Math Core

In BC you will differentiate parametric curves, polar curves and power series, and every one of those rests on the basic rules. This lesson reviews the limit definition, differentiability, and the full rule set, so that the mechanics are automatic before new material arrives.

The derivative as a limit

The derivative of $f$ at $x = a$ is

f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a},

when the limit exists. Geometrically it is the slope of the tangent line; in context it is the instantaneous rate of change, with units of (output units) per (input unit).

AP exams love to hand you a limit and ask you to recognize it as a derivative. Look for the pattern "change in $f$ over change in input" with the input change going to $0$ . Then identify the function and the point, and differentiate instead of grinding through algebra.

Worked example: Reading a limit as a derivative

Evaluate $\displaystyle\lim_{h \to 0} \frac{\sqrt[3]{8 + h} - 2}{h}$ and $\displaystyle\lim_{x \to \pi} \frac{\sin x}{x - \pi}$ .

Solution. The first is $f'(8)$ for $f(x) = x^{1/3}$ , since $f(8) = 2$ . Then $f'(x) = \dfrac{1}{3}x^{-2/3}$ and $f'(8) = \dfrac{1}{3} \cdot \dfrac{1}{4} = \dfrac{1}{12}$ .

The second is $g'(\pi)$ for $g(x) = \sin x$ , because $\sin \pi = 0$ lets you write the numerator as $\sin x - \sin \pi$ . So the limit is $\cos \pi = -1$ .

Differentiability

Differentiability implies continuity

If $f$ is differentiable at $a$ , then $f$ is continuous at $a$ . The converse is false.

So a discontinuity always kills differentiability. A continuous function can still fail to be differentiable at:

a corner, where the one-sided derivatives differ, like $|x|$ at $0$ ;
a cusp, where the slopes go to $+\infty$ on one side and $-\infty$ on the other, like $x^{2/3}$ at $0$ ;
a vertical tangent, where the slope goes to $\pm\infty$ from both sides, like $\sqrt[3]{x}$ at $0$ .

Continuous but not differentiable: a corner at x = 2 and a vertical tangent at x = −2.Open in grapher →

For a piecewise function to be differentiable at a seam $x = c$ , you need two conditions: the pieces must meet (continuity), and their derivatives must match at $c$ (equal one-sided derivatives). Two conditions let you solve for two unknown constants.

The rules

Derivative rules

Function	Derivative
$x^n$ (any real $n$ )	$nx^{n-1}$
$e^x$ , $a^x$	$e^x$ , $a^x \ln a$
$\ln x$ , $\log_a x$	$\dfrac{1}{x}$ , $\dfrac{1}{x \ln a}$
$\sin x$ , $\cos x$	$\cos x$ , $-\sin x$
$\tan x$ , $\cot x$	$\sec^2 x$ , $-\csc^2 x$
$\sec x$ , $\csc x$	$\sec x \tan x$ , $-\csc x \cot x$
$fg$	$f'g + fg'$
$\dfrac{f}{g}$	$\dfrac{f'g - fg'}{g^2}$

A memory aid for the trig table: every "co-" function ( $\cos$ , $\cot$ , $\csc$ ) has a minus sign in its derivative. For $\ln|x|$ , the derivative is still $\dfrac{1}{x}$ on its whole domain, which matters when you antidifferentiate later.

Before using the quotient rule, see whether you can simplify. $\dfrac{x^3 - 2\sqrt{x}}{x}$ is just $x^2 - 2x^{-1/2}$ , whose derivative is $2x + x^{-3/2}$ . Rewriting takes seconds and avoids errors.

Worked example: Product and quotient together

Find $f'(x)$ for $f(x) = \dfrac{x e^x}{x^2 + 1}$ , and evaluate $f'(0)$ .

Solution. The numerator's derivative is $(xe^x)' = e^x + xe^x = (x + 1)e^x$ by the product rule. Then by the quotient rule,

f'(x) = \frac{(x+1)e^x(x^2 + 1) - xe^x \cdot 2x}{(x^2+1)^2} = \frac{e^x\left(x^3 - x^2 + x + 1\right)}{(x^2 + 1)^2}.

At $x = 0$ : $f'(0) = \dfrac{1 \cdot 1}{1} = 1$ .

Working from tables

Many AP questions give only values of $f$ , $g$ and their derivatives at a few points. The rules still apply; you just plug in numbers instead of formulas.

Worked example: Rules with tabled values

Suppose $f(1) = 4$ , $f'(1) = -2$ , $g(1) = 3$ and $g'(1) = 5$ . Let $P(x) = x^2 f(x)$ and $Q(x) = \dfrac{f(x)}{g(x)}$ . Find $P'(1)$ and $Q'(1)$ .

Solution. $P'(x) = 2x f(x) + x^2 f'(x)$ , so $P'(1) = 2 \cdot 4 + 1 \cdot (-2) = 6$ .

$Q'(1) = \dfrac{f'(1)g(1) - f(1)g'(1)}{g(1)^2} = \dfrac{(-2)(3) - (4)(5)}{9} = -\dfrac{26}{9}$ .

Common mistake

The derivative of a product is not the product of the derivatives, and in the quotient rule the order matters: it is $f'g - fg'$ on top ("low d-high minus high d-low"). Reversing the terms flips the sign of your answer.

Higher derivatives

$f''$ measures how $f'$ changes: concavity for graphs, acceleration for motion. Keep differentiating with the same rules. A useful pattern to recognize: the derivatives of $\sin x$ cycle with period $4$ ( $\sin, \cos, -\sin, -\cos$ ), and the $n$ th derivative of $e^{kx}$ is $k^n e^{kx}$ . Patterns like these become the heart of Taylor series later in BC.

Tip

When a problem asks for a derivative at one point, you rarely need a simplified formula. Differentiate, substitute, and simplify the number.

Practice

Practice 1

Evaluate $\displaystyle\lim_{h \to 0} \frac{(2 + h)^5 - 32}{h}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Evaluate $\displaystyle\lim_{h \to 0} \frac{\cos\left(\frac{\pi}{3} + h\right) - \frac{1}{2}}{h}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find $\dfrac{d}{dx}\left(x^2 \ln x\right)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

Find $\dfrac{d}{dx}\left(\dfrac{x^2 + 1}{x - 3}\right)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 5

Let $f(x) = \sec x \tan x$ . Find $f'\left(\dfrac{\pi}{4}\right)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

(AP-style) Values of $g$ , $h$ and their derivatives at $x = 2$ are $g(2) = 3$ , $g'(2) = -1$ , $h(2) = -2$ , $h'(2) = 5$ . If $k(x) = \dfrac{g(x)}{h(x)}$ , what is $k'(2)$ ?

Practice 7

Which function is differentiable at $x = 0$ ?

Practice 8

Find the constants $a$ and $b$ that make $f$ differentiable at $x = 2$ , and give your answer as the ordered pair $(a, b)$ :

f(x) = \begin{cases} ax^3, & x \le 2 \\ bx^2 + 4, & x > 2. \end{cases}

Enter a point like (2, -3)