2.9 - The Quotient Rule

Introduction

When you need the derivative of two functions divided, use the quotient rule. It looks similar to the product rule but has some important differences.

The Formula

y = \dfrac{f(x)}{g(x)}

\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f' \cdot g - f \cdot g'}{g^2}

In words:

Derivative of the top

Times the bottom unchanged

Minus the top unchanged

Times the derivative of the bottom

All over the bottom squared

Memory aid: “low dee-high minus high dee-low, over low-low.”

Key Differences from the Product Rule

Feature	Product Rule	Quotient Rule
Operation	$f \cdot g$	$\frac{f}{g}$
Sign	Plus $(+)$	Minus $(-)$
Denominator	None	$g^2$
Formula	$f'g + fg'$	$\frac{f'g - fg'}{g^2}$

Critical points:

The quotient rule uses subtraction, not addition

The denominator is $g^2$ , not $(g')^2$

Order matters, you can’t swap numerator and denominator

Basic Examples

Example 1: Polynomial Over Polynomial

Find the derivative of

y = \dfrac{x^3}{2x + 5}

Top: $f = x^3$ , so $f' = 3x^2$

Bottom: $g = 2x + 5$ , so $g' = 2$

\frac{dy}{dx} = \frac{(3x^2)(2x + 5) - (x^3)(2)}{(2x + 5)^2}

Expand the numerator:

= \frac{6x^3 + 15x^2 - 2x^3}{(2x + 5)^2} = \frac{4x^3 + 15x^2}{(2x + 5)^2}

Factor if you want:

= \frac{x^2(4x + 15)}{(2x + 5)^2}

Example 2: Exponential Over Linear

Differentiate

g(x) = \dfrac{e^x}{3x - 1}

Top: $e^x$ , derivative $e^x$

Bottom: $3x - 1$ , derivative $3$

g'(x) = \frac{(e^x)(3x - 1) - (e^x)(3)}{(3x - 1)^2} = \frac{3xe^x - e^x - 3e^x}{(3x - 1)^2}

Combine and factor:

= \frac{3xe^x - 4e^x}{(3x - 1)^2} = \frac{e^x(3x - 4)}{(3x - 1)^2}

Example 3: Trig Function in a Quotient

Find the derivative of

h(x) = \dfrac{\sin x}{x^2 + 1}

Top: $\sin x$ , derivative $\cos x$

Bottom: $x^2 + 1$ , derivative $2x$

h'(x) = \frac{(\cos x)(x^2 + 1) - (\sin x)(2x)}{(x^2 + 1)^2} = \frac{x^2\cos x + \cos x - 2x\sin x}{(x^2 + 1)^2}

That’s as simplified as it gets. Not everything factors nicely.

Don’t Multiply Out the Denominator

When you have

(g(x))^2

on the bottom, leave it as a squared binomial.

\frac{dy}{dx} = \frac{2x + 5}{(x - 3)^2}

Correct: leave it as

(x - 3)^2

Wrong: expanding to

x^2 - 6x + 9

. It’s extra work and more chances to mess up.

When to Simplify First

Sometimes you can dodge the quotient rule entirely by simplifying first.

Example 4: Single Term in the Denominator

Find the derivative of

f(x) = \dfrac{6x^3 - 4x^2 + 2x}{2x^2}

You could grind through the quotient rule. But splitting the fraction is way easier.

Step 1: Split it up.

f(x) = \frac{6x^3}{2x^2} - \frac{4x^2}{2x^2} + \frac{2x}{2x^2}

Step 2: Simplify each term.

= 3x - 2 + x^{-1}

Step 3: Power rule.

f'(x) = 3 - x^{-2}

Step 4: Rewrite with positive exponents.

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

When to Look for This

Watch for:

A single term in the denominator

Common factors in the numerator and denominator

Any easy algebraic simplification

Rule of thumb: if the denominator is a single term like

x

x^2

, or

3x

, split the fraction before differentiating.

Example 5: Another Simplify-First Problem

Find

g'(x)

for

g(x) = \dfrac{10x^4 + 5x^2}{5x}

g(x) = \frac{10x^4}{5x} + \frac{5x^2}{5x} = 2x^3 + x

g'(x) = 6x^2 + 1

Using Tables with the Quotient Rule

Example 6: Table-Based Problem

Find

h'(2)

where

h(x) = \dfrac{f(x)}{g(x)}

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
2	6	3	4	-2

Step 1: Write the formula.

h'(x) = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}

Step 2: Substitute the table values at

x = 2

h'(2) = \frac{(3)(4) - (6)(-2)}{(4)^2}

Step 3: Simplify.

= \frac{12 + 12}{16} = \frac{24}{16} = \frac{3}{2}

Example 7: Constant Multiple with a Quotient

Using the same table, find

k'(2)

where

k(x) = \dfrac{2f(x)}{g(x) + 1}

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
2	6	3	4	-2

Factor out the constant, then apply the quotient rule:

k'(x) = 2 \cdot \frac{f'(x)(g(x) + 1) - f(x) \cdot g'(x)}{[g(x) + 1]^2}

Evaluate at

x = 2

k'(2) = 2 \cdot \frac{(3)(5) - (6)(-2)}{25} = 2 \cdot \frac{15 + 12}{25} = 2 \cdot \frac{27}{25} = \frac{54}{25}

Handling Negative Signs

If there’s a negative in front of the whole quotient, factor it out first.

Example 8: Negative Quotient

Find the derivative of

r(x) = -\dfrac{f(x)}{g(x)}

r'(x) = -\left[\frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\right] = \frac{-f'(x)g(x) + f(x)g'(x)}{[g(x)]^2}

or equivalently:

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

The negative just distributes through the numerator.

Example 9: Negative Sign with a Table

Find

p'(2)

where

p(x) = -\dfrac{3f(x)}{2g(x)}

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
2	6	3	4	-2

p'(x) = -\frac{3}{2} \cdot \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

p'(2) = -\frac{3}{2} \cdot \frac{(3)(4) - (6)(-2)}{(4)^2} = -\frac{3}{2} \cdot \frac{24}{16} = -\frac{3}{2} \cdot \frac{3}{2} = -\frac{9}{4}

Practice Problems

Problem 1. Find the derivative:

y = \dfrac{x^2 + 1}{x - 3}

$\frac{dy}{dx} = \frac{(2x)(x - 3) - (x^2 + 1)(1)}{(x - 3)^2} = \frac{x^2 - 6x - 1}{(x - 3)^2}$

Problem 2. Differentiate:

f(x) = \dfrac{\ln x}{x}

$f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$

Problem 3. Find

g'(x)

for

g(x) = \dfrac{4x^3 - 6x}{2x}

(simplify first)

Simplify: $g(x) = 2x^2 - 3$ , so $g'(x) = 4x$

Problem 4. Find

h'(1)

where

h(x) = \dfrac{f(x) + 2}{g(x)}

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
1	3	5	2	1

$h'(1) = \frac{(5)(2) - (3 + 2)(1)}{(2)^2} = \frac{10 - 5}{4} = \frac{5}{4}$

Common Mistakes

Mistake 1: Wrong sign.

Wrong:

\frac{f'g + fg'}{g^2}

. Right:

\frac{f'g - fg'}{g^2}

. Minus, not plus.

Mistake 2: Wrong denominator.

Wrong:

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

. Right:

\frac{f'g - fg'}{g^2}

. Square

g

, not

g'

Mistake 3: Expanding the denominator.

Don’t turn

(x + 1)^2

into

x^2 + 2x + 1

. Leave it alone.

Mistake 4: Forgetting to distribute the negative.

Be careful with signs when subtracting in the numerator:

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

Key Takeaways

Quotient rule: $\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}$

Use subtraction, unlike the product rule

Denominator is $g^2$ , not $(g')^2$

Order matters, you can’t swap top and bottom

Simplify first when the denominator is a single term

Don’t expand the $(g)^2$ in the denominator

Factor out constants and negatives before applying the rule

With tables, substitute carefully

Distribute negatives carefully in the numerator

Multiple simplified forms can be correct, so factor when you can

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