the Quotient Rule?

The formula for differentiating one function divided by another — $\dfrac{f'g - fg'}{g^2}$. Order matters: subtraction, not addition.

Simple Definition

When one function is divided by another, differentiating requires the quotient rule. Unlike addition and subtraction (where you just differentiate term by term), division introduces an interaction between numerator and denominator that you must account for explicitly.

The quotient rule states: if $h(x) = \frac{f(x)}{g(x)}$, then $h'(x) = \frac{f'g - fg'}{g^2}$. Note the minus sign — this distinguishes it from the product rule's plus sign, and getting the order wrong is the most common error.

The Formula

Quotient Rule

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

"Low d-High minus High d-Low, over Low squared"

Low = denominator g · High = numerator f · d = derivative
(g)(f′) − (f)(g′) all over g²

Intuition

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Derived from the product rule

The quotient rule isn't independent — it's just the product rule applied cleverly. Write $\frac{f}{g} = f \cdot g^{-1}$, then apply the product rule: $f' \cdot g^{-1} + f \cdot (-g^{-2} \cdot g') = \frac{f'}{g} - \frac{fg'}{g^2} = \frac{f'g - fg'}{g^2}$. So you don't even need to memorise it separately — it flows from the product rule and the chain rule.

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Why the minus sign?

If the denominator $g$ grows while the numerator stays constant, the fraction shrinks — so the derivative should be negative. The $-fg'$ term captures exactly this: when $g$ is growing ($g' > 0$) and $f$ is positive, the fraction is decreasing. The minus sign is not arbitrary — it's the geometry of fractions.

Step by Step — $\tan x = \frac{\sin x}{\cos x}$

Identify f and g

$f = \sin x$, \quad $g = \cos x$

Find f′ and g′

$f' = \cos x$, \quad $g' = -\sin x$

Apply: (f′g − fg′) / g²

$\dfrac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x}$

Simplify using Pythagorean identity

$= \dfrac{1}{\cos^2 x} = \sec^2 x$ — the derivative of $\tan x$ ✓

Quick Examples

Example 1

$\dfrac{x^2}{x+1}$

$f=x^2,\ g=x+1$
$= \dfrac{2x(x+1) - x^2(1)}{(x+1)^2}$
$= \dfrac{x^2+2x}{(x+1)^2}$

Example 2

$\dfrac{e^x}{x^2}$

$f=e^x,\ g=x^2$
$= \dfrac{e^x \cdot x^2 - e^x \cdot 2x}{x^4}$
$= \dfrac{e^x(x-2)}{x^3}$

Example 3

$\dfrac{\ln x}{x}$

$= \dfrac{\frac{1}{x}\cdot x - \ln x \cdot 1}{x^2} = \dfrac{1 - \ln x}{x^2}$

Example 4 — Trig

$\cot x = \dfrac{\cos x}{\sin x}$

$= \dfrac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x}$
$= \dfrac{-1}{\sin^2 x} = -\csc^2 x$

Common Mistakes

✗ Reversing the order: fg′ − f′g

The quotient rule is $f'g - fg'$, NOT $fg' - f'g$. The numerator's derivative comes first. Reversing gives the wrong sign throughout. The mnemonic "Low d-High minus High d-Low" keeps the order correct.

✗ Squaring only part of the denominator

The denominator of the quotient rule is $[g(x)]^2$ — the entire original denominator, squared. If $g = (x+1)$, then $g^2 = (x+1)^2$, not $x^2 + 1$ or $x + 1$.

✗ Using quotient rule when simpler methods work

$\frac{x^3 + 2x}{x}$ can be simplified to $x^2 + 2$ first, then differentiated with the power rule. Don't reach for the quotient rule when algebra can simplify the expression first — it's slower and more error-prone.

Related Concepts

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The quotient rule is tested in every Calculus I section on BUders.

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