The chain rule and logarithms

The chain rule is the rule for compositions, and once you have it almost every function in the course is in reach. The logarithm arrives with it: $\log x$ , which here as in the course text means the natural logarithm, differentiates to $1/x$ , and logarithmic differentiation turns an ugly product, quotient or variable power into a sum before any differentiating happens.

Start practising Browse all 30 questions

In this topic

Taught: Before the midterm
Questions: 30
Total marks: 85
Suggested time: 132 min
Difficulty: Introductory to Capstone
Show the 8 hardest

Key results

The chain rule: $\frac{d}{dx} f\bigl(g(x)\bigr) = f'\bigl(g(x)\bigr)\, g'(x)$
When $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ .
The derivative of the logarithm: $\frac{d}{dx}\log x = \frac{1}{x}, \qquad \frac{d}{dx}\log|x| = \frac{1}{x}$
The first for $x > 0$ , the second for every $x \neq 0$ .
The logarithm of a function: $\frac{d}{dx}\log u(x) = \frac{u'(x)}{u(x)}$
When $u$ is differentiable at $x$ and $u(x) > 0$ .
Laws of logarithms: $\log(ab) = \log a + \log b, \qquad \log\frac{a}{b} = \log a - \log b, \qquad \log\bigl(a^{r}\bigr) = r\log a$
For $a > 0$ , $b > 0$ and any constant $r$ .
Logarithmic differentiation: $f'(x) = f(x)\,\frac{d}{dx}\log f(x)$
When $f$ is differentiable at $x$ and $f(x) > 0$ ; if $f(x) < 0$ , use $\log|f(x)|$ instead.
Change of base: $a^{x} = e^{x\log a}, \qquad \log_{a} x = \frac{\log x}{\log a}$
For a constant base $a > 0$ with $a \neq 1$ , and $x > 0$ in the second.

Read the theoryUBC Math 100 §4.3 The chain ruleUBC Math 100 §4.4 Logarithmic differentiationCLP-1 §2.9 The chain ruleCLP-1 §2.10 The natural logarithm

Worked example

Standard3 marks · 4 min

Let $y = \bigl(3x^{2} + 1\bigr)^{5}$ . Find $\dfrac{dy}{dx}$ at $x = 1$ .

Strategy

Differentiate the outside function with the inside left alone, then multiply by the derivative of the inside. Substitute only at the end.

Working

$\frac{dy}{dx} = 5\bigl(3x^{2} + 1\bigr)^{4} \cdot \frac{d}{dx}\bigl(3x^{2} + 1\bigr)$
Outer derivative first. The bracket is carried through unchanged.
$= 5\bigl(3x^{2} + 1\bigr)^{4} \cdot 6x = 30x\bigl(3x^{2} + 1\bigr)^{4}$
The inner derivative is 6x. Multiply the constants.
$3(1)^{2} + 1 = 4$
Evaluate the inside at x = 1 before raising to the fourth power.
$\left.\frac{dy}{dx}\right|_{x=1} = 30(1)(4)^{4} = 30 \cdot 256 = 7680$
Substitute and compute. 4⁴ = 256.

Answer

\left.\frac{dy}{dx}\right|_{x=1} = 7680

Check

y(0.999) ≈ 1016.35 and y(1.001) ≈ 1031.71, a rise of about 15.36 over a step of 0.002, a slope of 7680.

Where this goes wrong

Stopping at 5(3x² + 1)⁴ and forgetting the inner derivative, which gives 1280. The chain rule is a product of two derivatives, and the second one is easy to leave out.

On the paper

Name the inner function, write the outer derivative times the inner derivative before simplifying, and substitute the point last.

Leans on

The power rule, from Differentiation rules

Your turn

Standard3 marks · 4 min

Let $f(x) = \log\bigl(x^{2} + 4\bigr)$ . Find $f'(2)$ .

Answer this question

Practice questions

5 of the 30, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
Let $f(x) = (2x + 1)^{4}$ . Find $f'(0)$ .
Strategy
The function is a power of a linear expression. The chain rule gives the outer derivative times the inner derivative; the inner derivative is the constant 2.
Working
$f'(x) = 4(2x + 1)^{3}\cdot 2 = 8(2x + 1)^{3}$
Outer derivative 4u³, inner derivative 2.
$f'(0) = 8(1)^{3} = 8$
Substitute x = 0.
Answer
$f'(0) = 8$
The answer is 8.
Check
Between x = −0.01 and x = 0.01, f goes from 0.98⁴ ≈ 0.9224 to 1.02⁴ ≈ 1.0824, a rise of about 0.16 over 0.02, so a slope of 8.
Where this goes wrong
Forgetting the inner derivative and answering 4. The factor of 2 from differentiating 2x + 1 is the whole point of the chain rule.
Leans on
The power rule, from Differentiation rules
Try a similar question
Let $y = \bigl(3x^{2} + 1\bigr)^{5}$ . Find $\dfrac{dy}{dx}$ at $x = 1$ .
Answer this question
Introductory2 marks · 3 min
Let $F(x) = \sin\!\left(x^{2}\right)$ . Which of the following is $F'(x)$ ?
Strategy
Identify the composition, differentiate the outer function at the inner one, and multiply by the inner derivative.
Working
$F(x) = \sin(u), \quad u = x^{2}$
Outer: sin. Inner: x².
$F'(x) = \cos(u)\cdot u' = \cos\!\left(x^{2}\right)\cdot 2x$
Derivative of sin is cos, evaluated at the inside; times the inner derivative 2x.
Answer
$F'(x) = 2x\cos\!\left(x^{2}\right)$
The answer is 2x times cosine of x squared.
Check
At x = √(π/2) the inside x² is π/2, where sine peaks at 1, so F′ must be 0 there, and 2x cos(x²) is.
Where this goes wrong
Choosing 2 sin x cos x, which is the derivative of (sin x)², a different function. Where the square sits, inside or outside the sine, changes everything.
Leans on
Parsing a composite formula, from Functions, graphs and inverses
Try a similar question
Let $f(x) = \tan^{3}x$ for $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ . Find $f'(x)$ .
Answer this question
Standard3 marks · 4 min
Let $g(x) = x^{2}\,h\!\left(x^{3}\right)$ , where $h$ is a differentiable function with $h(8) = 3$ and $h'(8) = -1$ . Find $g'(2)$ .
Strategy
You never learn a formula for h; you only need its value and derivative at 8, which is where x³ lands when x = 2. Product rule outside, chain rule inside.
Working
$g'(x) = 2x\,h(x^{3}) + x^{2}\cdot h'(x^{3})\cdot 3x^{2}$
Product rule; chain rule on h(x³).
$g'(2) = 2(2)\,h(8) + (4)\,h'(8)\,(3\cdot 4) = 4\cdot 3 + 4\cdot(-1)\cdot 12$
Substitute x = 2, so x³ = 8.
$= 12 - 48 = -36$
Arithmetic.
Answer
$g'(2) = -36$
The answer is negative 36.
Check
Take h(u) = 11 − u, which has h(8) = 3 and h′(8) = −1: then g = 11x² − x⁵ and g′(2) = 44 − 80 = −36.
Where this goes wrong
Writing the derivative of h(x³) as h′(x³) and dropping the 3x², or evaluating h′ at 2 rather than at 8. The chain rule evaluates the outer derivative at the inner value.
Leans on
The product rule, from Differentiation rules
Try a similar question
Let $g(x) = x^{2}\sin(3x)$ . Find $g'\!\left(\dfrac{\pi}{3}\right)$ .
Answer this question
Challenging4 marks · 6 min
Let $f(x) = \bigl(x^{2} + 1\bigr)^{x}$ . Use logarithmic differentiation to find $f'(1)$ .
Strategy
Take logarithms to move the variable exponent down as a multiplier, differentiate with the product and chain rules, then multiply through by the original function.
Working
$\log f(x) = x\log\!\left(x^{2} + 1\right)$
x² + 1 > 0, so f(x) > 0 and log f(x) is defined. The log of a power brings the exponent down.
$\frac{f'(x)}{f(x)} = \log\!\left(x^{2} + 1\right) + x\cdot\frac{2x}{x^{2} + 1}$
Differentiate both sides; the left is (log f)′ by the chain rule, the right by the product rule.
$\frac{f'(1)}{f(1)} = \log 2 + \frac{2}{2} = \log 2 + 1, \qquad f(1) = 2^{1} = 2$
Evaluate at x = 1.
$f'(1) = 2(\log 2 + 1) = 2\log 2 + 2 \approx 3.386$
Multiply by f(1).
Answer
$f'(1) = 2\log 2 + 2$
The answer is 2 natural log 2 plus 2, about 3.386.
Check
Holding the exponent at 1 gives a slope of 2, holding the base at 2 gives 2 log 2 ≈ 1.386, and the two effects add to 3.386.
Where this goes wrong
Treating the exponent as constant and writing x(x² + 1)^x−1·2x, or treating the base as constant. With both varying, only logarithmic differentiation (or rewriting as e^{x log(x²+1)}) is valid.
Leans on
Laws of logarithms, from Functions, graphs and inverses
Try a similar question
Let $f(x) = (2 + \cos x)^{x}$ . Find $f'\!\left(\dfrac{\pi}{2}\right)$ .
Answer this question
Capstone5 marks · 9 min
Let $f$ be a differentiable function with $f(x) > 0$ for every $x$ , $f(1) = e$ and $f'(1) = 3$ . Let $g$ be the function below. Find $g'(1)$ .
$g(x) = \bigl(f(x^{2})\bigr)^{x}$
Strategy
A variable base to a variable power calls for logarithmic differentiation. The logarithm brings the exponent down as a factor; the product rule and the chain rule, applied twice to log f(x²), do the rest, and only then are the values at x = 1 put in.
Working
$\log g(x) = x\log f(x^{2})$
The logarithm is defined because f > 0, and it brings the exponent down.
$\frac{g'(x)}{g(x)} = \log f(x^{2}) + x\cdot\frac{f'(x^{2})\cdot 2x}{f(x^{2})}$
The product rule on the right; log f(x²) needs the chain rule twice, for log of f and for f of x².
$g(1) = f(1)^{1} = e, \qquad \frac{g'(1)}{g(1)} = \log e + \frac{1\cdot 3\cdot 2}{e} = 1 + \frac{6}{e}$
At x = 1, x² = 1, so f(x²) = e and f′(x²) = 3.
$g'(1) = e\left(1 + \frac{6}{e}\right) = e + 6 \approx 8.718$
Multiply back by g(1).
Answer
$g'(1) = e + 6$
The answer is e plus 6, about 8.718.
Check
With f(t) = e · exp(3(t − 1)/e), which is positive everywhere and has f(1) = e and f′(1) = 3, the difference quotient (g(1.001) − g(0.999))/0.002 is about 8.7183.
Where this goes wrong
Using the power rule as if the exponent were fixed, which gives 6, or the rule for aˣ as if the base were fixed, which gives e. Both the base and the exponent vary, so logarithmic differentiation is needed, and the chain rule on f(x²) brings in the factor 2x.
Leans on
Laws of logarithms, from Functions, graphs and inverses
Try a similar question
Let $f(x) = \bigl(x^{2} + 1\bigr)^{x}$ . Use logarithmic differentiation to find $f'(1)$ .
Answer this question

Browse all 30 questions

Needs
Differentiation rules
Power, sum, product and quotient rules, tangent lines and higher-order derivatives.
23 questions
Needs
Exponential and trigonometric functions
Derivatives of eˣ, aˣ, sine, cosine and tangent, and the limits behind them.
22 questions
Leads to
Implicit differentiation and inverse trigonometric functions
Differentiating relations, tangent lines to curves, and the derivatives of inverse and inverse trigonometric functions.
30 questions
Leads to
L’Hôpital’s rule
Indeterminate forms 0/0, ∞/∞, 0·∞, ∞−∞, 1^∞, 0⁰ and ∞⁰, and when the rule does not apply.
26 questions
Leads to
Curve sketching
Increase, concavity, inflection points, asymptotes and full sketches.
29 questions
Leads to
Optimization
Building an objective function and justifying the extreme value.
18 questions
Leads to
Linear approximation
Tangent-line estimates, whether they run high or low, and approximating a value without a calculator.
14 questions
Leads to
Exponential growth and decay
The differential equations y′ = ky and y′ = a − by: half-lives, doubling times, Newton’s law of cooling and inflow and outflow.
11 questions
Leads to
Functions of two variables and partial derivatives
Evaluating and reading functions of two variables, level curves, partial derivatives as marginal rates, and second partials.
15 questions
Leads to
Antiderivatives
Reversing differentiation, the constant of integration and initial conditions.
11 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Related topics