How do you differentiate x squared times e to the x?

Let u equal x squared and v equal e to the x, so u prime is 2x and v prime is e to the x. The product rule gives 2x times e to the x plus x squared times e to the x, which factorises to x times e to the x times the bracket x plus 2.

Why is the derivative of a product not the product of the derivatives?

Because each factor keeps changing while the other is held fixed, so the change in the product uv has two contributions, u times the change in v plus v times the change in u. That is why the derivative is u prime v plus u v prime, not u prime times v prime.

How do you differentiate x squared times e to the x?

Let u equal x squared and v equal e to the x, so u prime is 2x and v prime is e to the x. The product rule gives 2x times e to the x plus x squared times e to the x, which factorises to x times e to the x times the bracket x plus 2.

Why is the derivative of a product not the product of the derivatives?

Because each factor keeps changing while the other is held fixed, so the change in the product uv has two contributions, u times the change in v plus v times the change in u. That is why the derivative is u prime v plus u v prime, not u prime times v prime.

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

The product rule

Q: What is the product rule?

The product rule differentiates a product of two functions. If y equals u times v, where u and v are both functions of x, then y prime equals u prime times v plus u times v prime. You differentiate each factor once while keeping the other unchanged, then add the two terms.

Q: How do you use the product rule with the chain rule?

When one factor is a composite function, differentiate that factor with the chain rule, keeping its inner factor. For y equals x squared times e to the 3x, the derivative of e to the 3x is 3 times e to the 3x, so y prime equals 2x times e to the 3x plus 3 x squared times e to the 3x, which factorises to x times e to the 3x times the bracket 3x plus 2.

Q: When do you need the product rule?

You need it whenever a function is written as two functions of x multiplied together, such as x times ln x, e to the x times sin x, or x squared times e to the minus x. If either factor is just a constant, you do not need the product rule.

Q: What does simplest and factorised form mean?

After applying the product rule you usually get two terms that share a common factor. The syllabus asks you to take out the highest common factor so the derivative is written as a single factorised expression, for example x e to the x times the bracket x plus 2 rather than 2x e to the x plus x squared e to the x.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the product rule differentiates a product of two functions: if \(y=u\,v\) then \(y'=u'v+uv'\). It is used for products such as \(x^{2}e^{x}\), \(x\ln x\) and \(e^{x}\sin x\), is combined with the chain rule when a factor is composite, and the answer is written in simplest and factorised form.

The product rule is the rule for differentiating a product of two functions of \(x\). If \(y=u\,v\), where \(u=u(x)\) and \(v=v(x)\), then

\(y'=u'v+uv'.\)

In words: differentiate the first factor and multiply by the second, add the first factor times the derivative of the second. Each factor is differentiated once while the other is held unchanged, and the two terms are added.

The rule is needed whenever two functions of \(x\) are multiplied — for example \(x^{2}e^{x}\), \(x\ln x\) or \(e^{x}\cos x\). It does not say \((uv)'=u'v'\): the derivative of a product is never simply the product of the derivatives. When one of the factors is itself a composite function (such as \(e^{3x}\) or \((2x+1)^{3}\)), the product rule is combined with the chain rule to differentiate that factor.

Key idea. \(y=u\,v\Rightarrow y'=u'v+uv'\). Differentiate each factor once, keep the other, and add. Never write \(u'v'\). Give the answer factorised.

Change in area \(uv\): a strip \(u\,dv\) plus a strip \(v\,du\), so \((uv)'=u'v+uv'\)

Product rule in action: \(f(x)=x e^{x}\), \(f'(x)=e^{x}(x+1)\), so \(f'(0)=1\) and the tangent at the origin is \(y=x\)

The product rule for \(y=u\,v\) (both \(u,v\) functions of \(x\)):

\[\dfrac{d}{dx}(uv)=u'v+uv'\]

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

The standard derivatives used with it:

\[\dfrac{d}{dx}e^{x}=e^{x},\quad \dfrac{d}{dx}\ln x=\dfrac{1}{x},\quad \dfrac{d}{dx}\sin x=\cos x,\quad \dfrac{d}{dx}\cos x=-\sin x\]

\frac{d}{d x} \ln x = \frac{1}{x}

When a factor is composite, differentiate it with the chain rule (keep the inner factor):

\[\dfrac{d}{dx}e^{kx}=k\,e^{kx},\qquad \dfrac{d}{dx}\big(g(x)\big)^{n}=n\big(g(x)\big)^{n-1}g'(x)\]

\frac{d}{d x} e^{k x} = k e^{k x}

Factorised form. After applying the rule you get two terms with a common factor; the syllabus asks you to take out the highest common factor. For example \(2x e^{x}+x^{2}e^{x}=x e^{x}(x+2)\).

How to differentiate a product

Name the factors. Write \(u\) and \(v\) for the two functions being multiplied.
Differentiate each. Find \(u'\) and \(v'\). If a factor is composite, use the chain rule and keep its inner factor.
Apply the rule. Write \(y'=u'v+uv'\) — two terms, one from each factor.
Factorise. Take out the highest common factor so the answer is in simplest and factorised form.

Applications. Once you have \(f'(x)\), the gradient at a point is \(f'(a)\) (used for a tangent), and the stationary points are the solutions of \(f'(x)=0\).

Example 1 — Power × exponential

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

Solution

Let \(u=x^{2}\), \(v=e^{x}\), so \(u'=2x\), \(v'=e^{x}\).

\(y'\)	\(=\)	\(u'v+uv'\)
	\(=\)	\(2x\,e^{x}+x^{2}e^{x}\)
	\(=\)	\(x e^{x}(x+2)\)

y^{'} = x e^{x} (x + 2)

Example 2 — Two functions

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

Solution

Both factors are functions of \(x\): \(u=e^{x}\), \(v=\sin x\).

\(y'\)	\(=\)	\(e^{x}\sin x+e^{x}\cos x\)
	\(=\)	\(e^{x}(\sin x+\cos x)\)

y^{'} = e^{x} (\sin x + \cos x)

Example 3 — Power × logarithm

Differentiate \(y=x\ln x\).

Solution

\(u=x\), \(v=\ln x\), so \(u'=1\), \(v'=\dfrac{1}{x}\).

\(y'\)	\(=\)	\(1\cdot\ln x+x\cdot\dfrac{1}{x}\)
	\(=\)	\(\ln x+1\)

y^{'} = \ln x + 1

Example 4 — Product + chain

Differentiate \(y=x^{2}e^{3x}\), in factorised form.

Solution

\(u=x^{2}\), \(v=e^{3x}\). By the chain rule \(v'=3e^{3x}\) (keep the inner \(3\)).

\(y'\)	\(=\)	\(2x\,e^{3x}+x^{2}\cdot 3e^{3x}\)
	\(=\)	\(x e^{3x}(2+3x)\)

y^{'} = x e^{3 x} (3 x + 2)

Common pitfalls

The derivative of a product is not the product of the derivatives. \((uv)'=u'v+uv'\), never \(u'v'\). For \(x^{2}e^{x}\) the answer is \(x e^{x}(x+2)\), not \(2x e^{x}\).

Keep both terms. The rule always gives two terms; writing only \(u'v\) or only \(uv'\) drops half the derivative.

Do not lose the chain-rule inner factor. If a factor is composite, keep its inner derivative — \(\dfrac{d}{dx}e^{3x}=3e^{3x}\), not \(e^{3x}\).

Frequently asked questions

What is the product rule?

If \(y=u\,v\), where \(u\) and \(v\) are both functions of \(x\), then \(y'=u'v+uv'\). Differentiate each factor once, keep the other unchanged, and add.

How do you differentiate \(x^{2}e^{x}\)?

Take \(u=x^{2}\), \(v=e^{x}\), so \(y'=2x e^{x}+x^{2}e^{x}=x e^{x}(x+2)\).

Why is \((uv)'\) not \(u'v'\)?

Because both factors change at once, so the change in \(uv\) has two parts, \(u\,dv\) and \(v\,du\). That gives \(u'v+uv'\), not the product of the derivatives.

How do you use the product rule with the chain rule?

Differentiate any composite factor with the chain rule. For \(x^{2}e^{3x}\), \(v'=3e^{3x}\), so \(y'=x e^{3x}(3x+2)\).

When do you need the product rule?

Whenever two functions of \(x\) are multiplied, such as \(x\ln x\), \(e^{x}\sin x\) or \(x^{2}e^{-x}\). A constant factor does not need it.

What does simplest and factorised form mean?

Take out the highest common factor of the two terms, e.g. write \(x e^{x}(x+2)\) rather than \(2x e^{x}+x^{2}e^{x}\).

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