Quotient Rule

The rule

For a quotient of two functions, "low d-high minus high d-low, over low squared":

Worked example

Differentiate $y = \dfrac{2x+1}{3x-4}$.

Solution: With $u=2x+1$ ($u'=2$) and $v=3x-4$ ($v'=3$):

A function's derivative at a point

The functions $f(x)$ and $g(x)$ are differentiable, and $h(x) = \dfrac{g(x)}{f(x)}$. Given $f(0)=-7$, $f'(0)=5$, $g(0)=3$, and $g'(0)=-6$, find $h'(0)$.

Solution: By the quotient rule, $h'(x) = \dfrac{g'(x)f(x) - g(x)f'(x)}{f(x)^2}$. Substituting the values at $x=0$:

Worked example: the derivative of tan x

Use the quotient rule to show that $\dfrac{d}{dx}\tan x = \sec^2 x$.

Solution: Write $\tan x = \dfrac{\sin x}{\cos x}$, so $u = \sin x$ ($u' = \cos x$) and $v = \cos x$ ($v' = -\sin x$):

Worked example: where the slope is zero

Find where $y = \dfrac{x^2+1}{x-1}$ has a horizontal tangent.

The derivative is zero where the numerator is: $x^2 - 2x - 1 = 0$, so $x = 1 \pm \sqrt2$. Both are allowed, since neither makes the denominator zero.

The same rule gives $\dfrac{d}{dx}\sec x = \dfrac{d}{dx}\dfrac{1}{\cos x} = \dfrac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \sec x\tan x$.

When not to use the quotient rule

If the numerator is a constant, rewriting is quicker: $\dfrac{1}{x^2} = x^{-2}$, so the power rule gives $-2x^{-3}$ at once. The quotient rule agrees, $\dfrac{0 \cdot x^2 - 1 \cdot 2x}{x^4} = -\dfrac{2}{x^3}$, but takes longer and gives more chances to slip.

Common mistakes

• Swapping the order in the numerator. It is $u'v - uv'$, derivative of the top first. Writing $uv' - u'v$ flips the sign of the whole answer. In the first example above, that would give $+\dfrac{11}{(3x-4)^2}$ instead of $-\dfrac{11}{(3x-4)^2}$.
• Forgetting to square the denominator. The bottom of the answer is $v^2$, not $v$.
• Losing a minus sign while expanding. Keep brackets around $uv'$ until the end: $2(3x-4) - (2x+1)(3)$ expands to $6x - 8 - 6x - 3$, with the minus applied to both terms.
• Using it when the denominator is a constant. $\dfrac{x^2}{5}$ is just $\dfrac{1}{5}x^2$, with derivative $\dfrac{2x}{5}$. No quotient rule needed.

Where this comes up in exams

JEE Main 2026 lists differentiation of the sum, difference, product and quotient of two functions, JEE Advanced 2026 lists the derivative of the sum, difference, product and quotient of two functions, and the 2025-26 CBSE Class XI syllabus lists the derivative of sum, difference, product and quotient of functions.

Checked against JEE (Main) 2026 syllabus (NTA), JEE (Advanced) 2026 syllabus and CBSE Mathematics curriculum, Classes XI-XII (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

• Fit it into your week with the free timetable generator.
• Mark it off as you revise in the syllabus tracker.
• See how long you have left: JEE Main 2027 (Session 1) countdown and JEE Advanced 2027 countdown.