How do I differentiate x squared times two x plus three to the fourth?

Let u be x squared and v be two x plus three to the fourth, so u prime is two x and, by the chain rule, v prime is eight times two x plus three cubed. Then apply u times v prime plus v times u prime and factorise.

How do I differentiate x squared times two x plus three to the fourth?

Let u be x squared and v be two x plus three to the fourth, so u prime is two x and, by the chain rule, v prime is eight times two x plus three cubed. Then apply u times v prime plus v times u prime and factorise.

Year 11 Methods (Unit 1 & 2) Further Differentiation

The Product Rule

Q: What is the product rule?

For y equals u times v, the derivative is u times v prime plus v times u prime. You differentiate one factor at a time and add the two results.

Q: Can I just multiply the two derivatives?

No. The derivative of u times v is not u prime times v prime. You must use u times v prime plus v times u prime.

20 practice questions 1 video lesson Theory + worked examples

Master the product rule for Queensland Year 11 Mathematical Methods (QCAA). When two functions are multiplied, its derivative is the first times the derivative of the second, plus the second times the derivative of the first.

You will learn to identify the two factors, apply the rule to products of power and polynomial functions, combine it with the chain rule, and give the derivative in simplest and factorised form.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), the product rule differentiates a product of two functions: \((uv)'=u\,v'+v\,u'\). This page applies it to products of power and polynomial factors, combines it with the chain rule for \(x^{m}(ax+b)^{n}\), and uses it to find gradients and where a curve has a horizontal tangent.

The product rule gives the derivative of a product \(y=u\,v\), where \(u\) and \(v\) are both functions of \(x\). It is \(\dfrac{dy}{dx}=u\,v'+v\,u'\): each factor is differentiated in turn while the other is held, and the results are added.

To use it, first identify \(u\) and \(v\), then write down \(u'\) and \(v'\). If a factor is itself a composite such as \((ax+b)^{n}\), its derivative uses the chain rule.

Answers are usually factorised: take out the highest common factor so the derivative can be set to zero to find horizontal tangents.

\((uv)^{\prime}=u\,v^{\prime}+v\,u^{\prime}\). Differentiate one factor at a time; you cannot just multiply the two separate derivatives together.

\(y=x^{2}(x-3)\): a product of two polynomial factors.

\(y=x(x-4)^{2}\): horizontal tangents where \((uv)'=0\), at \(x=\tfrac43\) and \(x=4\).

The product rule, with \(u\) and \(v\) functions of \(x\):

\[\dfrac{d}{dx}(uv)=u\,\dfrac{dv}{dx}+v\,\dfrac{du}{dx}=u\,v^{\prime}+v\,u^{\prime}\]

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

For a horizontal tangent, set the (factorised) derivative to zero:

\[u\,v^{\prime}+v\,u^{\prime}=0\]

u v^{'} + v u^{'} = 0

Factorise before solving: take out the common factor of \(u\,v'+v\,u'\), then each factor set to zero gives a horizontal tangent.

How to use the product rule

Split: write the function as \(u\times v\) and state \(u\) and \(v\).
Differentiate each factor: find \(u'\) and \(v'\) (use the chain rule if a factor is a composite).
Combine: substitute into \(u\,v'+v\,u'\).
Factorise / evaluate: take out the common factor and simplify, then evaluate at a point or solve \(=0\) as required.

Example 1 — Two polynomial factors

Differentiate \(y=(3x-1)(x^{2}+2)\).

Solution

Identify \(u,\,v\) and their derivatives:

\(u\)	\(=\)	\(3x-1,\quad u'=3\)
\(v\)	\(=\)	\(x^{2}+2,\quad v'=2x\)

Apply \((uv)'=u\,v'+v\,u'\):

\(\dfrac{dy}{dx}\)

\(=\)

\((3x-1)(2x)+(x^{2}+2)(3)\)

Expand and collect like terms:

	\(=\)	\(6x^{2}-2x+3x^{2}+6\)
	\(=\)	\(9x^{2}-2x+6\)

\(\dfrac{dy}{dx}=9x^{2}-2x+6\).

9 x^{2} - 2 x + 6

Example 2 — A power factor, then evaluate

For \(y=x^{3}(x^{2}-4)\), find \(\dfrac{dy}{dx}\) and its value at \(x=2\).

Solution

Identify \(u,\,v\) and their derivatives:

\(u\)	\(=\)	\(x^{3},\quad u'=3x^{2}\)
\(v\)	\(=\)	\(x^{2}-4,\quad v'=2x\)

Apply the product rule and simplify:

\(\dfrac{dy}{dx}\)	\(=\)	\(x^{3}(2x)+(x^{2}-4)(3x^{2})\)
	\(=\)	\(2x^{4}+3x^{4}-12x^{2}\)
	\(=\)	\(5x^{4}-12x^{2}\)

Evaluate at \(x=2\):

\(\left.\dfrac{dy}{dx}\right\|_{x=2}\)	\(=\)	\(5(2)^{4}-12(2)^{2}\)
	\(=\)	\(80-48=32\)

\(\dfrac{dy}{dx}=5x^{4}-12x^{2}\), and at \(x=2\) the gradient is \(32\).

5 x^{4} - 12 x^{2}

Example 3 — Product with the chain rule, factorised

Differentiate \(y=x^{2}(2x+3)^{4}\) and factorise the result.

Solution

Identify \(u,\,v\); use the chain rule for \(v'\):

\(u\)	\(=\)	\(x^{2},\quad u'=2x\)
\(v\)	\(=\)	\((2x+3)^{4},\quad v'=8(2x+3)^{3}\)

Apply the product rule:

\(\dfrac{dy}{dx}\)

\(=\)

\(x^{2}\cdot 8(2x+3)^{3}+(2x+3)^{4}\cdot 2x\)

Take out the common factor \(2x(2x+3)^{3}\):

	\(=\)	\(2x(2x+3)^{3}\big[4x+(2x+3)\big]\)
	\(=\)	\(2x(2x+3)^{3}(6x+3)\)
	\(=\)	\(6x(2x+3)^{3}(2x+1)\)

\(\dfrac{dy}{dx}=6x(2x+3)^{3}(2x+1)\).

6 x {(2 x + 3)}^{3} (2 x + 1)

Example 4 — Where the tangent is horizontal

Find where \(y=x(x-4)^{2}\) has a horizontal tangent.

Solution

Identify \(u,\,v\); chain rule for \(v'\):

\(u\)	\(=\)	\(x,\quad u'=1\)
\(v\)	\(=\)	\((x-4)^{2},\quad v'=2(x-4)\)

Apply the product rule and factorise:

\(\dfrac{dy}{dx}\)	\(=\)	\(x\cdot 2(x-4)+(x-4)^{2}\cdot 1\)
	\(=\)	\((x-4)\big[2x+(x-4)\big]\)
	\(=\)	\((x-4)(3x-4)\)

Set the derivative to \(0\):

\((x-4)(3x-4)\)	\(=\)	\(0\)
\(x\)	\(=\)	\(4\ \text{ or }\ \tfrac43\)

Horizontal tangents at \(x=\dfrac{4}{3}\) and \(x=4\).

(x - 4) (3 x - 4)

Common pitfalls

Multiplying the two derivatives. \((uv)^{\prime}\) is not \(u'\,v'\). You must use \(u\,v'+v\,u'\).

Forgetting the chain rule inside a factor. The derivative of \((2x+3)^{4}\) is \(8(2x+3)^{3}\), not \(4(2x+3)^{3}\); the inner \(\times 2\) is needed.

Not factorising. Leaving the derivative expanded makes solving \(=0\) far harder; take out the common factor first.

Mixing up which factor is differentiated. Keep \(u\), \(v\), \(u'\), \(v'\) in a short table so each term of \(u\,v'+v\,u'\) is correct.

Frequently asked questions

What is the product rule?

For \(y=uv\), the derivative is \(\dfrac{dy}{dx}=u\,v'+v\,u'\): differentiate one factor at a time and add.

When should I use the product rule?

When a function is written as a product of two factors that are each easier to differentiate on their own, such as \(x^{3}(x^{2}-4)\).

Can I just multiply the two derivatives?

No. \((uv)'\ne u'v'\). You must use \(u\,v'+v\,u'\); multiplying the derivatives gives the wrong answer.

How do I differentiate x squared times (2x+3) to the fourth?

Let \(u=x^{2}\) and \(v=(2x+3)^{4}\), so \(u'=2x\) and (by the chain rule) \(v'=8(2x+3)^{3}\), then apply \(u\,v'+v\,u'\) and factorise.

How do I find where a product function has a horizontal tangent?

Differentiate with the product rule, factorise, set the derivative to zero, and solve; each factor set to zero gives a horizontal tangent.

The Product Rule

📖 Theory

How to use the product rule

Common pitfalls

Frequently asked questions

Theory