How do the quotient rule and chain rule work together?

When the top or bottom is itself a composite function, differentiate that part with the chain rule. For x over e to the 2x, the bottom v = e to the 2x has v prime = 2 e to the 2x by the chain rule, then substitute into the quotient rule.

What is the derivative of x divided by (x+1)?

It is 1 over (x+1) squared. With u = x and v = x+1, the quotient rule gives (1 times (x+1) minus x times 1) over (x+1) squared, which simplifies to 1 over (x+1) squared.

How do the quotient rule and chain rule work together?

When the top or bottom is composite, differentiate it with the chain rule. For x over e to the 2x, the bottom e to the 2x differentiates to 2 e to the 2x, then substitute into the quotient rule.

What is the derivative of x divided by (x+1)?

It is 1 over (x+1) squared. With u = x and v = x+1, the quotient rule gives (1 times (x+1) minus x times 1) over (x+1) squared = 1 over (x+1) squared.

Year 12 Methods (Unit 3 & 4) Further differentiation and applications

The quotient rule

Q: What is the quotient rule?

The quotient rule differentiates a fraction of two functions. If y = u/v then dy/dx = (u prime v minus u v prime) over v squared: the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

Q: Why does the order in the numerator matter?

Because subtraction is not commutative. The numerator is u prime v minus u v prime, not u v prime minus u prime v. Reversing the order flips the sign of the derivative, the most common mistake.

Q: When should I use the quotient rule?

Use it when a function is one expression divided by another that both contain x, such as x/(x+1), e to the x over x, or sin x over x. If the bottom is constant, or the fraction splits into simpler terms, ordinary differentiation is quicker.

Q: How do you find a stationary point of a quotient?

Set dy/dx = 0. A fraction is zero only when its numerator is zero, so solve u prime v minus u v prime = 0. For e to the x over x the derivative e to the x times (x minus 1) over x squared is zero at x = 1.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the quotient rule differentiates a fraction of two functions: if \(y=\dfrac{u}{v}\) then \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}\). It applies to quotients of powers, exponentials, logarithms and trigonometric functions, and combines with the chain rule when the top or bottom is composite.

The quotient rule differentiates a function written as one expression divided by another. Writing \(y=\dfrac{u}{v}\), where \(u\) and \(v\) are both functions of \(x\), the rule is \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}\) — the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

Two features carry all the marks. First, the order in the numerator: it is \(u'v-uv'\), not \(uv'-u'v\). Because subtraction is not commutative, reversing the order flips the sign of the whole derivative — the single most common error. Second, the denominator is \(v^{2}\), the bottom squared, not just \(v\).

When the top or bottom is itself a composite function (for example \(e^{2x}\), \(\ln(3x-1)\) or \(\cos 2x\)), differentiate that part with the chain rule before substituting. The quotient rule is one of the three differentiation rules of Unit 3, alongside the product and chain rules, and its answers are written in simplest, factorised form.

Key idea. If \(y=\dfrac{u}{v}\) then \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}\). Keep the numerator order \(u'v-uv'\) and square the denominator.

\(y=\dfrac{e^{x}}{x}\): \(\dfrac{dy}{dx}=\dfrac{e^{x}(x-1)}{x^{2}}=0\) at \(x=1\), a minimum

\(y=\dfrac{x}{x^{2}+1}\): a stationary point where \(\dfrac{dy}{dx}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}=0\)

The quotient rule, in Leibniz and function form:

\[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^{2}} \qquad\text{where } u=u(x),\ v=v(x)\]

\frac{d}{d x} (\frac{u}{v}) = \frac{u^{'} v - u v^{'}}{v^{2}}

A memory form — "bottom d(top) minus top d(bottom), over bottom squared":

\[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\,\dfrac{du}{dx}-u\,\dfrac{dv}{dx}}{v^{2}}\]

When the bottom is composite, use the chain rule for \(v'\). For example \(v=e^{2x}\Rightarrow v'=2e^{2x}\), and \(v=(x^{2}-1)^{1/2}\Rightarrow v'=\dfrac{x}{\sqrt{x^{2}-1}}\):

Ready-made results. \(\dfrac{d}{dx}\dfrac{x}{x+1}=\dfrac{1}{(x+1)^{2}}\). \(\;\dfrac{d}{dx}\dfrac{\ln x}{x}=\dfrac{1-\ln x}{x^{2}}\). \(\;\dfrac{d}{dx}\dfrac{\cos x}{\sin x}=-\operatorname{cosec}^{2}x\).

How to use the quotient rule

Identify \(u\) and \(v\). Let \(u\) be the top (numerator) and \(v\) be the bottom (denominator).
Differentiate each part. Find \(u'\) and \(v'\), using the chain rule if the top or bottom is composite (e.g. \(e^{2x}\), \(\ln(3x-1)\), \(\cos 2x\)).
Substitute into the rule. Write \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}\), keeping the numerator order and squaring the denominator.
Simplify and factorise. Expand the numerator, collect like terms, and factor where possible; leave the answer in simplest form.

Application. The derivative is a gradient function: substitute an \(x\)-value for the gradient of a tangent, set \(\dfrac{dy}{dx}=0\) (numerator \(=0\)) for stationary points, and use the point-gradient form \(y-y_1=m(x-x_1)\) for a tangent line.

Example 1 — a polynomial quotient

Differentiate \(y=\dfrac{x}{x+1}\).

Solution

Let \(u=x,\ v=x+1\), so \(u'=1,\ v'=1\).

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{u'v-uv'}{v^{2}}=\dfrac{1\cdot(x+1)-x\cdot 1}{(x+1)^{2}}\)
	\(=\)	\(\dfrac{1}{(x+1)^{2}}\)

\frac{1}{{(x + 1)}^{2}}

Example 2 — with the chain rule

Differentiate \(y=\dfrac{x}{e^{2x}}\).

Solution

Let \(u=x,\ v=e^{2x}\); the chain rule gives \(v'=2e^{2x}\).

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1\cdot e^{2x}-x\cdot 2e^{2x}}{(e^{2x})^{2}}\)
	\(=\)	\(\dfrac{e^{2x}(1-2x)}{e^{4x}}=\dfrac{1-2x}{e^{2x}}\)

\frac{1 - 2 x}{e^{2 x}}

Example 3 — a trigonometric quotient

Differentiate \(y=\dfrac{\cos x}{\sin x}\).

Solution

Let \(u=\cos x,\ v=\sin x\), so \(u'=-\sin x,\ v'=\cos x\), and use \(\sin^{2}x+\cos^{2}x=1\).

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{-\sin x\sin x-\cos x\cos x}{\sin^{2}x}\)
	\(=\)	\(\dfrac{-1}{\sin^{2}x}=-\operatorname{cosec}^{2}x\)

- {cosec}^{2} x

Example 4 — gradient of a tangent

Find the gradient of \(y=\dfrac{x}{x+1}\) at \(x=1\).

Solution

From Example 1, \(\dfrac{dy}{dx}=\dfrac{1}{(x+1)^{2}}\); substitute \(x=1\).

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1}{(x+1)^{2}}\)
at \(x=1\)	\(=\)	\(\dfrac{1}{(1+1)^{2}}=\dfrac{1}{4}\)

\frac{1}{4}

Common pitfalls

Keep the numerator order. It is \(u'v-uv'\), not \(uv'-u'v\). Reversing the two terms flips the sign of the whole derivative — the most common quotient-rule mistake.

Square the denominator. The bottom of the answer is \(v^{2}\), not \(v\). For \(y=\dfrac{x}{x+1}\) the derivative is \(\dfrac{1}{(x+1)^{2}}\).

Use the chain rule on \(v'\). If the bottom is composite, differentiate it fully: \(v=e^{2x}\Rightarrow v'=2e^{2x}\), not \(e^{2x}\). Dropping the inner factor is a hidden slip.

Frequently asked questions

What is the quotient rule?

If \(y=\dfrac{u}{v}\) then \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^{2}}\): the derivative of the top times the bottom, minus the top times the derivative of the bottom, over the bottom squared.

Why does the order in the numerator matter?

Subtraction is not commutative, so \(u'v-uv'\) and \(uv'-u'v\) differ by a sign. Reversing the order gives the negative of the correct derivative.

When should I use the quotient rule?

When a function is one expression divided by another that both contain \(x\), such as \(\dfrac{x}{x+1}\), \(\dfrac{e^{x}}{x}\) or \(\dfrac{\sin x}{x}\). If the bottom is constant, or the fraction splits into simpler terms, differentiate directly instead.

How do the quotient and chain rules work together?

Differentiate a composite top or bottom with the chain rule first. For \(\dfrac{x}{e^{2x}}\), the bottom \(e^{2x}\) has \(v'=2e^{2x}\), then substitute into the quotient rule.

How do you find a stationary point of a quotient?

Set \(\dfrac{dy}{dx}=0\). A fraction is zero only when its numerator is zero, so solve \(u'v-uv'=0\). For \(\dfrac{e^{x}}{x}\) the derivative \(\dfrac{e^{x}(x-1)}{x^{2}}\) is zero at \(x=1\).

What is the derivative of \(\dfrac{x}{x+1}\)?

\(\dfrac{1}{(x+1)^{2}}\). With \(u=x,\ v=x+1\), the quotient rule gives \(\dfrac{1\cdot(x+1)-x\cdot 1}{(x+1)^{2}}=\dfrac{1}{(x+1)^{2}}\).

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