7.7 The product, quotient and chain rules — Learn

Prerequisites: the standard derivatives from Section 7.3, including $e^{kx}$ , $\sin kx$ , $\cos kx$ and $\ln x$ ; multiplying and dividing algebraic fractions.

By the end of this section you can: differentiate a product of two functions; differentiate a quotient of two functions; differentiate a function of a function with the chain rule; and choose the right rule for an expression.

1. The product rule

When a function is a product of two simpler functions, write the first as $u$ and the second as $v$ . Then

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.

In words: differentiate the first and keep the second, then keep the first and differentiate the second, and add the two results.

Example 1. Differentiate $y = x^2\sin x$ .

Take $u = x^2$ and $v = \sin x$ , so $\dfrac{du}{dx} = 2x$ and $\dfrac{dv}{dx} = \cos x$ . Then

\frac{dy}{dx} = x^2\cos x + (\sin x)(2x) = x^2\cos x + 2x\sin x.

Example 2. Differentiate $y = x e^{3x}$ .

Take $u = x$ and $v = e^{3x}$ . Then $\dfrac{du}{dx} = 1$ and $\dfrac{dv}{dx} = 3e^{3x}$ , so

\frac{dy}{dx} = x(3e^{3x}) + e^{3x}(1) = e^{3x}(3x + 1).

Taking out the common factor $e^{3x}$ tidies the answer.

2. The quotient rule

For a quotient $\dfrac{u}{v}$ ,

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

The order matters in the numerator: the denominator times the derivative of the top, minus the top times the derivative of the bottom.

Example 3. Differentiate $y = \dfrac{\sin x}{x}$ .

Take $u = \sin x$ and $v = x$ . Then $\dfrac{du}{dx} = \cos x$ and $\dfrac{dv}{dx} = 1$ , so

\frac{dy}{dx} = \frac{x\cos x - \sin x \times 1}{x^2} = \frac{x\cos x - \sin x}{x^2}.

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

Take $u = 2x + 1$ and $v = x - 3$ . Then $\dfrac{du}{dx} = 2$ and $\dfrac{dv}{dx} = 1$ , so

\frac{dy}{dx} = \frac{(x - 3)(2) - (2x + 1)(1)}{(x - 3)^2} = \frac{2x - 6 - 2x - 1}{(x - 3)^2} = \frac{-7}{(x - 3)^2}.

3. The chain rule

When a function is applied to another function, differentiate the outside and multiply by the derivative of the inside:

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx},

where $u$ is the inside function.

Example 5. Differentiate $y = (3x + 1)^5$ .

Let $u = 3x + 1$ , so $y = u^5$ . Then

\frac{dy}{du} = 5u^4, \qquad \frac{du}{dx} = 3,

and

\frac{dy}{dx} = 5u^4 \times 3 = 15(3x + 1)^4.

Example 6. Differentiate $y = \sin(x^2)$ .

The outside function is sine and the inside function is $x^2$ :

\frac{dy}{dx} = \cos(x^2) \times 2x = 2x\cos(x^2).

Example 7. Differentiate $y = e^{x^2}$ .

\frac{dy}{dx} = e^{x^2} \times 2x = 2xe^{x^2}.

Example 8. Differentiate $y = \ln(2x + 1)$ .

\frac{dy}{dx} = \frac{1}{2x + 1} \times 2 = \frac{2}{2x + 1}.

4. Choosing the rule

Read the structure of the expression before you start.

Structure	Rule	Example
A product, $u \times v$	product rule	$x^2\sin x$
A quotient, $\dfrac{u}{v}$	quotient rule	$\dfrac{\sin x}{x}$
A function of a function	chain rule	$\sin(x^2)$ , $(3x+1)^5$
A simple sum or multiple	power rule, term by term	$4x^3 - 2x$

A chain-rule expression can be a product as well. Then use the chain rule inside the product rule, and name each part.

Watch out. Four mistakes are common with these rules.

Leaving a term out of the product rule. The rule has two terms, joined by a plus sign. The derivative of $uv$ is not $u'v'$ . Check with $u = x$ and $v = x$ : the product rule gives $1 \cdot x + x \cdot 1 = 2x$ , which matches the derivative of $x^2$ .

Reversing the quotient rule. The numerator starts with the bottom times the derivative of the top. Reversing the two products changes the sign of the answer.

Forgetting the inside derivative in the chain rule. The derivative of $\sin(x^2)$ is $2x\cos(x^2)$ , not $\cos(x^2)$ . Always multiply by the derivative of the inside.

Using the quotient rule on a constant denominator. A fraction such as $\dfrac{x^2}{3}$ has a constant bottom, so divide by $3$ and use the power rule. Re-check: after differentiating, look for a common factor and take it out.

You should now be able to

differentiate a product with the product rule;
differentiate a quotient with the quotient rule;
differentiate a function of a function with the chain rule;
choose the right rule from the structure of the expression;
take out a common factor to simplify an answer.