Selecting Procedures for Calculating Derivatives

By topic 3.4 you have a rule for every way a function can be built: sums, products, quotients, compositions, inverses, and equations that never get solved for $y$. Knowing each rule is not the hard part of a derivative question. The hard part is deciding, in a few seconds, which rule goes first, whether to rewrite the function before using any rule, and whether the question is asking for a derivative at all.

The derivatives every procedure ends in

Every procedure below breaks a function down until only basic pieces are left, so these have to be automatic.

The derivatives of the three "co" functions (cosine, cotangent, cosecant) all carry a minus sign, which is a quick check on your memory. The power rule needs a variable base and a constant exponent ($x^5$). The exponential rule needs a constant base and a variable exponent ($5^x$, whose derivative is $5^x\ln 5$). When both base and exponent are constants ($5^5$, $\pi^2$, $e^3$, $\ln 7$), the expression is a number, and its derivative is 0.

The last operation decides the first rule

To pick the first rule, ask what you would do last if you evaluated the function at a number by hand. That operation is the function's outer structure, and its rule is the one you apply first. Every other rule shows up inside it, when you differentiate the pieces.

Look-alike functions can have different last operations. In $\sin x\cos x$ the last step multiplies two separately computed values, so it is a product. In $\sin(x\cos x)$ the last step is the sine, so it is a chain rule problem whose inside, $x\cos x$, needs the product rule. Likewise $\ln\left(x^2\right)$ takes a logarithm last, while $(\ln x)^2$ squares last. Where notation is ambiguous to you, insert the parentheses before differentiating.

Rewrite before you differentiate

A rewrite that is algebraically valid gives the same derivative with fewer chances for error. These are the ones worth doing:

• Roots and reciprocals become powers. $\frac{4}{\sqrt[3]{x}} = 4x^{-1/3}$, so its derivative is $-\frac{4}{3}x^{-4/3}$, with no quotient rule.
• A one-term denominator splits the fraction. $\frac{6x^2 - 5}{2x} = 3x - \frac{5}{2}x^{-1}$, whose derivative is $3 + \frac{5}{2}x^{-2}$.
• A constant denominator is a constant multiple. $\frac{x^2 + \sin x}{7}$ has derivative $\frac{2x + \cos x}{7}$.
• Logarithm properties turn products and powers into sums. For $x > 0$, $\ln\left(x^3\sqrt{x+1}\right) = 3\ln x + \frac{1}{2}\ln(x+1)$, whose derivative is $\frac{3}{x} + \frac{1}{2(x+1)}$.
• Identities shrink trig expressions. $\sin^2 x + \cos^2 x = 1$, $\frac{\sin x}{\cos x} = \tan x$, and $2\sin x\cos x = \sin 2x$.

A rewrite is only allowed if it is true for every $x$ in the domain. $\sqrt{x^2+16}$ is not $x + 4$, $\ln(x+5)$ is not $\ln x + \ln 5$, and $\frac{1}{x+x^2}$ does not split into $\frac{1}{x} + \frac{1}{x^2}$; a denominator can be split off only when it is a single term.

Rewriting also matters after you differentiate. A correct derivative often doesn't match any answer choice until you factor it or combine fractions. If you can't make two forms agree by algebra, evaluate both at a convenient number, such as $x = 1$, and compare.

When the function isn't a formula in $x$

Some questions don't give you $y = f(x)$ at all. The form of what you are given picks the procedure:

• An equation that was never solved for one variable. When $x$ and $y$ are mixed in an equation, differentiate both sides implicitly (topic 3.2), remembering that each $y$ term needs a factor $\frac{dy}{dx}$. For $y + \sin y = x$, that gives $\frac{dy}{dx}(1 + \cos y) = 1$, so $\frac{dy}{dx} = \frac{1}{1 + \cos y}$.
• An inverse function at a point. Don't try to find a formula for $f^{-1}$. Use $(f^{-1})'(a) = \frac{1}{f'(b)}$, where $f(b) = a$ (topic 3.3). For $f(x) = x^5 + x$, $f(1) = 2$ and $f'(1) = 6$, so $(f^{-1})'(2) = \frac{1}{6}$.
• Functions known only from a table or a graph. The rules don't change; each value of $f$ or $f'$ is looked up instead of computed (topic 3.1 shows the lookup order for compositions).
• A function defined in pieces, or with an absolute value. Differentiate each piece's formula on its own open interval. At a break point the rules say nothing: check continuity there, then compare the slopes from the two sides (topic 2.4).
• A limit that is a derivative. A limit of the form $\lim\limits_{h \to 0}\frac{f(a+h) - f(a)}{h}$ or $\lim\limits_{x \to a}\frac{f(x) - f(a)}{x - a}$ is $f'(a)$ by definition (topic 2.2). Identify $f$ and $a$, then use a rule. For instance, $\lim\limits_{h \to 0}\frac{(2+h)^5 - 32}{h}$ is $f'(2)$ for $f(x) = x^5$, since $2^5 = 32$, so it equals $5\cdot 2^4 = 80$. Expanding $(2+h)^5$ would also work, but much more slowly.

On a calculator-active question, the calculator's numerical derivative gives $f'(a)$ at a number directly. When a question asks for $f'(x)$ as an expression, or appears in a no-calculator section, the procedure is yours to choose and carry out by hand.

Worked examples

Where students lose points

On the exam

Multiple-choice answers are given in simplified form, and the wrong choices are the results of slips like the ones above: a missing chain factor, a wrong sign on a "co" function, the power rule used on $b^x$. If your answer is not among the choices, simplify it or evaluate it at a number before redoing the work. On free-response questions, write the general rule with function names first (for example, $h'(2) = \frac{f'(2)g(2) - f(2)g'(2)}{[g(2)]^2}$) and then substitute; that line shows which procedure you chose, and an unsimplified numerical answer is acceptable without a calculator.