SL 5.4 Apply chain, product, and quotient rules | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

IB Mathematics: Analysis and Approaches SL — Study topic SL 5.4

Differentiation describes how quickly a function changes. Before using these rules, recall that the derivative of a function

y=f(x)

is written

f'(x)

or

\frac{dy}{dx}

, and gives the gradient of the graph at a point. You may already know basic derivatives such as

\frac{d}{dx}(x^n)=nx^{n-1}

and

\frac{d}{dx}(\sin x)=\cos x

, where angles are measured in radians. The rules in this lesson extend that work when expressions are built from functions that are composed, multiplied, or divided. In each case, first identify the structure; then apply the matching rule and simplify carefully.

What you will learn

Recognise when a function is composed, multiplied, or divided and select the matching differentiation rule.
Apply the chain, product, and quotient rules accurately, using appropriate prerequisite derivatives.
Check a derivative using a graphing calculator or numerical estimate and interpret it in context.

1. Recognise the structure before differentiating

A function can be built in different ways. In

y=(3x-2)^4

, one function is placed inside another: the inner expression is

3x-2

, and the outer operation is raising that result to the fourth power. This is a composition, so use the chain rule.

In

y=x^2\sin x

, two functions are multiplied, so use the product rule. In

y=\frac{x+1}{x^2+4}

, one function is divided by another, so use the quotient rule. These rules are not interchangeable: choosing based on the visible structure helps prevent errors.

The chain rule connects the rate of change of an outer function to the rate of change of its inner input. The product rule accounts for both factors changing. The quotient rule accounts for changes in both the numerator and denominator.

Composition: one expression is an input to another; use the chain rule.
Multiplication: both factors may vary; use the product rule.
Division: both numerator and denominator may vary; use the quotient rule.

2. The three rules and how to use them

For a composition

y=f(g(x))

, differentiate the outer function while keeping the inner expression in place, then multiply by the derivative of the inner expression. This is the chain rule. For example, the power rule combined with the chain rule gives the derivative of

(g(x))^n

as

n(g(x))^{n-1}g'(x)

.

For a product

y=u(x)v(x)

, differentiate the first factor and multiply by the second, then add the first factor multiplied by the derivative of the second. Keep the two terms: differentiating only one factor misses part of the change.

For a quotient

y=\frac{u(x)}{v(x)}

, subtract the numerator function multiplied by the derivative of the denominator from the denominator function multiplied by the derivative of the numerator. Divide by the square of the original denominator. The rule applies only where

v(x)\ne 0

.

These rules often appear together. For instance, a product may contain a composite power. Differentiate the overall product with the product rule, then apply the chain rule within the derivative of a factor. Parentheses help keep track of each function and its derivative.

\frac{d}{dx}f(g(x))=f'(g(x))g'(x),\quad (uv)'=u'v+uv',\quad \left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}

Chain: outer derivative, then multiply by inner derivative.
Product: first derivative times second, plus first times second derivative.
Quotient: denominator times numerator derivative minus numerator times denominator derivative, all over denominator squared.

3. Check meaning and use technology purposefully

The derivative is also a gradient. If a graphing calculator displays a tangent at a chosen input, its gradient should agree with the derivative evaluated at that input. A numerical estimate can provide a useful check, but it does not replace showing the rule and algebra that produce the derivative.

A practical check is to graph the original function and its derivative, then compare the derivative value at a selected input with the tangent gradient on the original graph. Alternatively, use a calculator's numerical derivative feature at that same input. Record the input and retain enough displayed digits before rounding.

When a function describes a context, the derivative's units are output units per input unit. For example, if distance is measured in metres and time in seconds, its derivative has units of metres per second. The rules do not change; the interpretation and units come from the model.

A derivative value should match the local tangent gradient.
Use technology to check a result, not as an unexplained substitute for differentiation.
Check any denominator restriction and state units when the function models a quantity.

Worked example

Chain rule with a power

Differentiate

y=(3x-2)^4

and find the gradient when

x=1

.

Identify the inner expression
The outer operation raises an input to the fourth power, and the input is $3x-2$ . Differentiate the outer power first, then multiply by the derivative of the inner expression.
$y'=4(3x-2)^3\cdot 3$
Simplify and evaluate
The constant factor from the inner derivative is $3$ , so the derivative simplifies to $12(3x-2)^3$ . At $x=1$ , the inner expression is $1$ .
$y'(1)=12(3-2)^3=12$

Answer: The derivative is

y'=12(3x-2)^3

, and the gradient at

x=1

is

12

.

Check: A graphing calculator should show a tangent gradient of

12

for the original curve at

x=1

. The chain-rule factor of

3

is essential because the inner expression changes three times as fast as

x

.

Worked example

Product rule with a trigonometric factor

Differentiate

y=x^2\sin x

and find y'

at

x=

\frac{\pi}{2}

.

Assign the factors
Set $u=x^2$ and $v=\sin x$ . Their derivatives are $u'=2x$ and $v'=\cos x$ . Apply the product rule because the two expressions are multiplied.
$y'=u'v+uv'=2x\sin x+x^2\cos x$
Evaluate at the stated input
At $x=\frac{\pi}{2}$ , the sine value is $1$ and the cosine value is $0$ . Substitution leaves the first term only.
$y'\left(\frac{\pi}{2}\right)=2\left(\frac{\pi}{2}\right)(1)+\left(\frac{\pi}{2}\right)^2(0)=\pi$

Answer: The derivative is

y'=2x\sin x+x^2\cos x

, and its value at

x=\frac{\pi}{2}

is

\pi

.

Check: In radians, a numerical derivative near

x=1.571

should be close to

3.142

. This is consistent with

\pi

to three significant figures.

Worked example

Quotient rule and domain

Differentiate

f(x)=\frac{x+1}{x^2+4}

and find the derivative at

x=0

.

Set numerator and denominator
Let $u=x+1$ and $v=x^2+4$ . Then $u'=1$ and $v'=2x$ . The denominator $x^2+4$ is positive for every real $x$ , so the function is defined on the real numbers.
$f'(x)=\frac{(x^2+4)(1)-(x+1)(2x)}{(x^2+4)^2}$
Simplify and evaluate
Expand only the numerator and combine like terms. At $x=0$ , the numerator is $4$ and the denominator is $16$ .
$f'(x)=\frac{4-x^2-2x}{(x^2+4)^2},\qquad f'(0)=\frac{1}{4}$

Answer: The derivative is

f'(x)=\frac{4-x^2-2x}{(x^2+4)^2}

, and

f'(0)=\frac14

.

Check: The function is increasing locally at

x=0

because its derivative there is positive. A calculator's tangent-gradient or numerical-derivative value at zero should be approximately

0.25

.

Common mistakes and how to avoid them

For a composite power, differentiating the outer power but forgetting the derivative of the inside expression.

Correction: After differentiating the outer function, multiply by the derivative of its entire inner expression.

Differentiating a product by multiplying the derivatives of its factors.

Correction: Use

u'v+uv'

; the product rule has two terms.

Reversing the subtraction in the quotient rule or forgetting to square the denominator.

Correction: Use

vu'-uv'

in the numerator and

v^2

in the denominator, then check that the original denominator is nonzero.

Treating a calculator's decimal derivative as the full solution.

Correction: Show the applicable rule and algebra first; use the calculator value as a check and round only at the end.

Lesson summary

Identify whether the function is composed, multiplied, or divided before differentiating.
Apply the chain rule to a composition, the product rule to a product, and the quotient rule to a quotient.
Simplify with care, respect denominator restrictions, and check a derivative against a tangent gradient or numerical estimate.

Check your understanding

Question 1

What is the derivative of

y=(2x+1)^3

?

$6(2x+1)^2$
$3(2x+1)^2$
$6(2x+1)^3$
$3(2x+1)^2+2$

Show answer and explanation

6(2x+1)^2

The outer derivative is

3(2x+1)^2

and the inner derivative is

2

, giving

6(2x+1)^2

.

Question 2

For

y=x\cos x

, which expression is the derivative?

$\cos x-x\sin x$
$-\sin x$
$x(-\sin x)$
$\cos x+x\sin x$

Show answer and explanation

\cos x-x\sin x

The product rule gives

1\cdot\cos x+x\cdot(-\sin x)

, which is

\cos x-x\sin x

.

Question 3

For

g(x)=\frac{x}{x+2}

, what is

g'(x)

?

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

Show answer and explanation

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

The quotient rule gives

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

, for

x\ne -2

.

Key terms

Derivative: The rate of change of a function with respect to its input; graphically, it is the gradient of the tangent at a point.
Composition: A function formed by using one function's output as the input to another, such as $(3x-2)^4$ .
Inner function: The expression supplied as the input to the outer function in a composition.
Tangent: A line that gives the local direction of a curve at a point; its gradient equals the derivative there.

Continue through IB AA SL

View the complete IB AA SL International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL, study topic SL 5.4. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

SL 5.4 · Apply chain, product, and quotient rules

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — Study topic SL 5.4

What you will learn

1. Recognise the structure before differentiating

2. The three rules and how to use them

3. Check meaning and use technology purposefully

Chain rule with a power

Product rule with a trigonometric factor

Quotient rule and domain

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review