Selecting Procedures for Derivatives

Every explicit formula on these pages is built from a few basic functions by adding, multiplying, dividing, and composing. Before differentiating, check whether a quick rewrite applies: the first three rows of the table. Then find the last operation used to build the function, the one you'd do last when evaluating it. That operation names the first rule. The rule hands you smaller pieces, and each piece gets the same treatment. Curves given by an equation, and inverses, are the last two rows.

What you see	What to do first
A sum of powers of $x$ over a single power of $x$	Rewrite with negative exponents
A log of a product, quotient, or power, with each piece positive	Log laws, then differentiate
$x$ in both a base and an exponent, with a positive base	Write as $e^{(\text{exponent})\ln(\text{base})}$
A sum, difference, or constant multiple	Differentiate term by term
A product $f \cdot g$ (or of more factors)	Product rule
Any other quotient	Quotient rule
A function of an inside expression, $f(u)$	Chain rule
An equation not solved for $y$	Implicit differentiation
The inverse of a known $f$	$\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}$ , where $f(a) = b$ and $f'(a) \ne 0$

Worked choices

Example 5

Differentiate

y = x^x

for

x > 0

, and find where its tangent line is horizontal.

Choice:

x

is in both the base and the exponent, so neither the power rule nor the exponential rule applies. Rewrite with base

e

y = e^{x\ln x}.

The chain rule, with the product rule for the exponent

x\ln x

, gives

\begin{aligned} \frac{dy}{dx} &= e^{x\ln x}\left(\ln x + 1\right) \\ &= x^x(\ln x + 1). \end{aligned}

Since

x^x > 0

, the slope is

0

only when

\ln x = -1

, at

x = \frac{1}{e}

, where

y = e^{-1/e} \approx 0.69

y = xˣ for x > 0 has a horizontal tangent at (1/e, e^(−1/e)) ≈ (0.37, 0.69). Near x = 0 the graph appears to approach height 1.

Example 7

Differentiate

y = \arctan\sqrt{x^2 - 1}

for

x > 1

Choice: a composite, three layers deep: arctan, the square root, and

x^2 - 1

. The arctan rule, then the chain rule twice:

\begin{aligned} \frac{dy}{dx} &= \frac{1}{1 + \left(x^2 - 1\right)} \cdot \frac{2x}{2\sqrt{x^2 - 1}} \\[4pt] &= \frac{1}{x^2} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{x\sqrt{x^2 - 1}}. \end{aligned}

That's the derivative of

\operatorname{arcsec} x

for

x > 1

, and in fact

\arctan\sqrt{x^2 - 1} = \operatorname{arcsec} x

there: both are the angle in a right triangle with hypotenuse

x

and adjacent side

1