The counterpart to the product rule for fractions of functions — and the tool that finally proves the reciprocal trig derivatives from two lessons ago.
7 min read3 worked examples
The quotient rule
When $y=\dfrac{u}{v}$ is a quotient of two functions of $x$, the quotient rule states
The order in the numerator matters — it isn't symmetric like the product rule, so it's worth fixing the pattern "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared" firmly in memory.
Choosing the right rule
A single expression can need more than one rule at once. Differentiate the "outer shape" of the expression first: is it fundamentally a sum, a product, a quotient, or a composite function? That outer shape decides which rule to apply first — and any inner piece that is itself composite will need the chain rule as you go.
Worked example 1
Differentiating a quotient
Differentiate $y=\dfrac{x^2}{2x+3}$.
MethodApply the quotient rule
1
Let $u=x^2$ and $v=2x+3$, so $\dfrac{du}{dx}=2x$ and $\dfrac{dv}{dx}=2$.
Identify the top and bottom, and differentiate each.
At $x=1$: $\dfrac{dy}{dx}=\dfrac{3e^1(1-1)}{1^2}=0$.
Substitute $x=1$ — the numerator vanishes since $x-1=0$.
▪ solved
Look deeper
Because $e^x>0$ for all $x$, the sign of $\dfrac{dy}{dx}$ is controlled entirely by the factor $(x-1)$ — negative for $x<1$, positive for $x>1$. That confirms $x=1$ is a minimum of the curve, without needing a second derivative.
Worked example 3
Deriving the derivative of sec x · Challenge
Using $\sec x=\dfrac{1}{\cos x}$, show that $\dfrac{d}{dx}(\sec x)=\sec x\tan x$ — the standard result used without proof in an earlier lesson.
ProofApply the quotient rule to $\dfrac{1}{\cos x}$
1
Write $\sec x=\dfrac{u}{v}$ with $u=1$ and $v=\cos x$, so $\dfrac{du}{dx}=0$ and $\dfrac{dv}{dx}=-\sin x$.
Rewrite the fraction as a product of two familiar ratios.
4
$\dfrac{d}{dx}(\sec x)=\sec x\tan x$, as required.
Since $\dfrac{1}{\cos x}=\sec x$ and $\dfrac{\sin x}{\cos x}=\tan x$.
▪ proven
The same method proves all three reciprocal derivatives
Identical reasoning, starting from $\csc x=\dfrac{1}{\sin x}$ and $\cot x=\dfrac{1}{\tan x}$, gives $\dfrac{d}{dx}(\csc x)=-\csc x\cot x$ and $\dfrac{d}{dx}(\cot x)=-\csc^2x$ — the two other standard results from two lessons ago, now proven properly rather than just stated.
Common mistake
Applying the product rule to a quotient (or vice versa), and swapping the order of terms in the quotient rule's numerator. Since $\dfrac{d}{dx}\left(\dfrac{u}{v}\right)\ne\dfrac{d}{dx}\left(\dfrac{v}{u}\right)$, writing $u\frac{dv}{dx}-v\frac{du}{dx}$ instead of $v\frac{du}{dx}-u\frac{dv}{dx}$ produces the negative of the correct answer.
Check yourself
Differentiate $y=\dfrac{(x+1)^3}{x^2}$. (Hint: use the quotient rule, remembering the chain rule for the numerator's derivative.)