Choosing a Differentiation Rule — AP® Calculus AB Notes

By now you know every differentiation rule on the AP® Calculus AB exam: the power rule, the derivatives of sine, cosine, $e^x$ and $\ln x$ , the product, quotient and chain rules, implicit differentiation and the inverse rules. The hard part is no longer a rule. It is deciding, in front of a long expression, which rule goes first. This page gives one question that always answers that, and practices it on the kinds of expressions the exam likes.

The one question: what is the last operation?

Imagine evaluating the expression at a number with a calculator. The operation you would do last is the outermost layer, and it names the first rule.

Decision diagram: what is the last operation? A sum or difference: differentiate term by term. A product: product rule u′v + uv′. A quotient: quotient rule or rewrite with a power. A function of a function: chain rule f′(g(x))·g′(x). Then repeat inside each piece. — **Figure 1.** The last operation names the first rule; then ask the same question of each piece. LevlPrep original diagram.

Expression	Last operation	First rule
$x^4 - 3\cos x + e^x$	adding and subtracting	term by term
$x^2\sin(3x)$	multiplying $x^2$ by $\sin(3x)$	product rule
$\frac{e^{2x}}{x + 1}$	dividing	quotient rule
$\left(xe^x\right)^3$	cubing	chain rule
$\sin^2\left(x^3\right)$	squaring	chain rule
$\ln\left(x^2 + 1\right)$	taking the log	chain rule

After the first rule, each piece is a smaller expression. Ask the same question of each piece. Keep going until only basic derivatives remain. Each layer gets exactly one rule.

Worked examples

Worked example 1. Differentiate $y = x^2\sin(3x)$ .

Last operation: a product. Product rule: $y'$ $= (x^2)'\sin(3x) + x^2\left(\sin(3x)\right)'$ .

The piece $\sin(3x)$ is a composition: chain rule, $\left(\sin(3x)\right)' = 3\cos(3x)$ .

So $y'$ $= 2x\sin(3x) + 3x^2\cos(3x)$ .

Worked example 2. Differentiate $y = \left(xe^x\right)^3$ .

Last operation: cubing. Chain rule: $y' = 3\left(xe^x\right)^2 \cdot \left(xe^x\right)'$ .

The inside $xe^x$ is a product: $\left(xe^x\right)' = e^x + xe^x$ .

So $y'$ $= 3\left(xe^x\right)^2\left(e^x + xe^x\right)$ . There is no need to simplify further; the exam accepts this form, and extra algebra only adds risk.

Worked example 3. Values from a table: $f(1) = 2$ , $f'(1) = 3$ . Find $D'(1)$ for $D(x) = x \cdot f(x^2)$ .

Last operation: multiplying $x$ by $f(x^2)$ . Product rule: $D'(x)$ $= 1 \cdot f(x^2) + x \cdot \left(f(x^2)\right)'$ .

The piece $f(x^2)$ is a composition: $\left(f(x^2)\right)' = f'(x^2) \cdot 2x$ .

At $x = 1$ : $D'(1)$ $= f(1)$ $+ 1 \cdot f'(1) \cdot 2$ $= 2$ $+ 6$ $= 8$ .

Rewrite first, when it removes a rule

Some quotients are really sums of powers in disguise. Before reaching for the quotient rule, check whether the bottom is a single power of $x$ or a constant:

$\frac{5}{x^3} = 5x^{-3}$ , so its derivative is $-15x^{-4}$ . No quotient rule.
$\frac{x^2 + 3}{\sqrt{x}}$ $= x^{3/2}$ $+ 3x^{-1/2}$ , so its derivative is $\frac{3}{2}x^{1/2} - \frac{3}{2}x^{-3/2}$ .
$\frac{\sin x}{4} = \frac{1}{4}\sin x$ , so its derivative is $\frac{1}{4}\cos x$ .

Logarithm rules can help too: $\ln\left(x^3\right) = 3\ln x$ . But do not rewrite just to rewrite. If the quotient rule is quick, use it.

Beyond formulas

Two more situations name their own method. An equation that mixes $x$ and $y$ , such as $x^2 + xy + y^2 = 7$ , calls for implicit differentiation. A question about the slope of an inverse, $\left(f^{-1}\right)'(a)$ , calls for $\frac{1}{f'\left(f^{-1}(a)\right)}$ . And when the functions come as a table or a graph, the rules are the same; you simply read each value at the right input. A corner on a graph (where the function has no derivative) means a rule that needs $f'$ there cannot be used.

Slips that cost points

Multiplying derivatives. $(fg)' \ne f'g'$ and $\left(\frac{f}{g}\right)' \ne \frac{f'}{g'}$ .
Reversing the quotient rule's numerator. It is $f'g - fg'$ ; the derivative of the top comes first.
Missing a layer. Count the layers and check that your answer has one factor or term for each.
Three-factor products. $(fgh)'$ $= f'gh + fg'h + fgh'$ : each factor takes a turn being differentiated.