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Calculus BC · Unit 2 · 2.9
AP Calculus · Differentiation: Definition and Fundamental Properties
The Quotient Rule
Bottom times derivative of top, minus top times derivative of bottom, over bottom squared — in that order.
key idea$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}$
The rule
Unlike the product rule, order matters here — the numerator is a difference, so swapping the two terms flips the sign of the whole answer. Get the low-d-high mnemonic straight before anything else.
The Quotient Rule
\[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)g(x)-f(x)g'(x)}{\bigl[g(x)\bigr]^{2}}\]
In words: bottom times derivative of top, minus top times derivative of bottom, all over bottom squared. The denominator is squared, never differentiated as a whole.
Why the minus is not symmetric
Test on $\dfrac{x}{x^{2}}=\dfrac1x$, whose derivative is $-\dfrac{1}{x^{2}}$. The rule gives $\dfrac{1\cdot x^{2}-x\cdot 2x}{x^{4}}=\dfrac{-x^{2}}{x^{4}}=-\dfrac{1}{x^{2}}$. Reversing the numerator terms would give $+\dfrac{1}{x^{2}}$ — wrong sign, wrong graph.
| If you see… | Do this instead | Why |
|---|
| $\dfrac{x^{2}+3x}{x}$ | divide first → $x+3$ | one power rule beats a quotient rule |
| $\dfrac{5}{x^{3}}$ | rewrite as $5x^{-3}$ | constant over a power is just a power |
| $\dfrac{\sin x}{\cos x}$ | recognise $\tan x$ | the derivative $\sec^{2}x$ is already known |
| $\dfrac{f(x)}{7}$ | constant multiple $\tfrac17 f'(x)$ | a constant denominator is not a quotient |
| $\dfrac{x^{2}}{\sin x}$ | quotient rule | neither part simplifies away |
Four of these five rows avoid the quotient rule entirely. Check for a simplification before committing to it.
Pattern recognition
before you start
- Numerator is a difference — order is not interchangeable.
- The bottom function leads the first term.
- Denominator is squared, never differentiated as a unit.
- Check for a simplification before applying the rule at all.
Worked examples
EX 1
Straight application
$\dfrac{x^{2}}{\sin x}$Differentiate $y=\dfrac{x^{2}}{\sin x}$.
1
Name top and bottom
$f=x^{2}$ (top), $g=\sin x$ (bottom).
2
Differentiate each
$f'=2x$ and $g'=\cos x$.
3
Assemble in order
$y'=\dfrac{f'g-fg'}{g^{2}}=\dfrac{2x\sin x-x^{2}\cos x}{\sin^{2}x}$.
4
Check the shape
Bottom-times-derivative-of-top comes first. If your first term contains $\cos x$, the order is reversed.
EX 2
Simplify instead
$\dfrac{x^{3}+2x}{x}$Differentiate $y=\dfrac{x^{3}+2x}{x}$ two ways.
1
Divide through first
$y=x^{2}+2$ for $x\neq0$, so $y'=2x$.
2
Now try the quotient rule
$y'=\dfrac{(3x^{2}+2)x-(x^{3}+2x)(1)}{x^{2}}$.
3
Expand the numerator
$3x^{3}+2x-x^{3}-2x=2x^{3}$.
4
Simplify
$\dfrac{2x^{3}}{x^{2}}=2x$ — the same answer, after considerably more work.
5
The lesson
Both are correct. The first took one line. Always check whether the fraction simplifies.
EX 3
Derive a known result
$\tan x$Use the quotient rule to show $\dfrac{d}{dx}[\tan x]=\sec^{2}x$.
1
Write as a quotient
$\tan x=\dfrac{\sin x}{\cos x}$, so $f=\sin x$, $g=\cos x$.
2
Differentiate each
$f'=\cos x$ and $g'=-\sin x$.
3
Apply the rule
$\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^{2}x}=\dfrac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}$.
4
Use the identity
$\cos^{2}x+\sin^{2}x=1$, so this is $\dfrac{1}{\cos^{2}x}=\sec^{2}x$.
5
Note the double negative
The $-$ of the rule met the $-\sin x$ from $g'$, giving $+\sin^{2}x$. That is what produces the identity.
EX 4
Evaluate at a point from a table
structuralGiven $f(3)=2$, $f'(3)=-4$, $g(3)=1$, $g'(3)=5$, find $\left(\dfrac{f}{g}\right)'(3)$.
1
Write the rule with values in place
$\dfrac{f'(3)g(3)-f(3)g'(3)}{\left[g(3)\right]^{2}}$.
2
Substitute
$\dfrac{(-4)(1)-(2)(5)}{1^{2}}$.
3
Evaluate the numerator
$-4-10=-14$.
4
Divide
$\dfrac{-14}{1}=-14$.
5
Guard against sign errors
With a negative $f'$ and a positive $g'$, both numerator terms are negative. Getting $-4+10=6$ means the minus was applied to only one factor.
Watch out
Order matters — unlike the product rule
$f'g-fg'$ and $fg'-f'g$ differ by a sign. Fix the mnemonic: bottom × d(top) − top × d(bottom).
The denominator is squared, not differentiated
It is $\left[g(x)\right]^{2}$ in the denominator. Writing $g'(x)^{2}$ or $\left[g'(x)\right]^{2}$ is a different expression entirely.
Simplify before you commit
Example 2 takes one line by dividing and five by rule-following. Most exam quotients simplify.
Do not cancel across the minus
In $\dfrac{2x\sin x-x^{2}\cos x}{\sin^{2}x}$ you may factor an $x$ from the numerator, but you cannot cancel $\sin x$ against $\sin^{2}x$ — it is not a factor of the whole numerator.