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Method · Differentiation

How to differentiate using the quotient rule

The quotient rule differentiates one function divided by another. If y=uvy=\frac{u}{v}, then dydx=vdudxudvdxv2\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}: the order of the two top terms matters, and the denominator is squared.

What the quotient rule is for

In Add Math, many expressions are one function divided by another, such as y=2x+1x3y=\frac{2x+1}{x-3}, y=x22x+1y=\frac{x^{2}}{2x+1} or y=3xx2+1y=\frac{3x}{x^{2}+1}. The quotient rule is the tool for differentiating a fraction where both the numerator and the denominator contain xx.

You cannot simply differentiate the top and bottom separately, and dividing the two derivatives is wrong. The quotient rule gives a single reliable formula that keeps track of both parts and, crucially, the order in which they are combined.

It appears throughout the Differentiation chapter and is exactly what you need for gradients, tangents, normals and stationary points of curves written as fractions. A neat, well-set-out application also earns full method marks, which the analytic marking scheme rewards.

When to reach for it

Reach for the quotient rule whenever yy is written as a fraction and both the numerator and denominator are functions of xx, such as x+22x1\frac{x+2}{2x-1} or x21x+4\frac{x^{2}-1}{x+4}. The dividing line is the signal.

One check first: if the denominator is just a constant, you do not need the rule, 2x+15\frac{2x+1}{5} is simply 15(2x+1)\tfrac{1}{5}(2x+1). And if the fraction can be split into simple powers, for example x2+xx=x+1\frac{x^{2}+x}{x}=x+1, simplify first and differentiate directly.

Use the quotient rule when the fraction genuinely cannot be simplified away. Tell it apart from the other rules: quotient for division, product for multiplication, and chain for one function wrapped inside another.

The method, step by step

Quotient ruleMust memorise
dydx=vdudxudvdxv2\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}
  1. 1

    Name top and bottom

    Let uu be the numerator and vv be the denominator, so y=uvy=\frac{u}{v}.

  2. 2

    Differentiate each part

    Find dudx\frac{du}{dx} and dvdx\frac{dv}{dx} separately.

  3. 3

    Build the numerator

    Write vdudxudvdxv\frac{du}{dx}-u\frac{dv}{dx}. The order matters: it is bottom times derivative of top, minus top times derivative of bottom.

  4. 4

    Square the denominator

    The whole thing sits over v2v^{2}.

  5. 5

    Simplify the top

    Expand and collect like terms in the numerator; leave the denominator as v2v^{2}.

  6. 6

    Use it if a value is asked

    Substitute the given xx-value for a gradient, or set the numerator equal to 00 for a stationary point.

Worked example

Q1[4 marks]

Given y=2x+1x3y=\frac{2x+1}{x-3}, use the quotient rule to find dydx\frac{dy}{dx}, and hence the gradient of the curve at the point where x=4x=4.

Show worked solution

Name the parts. Let u=2x+1u=2x+1 (top) and v=x3v=x-3 (bottom), so y=uvy=\frac{u}{v}.

Differentiate each part: dudx=2\frac{du}{dx}=2 and dvdx=1\frac{dv}{dx}=1.

Apply the quotient rule:

dydx=vdudxudvdxv2=(x3)(2)(2x+1)(1)(x3)2\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}=\frac{(x-3)(2)-(2x+1)(1)}{(x-3)^{2}}

Expand the numerator: (x3)(2)=2x6(x-3)(2)=2x-6 and (2x+1)(1)=2x+1(2x+1)(1)=2x+1.

Subtract carefully, keeping the bracket: dydx=2x6(2x+1)(x3)2=2x62x1(x3)2=7(x3)2\frac{dy}{dx}=\frac{2x-6-(2x+1)}{(x-3)^{2}}=\frac{2x-6-2x-1}{(x-3)^{2}}=\frac{-7}{(x-3)^{2}}.

For the gradient at x=4x=4, substitute: (43)2=12=1(4-3)^{2}=1^{2}=1, so dydx=71=7\frac{dy}{dx}=\frac{-7}{1}=-7.

The gradient of the curve at x=4x=4 is 7-7. The gradient is negative everywhere, which fits a curve that is always decreasing away from its asymptote.

Common mistakes to avoid

  • Getting the order wrong in the numerator: it is vdudxudvdxv\frac{du}{dx}-u\frac{dv}{dx}, not udvdxvdudxu\frac{dv}{dx}-v\frac{du}{dx}. The wrong order flips the sign of the whole answer.
  • Forgetting to square the denominator, or writing vv instead of v2v^{2}.
  • Dropping the bracket when subtracting, so (2x+1)-(2x+1) becomes 2x+1-2x+1 instead of 2x1-2x-1.
  • Trying to differentiate the top and bottom separately and divide, the quotient rule does not work that way.
  • Not simplifying, or cancelling terms inside the squared denominator that cannot be cancelled.

How one-to-one teaching helps

The step students most often get wrong is the order and the subtraction sign in the numerator, one small slip turns 7-7 into +7+7 and loses the marks. In a one-to-one lesson our teachers get you to write uu, vv and their derivatives first, then say the rule aloud as 'bottom d-top minus top d-bottom, all over bottom squared', so the order is locked in.

Because your working is shown line by line, the bracket in the subtraction stays in place and every method mark is easy to award. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To see how we teach the quotient rule, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

How do I remember the order in the quotient rule?

Say it as 'bottom times derivative of the top, minus top times derivative of the bottom, all over bottom squared', i.e. vdudxudvdxv2\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}. The order matters: swapping the two top terms flips the sign of the whole answer, so keep the vdudxv\frac{du}{dx} term first.

Can I use the product rule instead of the quotient rule?

Yes. A fraction uv\frac{u}{v} can be rewritten as uv1u\,v^{-1} and differentiated with the product rule and chain rule.

Both give the same answer, so use whichever you find clearer. Many students find the quotient rule tidier when the denominator is a simple linear expression.

Do I always need the quotient rule for a fraction?

No. If the denominator is a constant, or the fraction simplifies to simple powers, for example x2+xx=x+1\frac{x^{2}+x}{x}=x+1, simplify first and differentiate directly.

Use the quotient rule only when both top and bottom contain xx and the fraction cannot be reduced.

How do I find a stationary point of a fraction?

After applying the quotient rule, a fraction equals zero only when its numerator is zero (and the denominator is not). So set the top of dydx\frac{dy}{dx} equal to 00 and solve, then check the point lies on the curve.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

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