.
Solution
$f'(1) \approx \dfrac{f(1.1) - f(0.9)}{0.2} = \dfrac{4.63 - 3.42}{0.2} = \mathbf{6.05}$.
$P(t)$ is a town's population (in thousands) $t$ years after 2020, and $P'(4) = -1.5$. Interpret this sentence in plain English with units.
Solution
In 2024, the population is decreasing at a rate of 1{,}500 people per year.
Using the limit definition, find $f'(2)$ for $f(x) = x^3$ via the alternate form $\lim\limits_{x\to 2}\frac{x^3 - 8}{x - 2}$.
Solution
Factor: $x^3 - 8 = (x-2)(x^2 + 2x + 4)$. Cancel and substitute: $4 + 4 + 4 = \mathbf{12}$.
Find the point(s) on $y = x^2$ where the tangent line is parallel to $y = 4x - 7$.
Solution
Parallel means equal slopes: $2x = 4 \Rightarrow x = 2$. Point: $\mathbf{(2, 4)}$.
For $f(x) = x^{1/3}$, describe what happens to the derivative at $x = 0$.
Solution
$\dfrac{f(h) - f(0)}{h} = \dfrac{h^{1/3}}{h} = h^{-2/3} \to +\infty$. The graph has a vertical tangent at $0$ — not differentiable there.
The tangent line to $f$ at $x = 5$ is $y = 3x - 4$. Find $f(5)$ and $f'(5)$.
Solution
The tangent touches the curve at the point of tangency: $f(5) = 3(5) - 4 = \mathbf{11}$; its slope is the derivative: $f'(5) = \mathbf{3}$.
An object's position is $s(t) = t^2 - 6t + 8$ m. When is it moving left (negative velocity), and what is its speed at $t = 1$?
Solution
$v(t) = 2t - 6 < 0$ for $\mathbf{t < 3}$. At $t = 1$: $v = -4$, so speed $= |v| = \mathbf{4 \text{ m/s}}$.
Challenge: use the definition to differentiate $f(x) = \dfrac{x}{x+2}$.
Solution
$\dfrac{1}{h}\left[\dfrac{x+h}{x+h+2} - \dfrac{x}{x+2}\right] = \dfrac{(x+h)(x+2) - x(x+h+2)}{h(x+h+2)(x+2)} = \dfrac{2h}{h(x+h+2)(x+2)}$.
Cancel $h$, let $h \to 0$: $\mathbf{\dfrac{2}{(x+2)^2}}$.