Math Core

AB spends three full units applying derivatives: rates in context, related rates, linear approximation, L'Hôpital's rule, the big existence theorems, and curve analysis. On the BC exam these ideas reappear inside parametric motion, polar curves and series, so this review condenses them into the facts and moves you must have ready.

Rates in context and motion

A derivative is a rate: if $V(t)$ is volume in liters and $t$ is minutes, then $V'(t)$ is in liters per minute. On free response, interpret a value with units, the input value, and increasing or decreasing, for example "at $t = 4$ minutes, the volume is decreasing at 3 liters per minute."

For straight-line motion with position $x(t)$ : velocity $v = x'$ , acceleration $a = v' = x''$ , and speed $= |v|$ .

Speeding up or slowing down

Speed is increasing when $v$ and $a$ have the same sign, and decreasing when they have opposite signs. The particle changes direction where $v$ changes sign.

Total distance traveled is not the net change in position: split the time interval at every direction change and add the absolute displacements.

Related rates

Draw and label, using variables for anything that changes.
Write an equation relating the quantities (geometry, similar triangles, Pythagoras).
Differentiate both sides with respect to $t$ (implicitly, every variable gets a $\frac{d}{dt}$ ).
Substitute the known values only after differentiating, then solve.

Worked example: A sliding ladder

A 13-foot ladder leans against a wall. The bottom slides away from the wall at 2 ft/s. How fast is the angle $\theta$ between the ladder and the ground changing when the bottom is 5 ft from the wall?

Solution. With $x$ the distance from the wall, $\cos\theta = \dfrac{x}{13}$ . Differentiate: $-\sin\theta\,\dfrac{d\theta}{dt} = \dfrac{1}{13}\dfrac{dx}{dt}$ . When $x = 5$ the height is $12$ , so $\sin\theta = \dfrac{12}{13}$ . Then

-\frac{12}{13}\frac{d\theta}{dt} = \frac{2}{13} \quad\Longrightarrow\quad \frac{d\theta}{dt} = -\frac{1}{6} \text{ radian per second}.

Linearization

Near $x = a$ , a differentiable function is close to its tangent line:

f(x) \approx L(x) = f(a) + f'(a)(x - a).

If $f$ is concave up near $a$ , the tangent line lies below the curve and $L$ underestimates; if concave down, it overestimates. This local linear model is the first Taylor polynomial, which BC extends to higher degrees with an error bound.

L'Hôpital's rule

If $\displaystyle\lim \frac{f(x)}{g(x)}$ has the form $\dfrac{0}{0}$ or $\dfrac{\pm\infty}{\pm\infty}$ , then

\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)},

provided the right-hand limit exists (or is infinite). Differentiate top and bottom separately, not as a quotient.

Other indeterminate forms must be rewritten first. For $0 \cdot \infty$ , move one factor into the denominator. For $1^\infty$ , $0^0$ and $\infty^0$ , take the natural log, find the limit of the log, then exponentiate. BC uses these constantly for improper integrals and series tests.

Worked example: A 1-to-the-infinity limit

Evaluate $\displaystyle\lim_{x \to \infty}\left(1 + \frac{3}{x}\right)^{x}$ .

Solution. Let $y = \left(1 + \frac{3}{x}\right)^x$ , so $\ln y = x\ln\left(1 + \frac{3}{x}\right) = \dfrac{\ln(1 + 3/x)}{1/x}$ , which is $\dfrac{0}{0}$ . By L'Hôpital,

\lim_{x\to\infty} \frac{\frac{1}{1 + 3/x}\cdot\left(-\frac{3}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x\to\infty}\frac{3}{1 + 3/x} = 3.

So $\ln y \to 3$ and $y \to e^3$ .

Common mistake

Check the form before every application of L'Hôpital. Applying it to a limit that is not $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ gives wrong answers, and the AP exam requires you to show the indeterminate form (for example, by stating that numerator and denominator both approach $0$ ).

The existence theorems

Extreme Value Theorem: a function continuous on a closed interval $[a, b]$ attains an absolute maximum and an absolute minimum there.
Mean Value Theorem: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then some $c$ in $(a, b)$ has $f'(c) = \dfrac{f(b) - f(a)}{b - a}$ : the instantaneous rate equals the average rate somewhere.

As with the IVT, state the hypotheses explicitly when you invoke a theorem on free response.

MVT for x³ − 3x on [0, 3]: the secant (dashed) has slope 6, and the tangent at c = √3 is parallel to it.Open in grapher →

Curve analysis

Critical points: where $f'(x) = 0$ or $f'$ is undefined (inside the domain).
First derivative test: $f'$ changes $+$ to $-$ : relative max; $-$ to $+$ : relative min; no sign change: neither.
Second derivative test: if $f'(c) = 0$ and $f''(c) > 0$ , relative min; $f''(c) < 0$ , relative max; $f''(c) = 0$ , inconclusive.
Inflection point: where $f''$ changes sign (equivalently, where $f'$ changes from increasing to decreasing or vice versa). $f''(c) = 0$ alone is not enough.
Candidates test for absolute extrema on $[a, b]$ : evaluate $f$ at every critical point in the interval and at both endpoints; the largest and smallest values win.

Worked example: Absolute extrema on a closed interval

Find the absolute maximum and minimum of $f(x) = x - 2\sin x$ on $[0, \pi]$ .

Solution. $f'(x) = 1 - 2\cos x = 0$ when $\cos x = \dfrac{1}{2}$ , so $x = \dfrac{\pi}{3}$ . Candidates: $f(0) = 0$ , $f\left(\frac{\pi}{3}\right) = \dfrac{\pi}{3} - \sqrt{3} \approx -0.685$ , $f(\pi) = \pi$ . The absolute maximum is $\pi$ at $x = \pi$ and the absolute minimum is $\dfrac{\pi}{3} - \sqrt{3}$ at $x = \dfrac{\pi}{3}$ .

Tip

On the exam, a graph of $f'$ is a common prompt. Read it directly: $f$ increases where $f'$ is above the axis, $f$ is concave up where $f'$ is rising, and inflection points of $f$ sit at the peaks and valleys of $f'$ .