Lesson 4 of 6Derivative Rules

Unit 2 · Lesson 4 of 6

Derivatives of Rational Functions

A quotient can be differentiated two ways: rewrite it as a product with a negative exponent and use the product and chain rules, or use the quotient rule directly. Both give the same answer.

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Key things to remember

1Quotient rule: $(\frac{f}{g})^{'} = \frac{f ^{'} g - f g ^{'}}{g ^{2}}$ , where $g \neq = 0$ .
2Alternative: $\frac{f}{g} = f \cdot g^{- 1}$ , then use the product and chain rules.
3Constant numerator: $\frac{d}{d x} [\frac{k}{g ( x )}] = - \frac{k g ^{'} ( x )}{[ g ( x ) ] ^{2}} .$
4 $(\frac{f}{g})^{'} \neq = \frac{f ^{'}}{g ^{'}}$ , and the numerator order $f^{'} g - f g^{'}$ matters.
5A rational function's derivative does not exist where its denominator is $0$ ; horizontal tangents occur where the numerator of $f^{'}$ is $0$ .

Rational functions, one polynomial divided by another, model many real situations: the average cost per item when fixed costs are shared over $x$ items, or the concentration of a medication in the blood, which rises and then falls, as in $c (t) = \frac{5 t}{t ^{2} + 1}$ . Just as with products, the derivative of a quotient is NOT the quotient of the derivatives. You already have every tool you need to find the right answer; this lesson shows two equivalent ways to organize the work.

Method 1: product and chain rules

Dividing by $g (x)$ is the same as multiplying by $[g (x)]^{- 1}$ . So $f (x) = \frac{x ^{2} + 1}{x - 1}$ can be rewritten as the product $(x^{2} + 1) (x - 1)^{- 1}$ . The product rule handles the multiplication, and the chain rule (with the power rule for a negative exponent) differentiates $(x - 1)^{- 1}$ as $- 1 (x - 1)^{- 2} \cdot 1$ .

f^{'} (x) = 2 x (x - 1)^{- 1} + (x^{2} + 1) (- 1) (x - 1)^{- 2} = \frac{2 x ( x - 1 ) - ( x ^{2} + 1 )}{( x - 1 ) ^{2}} = \frac{x ^{2} - 2 x - 1}{( x - 1 ) ^{2}}

Rewrite with a common denominator of (x − 1) squared to combine the two terms.

Fig 1The power rule table. Look at the x to the negative one row in the dashed box at the bottom: its derivative, negative x to the negative two, is the step used whenever a denominator is rewritten with a negative exponent.

Method 2: the quotient rule

(\frac{f}{g})^{'} = \frac{f ^{'} g - f g ^{'}}{g ^{2}}, g (x) \neq = 0

The quotient rule: derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

The quotient rule is just Method 1 done once in general. Write $\frac{f}{g} = f \cdot g^{- 1}$ . The product and chain rules give $f^{'} g^{- 1} + f (- 1) g^{- 2} g^{'} = \frac{f ^{'}}{g} - \frac{f g ^{'}}{g ^{2}},$ and over the common denominator $g^{2}$ that is $\frac{f ^{'} g - f g ^{'}}{g ^{2}}$ . Because the numerator is a subtraction, the order matters, unlike in the product rule.

A memory aid

Low d-high minus high d-low, over low squared. "Low" is the denominator

g

and "high" is the numerator

f

, so the numerator of the derivative is

g f^{'} - f g^{'}

. Start with the derivative of the top and you cannot get the sign backwards.

AProduct and chain rules

Best when the numerator is a constant: $\frac{6}{x ^{2} + 2} = 6 (x^{2} + 2)^{- 1}$ needs only the chain rule.
Uses rules you already know, so there is no new formula to remember.
Needs a common denominator at the end to simplify.

BQuotient rule

Best when both the top and bottom contain $x$ .
Produces one fraction directly, with the denominator $g^{2}$ left factored.
The order in the numerator matters: $f^{'} g - f g^{'}$ , never the reverse.

VideoJensenMath works examples with the quotient rule, written in the same numerator-minus pattern over the denominator squared as this section.Watch on YouTube

Where the derivative exists

The derivative of a rational function has denominator $g^{2}$ , which is $0$ exactly where $g$ is $0$ . At those $x$ -values the original function is undefined (there is a vertical asymptote or a hole), so the derivative does not exist there either. Everywhere else, $f^{'} (x) = 0$ only when the NUMERATOR of $f^{'} (x)$ is $0$ , which is how you find horizontal tangents.

Cancel factors, not terms

\frac{x ^{2} - 2 x - 1}{( x - 1 ) ^{2}}

nothing cancels:

x^{2} - 2 x - 1

does not have

(x - 1)

as a factor. You may only cancel a factor that multiplies the WHOLE numerator and the WHOLE denominator. When you only need a value such as

f^{'} (3)

, it is safest to substitute before simplifying at all.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

For

f (x) = \frac{x ^{2} + 1}{x - 1}

, find

f^{'} (3)

using the quotient rule.

Example 2

Find the equation of the tangent line to

f (x) = \frac{2 x + 1}{x - 3}

x = 4

Example 3

Find the points where

f (x) = \frac{x}{x ^{2} + 1}

has a horizontal tangent.

Example 4

For

g (x) = \frac{6}{x ^{2} + 2}

, find

g^{'} (1)

without the quotient rule.

Where marks get lost

Reversing the numerator to $f g^{'} - f^{'} g$ . That gives exactly the negative of the correct derivative. Always start with the derivative of the top.
Forgetting to square the denominator, or squaring the numerator as well.
Writing $(\frac{f}{g})^{'} = \frac{f ^{'}}{g ^{'}}$ . Test it on $\frac{x ^{2}}{x} = x$ : it would give $\frac{2 x}{1} = 2 x$ , but the derivative of $x$ is $1$ .
Cancelling terms instead of factors, for example crossing out an $x^{2}$ that appears in a sum on top and in $(x^{2} + 1)^{2}$ on the bottom.
Reporting a derivative at a value that makes the denominator $0$ . The function itself is undefined there, so the derivative does not exist.

Flashcards 6

Quotient rule

Memory aid for the quotient rule

Rewrite

\frac{f}{g}

as a product

\frac{d}{d x} [\frac{1}{g ( x )}]

+ 2 more

Practice

For

f (x) = \frac{3 x}{x + 2}

, find

f^{'} (1)

. Give your answer to 3 decimal places.

Which is the quotient rule?

Find the slope of the tangent to

y = \frac{1}{x ^{2} + 1}

x = 1

. Give the exact decimal value.

Match each quotient to the way you would rewrite it as a product (or a single power).

A. $5 (2 x + 1)^{- 1}$
B. $x (x + 4)^{- 1}$
C. $x^{- \frac{1}{2}}$
D. $x^{- 3}$

\frac{1}{x ^{3}}

\frac{5}{2 x + 1}

\frac{x}{x + 4}

\frac{1}{x}

f (x) = \frac{x ^{2}}{x - 2}

has two horizontal tangents. One is at

x = 0

. At what other

x

-value is the tangent horizontal?

For

f (x) = \frac{x + 1}{x - 5}

, where does

f^{'} (x)

fail to exist?

Put the steps in order to find the points where a rational function

f

has a horizontal tangent.

1Set the numerator of $f^{'} (x)$ equal to $0$
2Substitute each $x$ -value into $f (x)$ to find the point
3Solve for $x$ , keeping only values where the denominator is not $0$
4Differentiate $f$ with the quotient rule

For

f (x) = \frac{2 x + 1}{x - 3}

, find

f^{'} (5)

. Give the exact decimal value.

Name the rule

(\frac{f}{g})^{'} = \frac{f ^{'} g - f g ^{'}}{g ^{2}}

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