Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q5: Second derivative of $x^3\log x$

Exercise 5.7Second order derivative · Section 5.7

The question

Find the second order derivatives of the functions given in Exercises 1 to 10.

5. $x^3\log x$

Exam answer

Let $y = x^3\log x$ ( $x > 0$ ⁠). By the product rule,

\begin{aligned} \frac{dy}{dx} &= 3x^2\log x + x^3\cdot\frac1x \\ &= 3x^2\log x + x^2, \\ \frac{d^2y}{dx^2} &= 6x\log x + 3x^2\cdot\frac1x + 2x \\ &= 6x\log x + 5x \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} = x(5 + 6\log x)$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

$x^3\log x$ is a product, so the product rule gives the first derivative. That derivative, $3x^2\log x + x^2$ ⁠, still contains the product $3x^2\log x$ ⁠, so the product rule is used again. Along the way $x^3\cdot\frac1x$ simplifies to $x^2$ ⁠, which is allowed because $x \gt 0$ ⁠.

Given, to find, and the plan

Given: $y = x^3\log x$ (natural logarithm), defined for $x \gt 0$ ⁠.
To find: $\dfrac{d^2y}{dx^2}$ ⁠.
Plan: (1) State the domain. (2) Product rule for $\frac{dy}{dx}$ and simplify. (3) Sum rule, product rule on $3x^2\log x$ ⁠, power rule on $x^2$ ⁠. (4) Collect and factor; compare with the textbook form.

Basics you need

1Product rule (Leibnitz rule)

$(uv)' = u'v + uv'$ (Section 5.3, rule (2)). Basics: derivative rules

2Logarithm

$\log x$ exists only for $x \gt 0$ ⁠, and $\frac{d}{dx}\log x = \frac1x$ (Theorem 5(2), Section 5.4). Basics: exponentials and logarithms

3Power rule

$\frac{d}{dx}x^n = nx^{n-1}$ (Table 5.3); constants stay in front.

4Second order derivative (Section 5.7)

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ ⁠. Lesson: second order derivative

Solution

Name the function and its domain

Let $y = x^3\log x$ ⁠. The factor $\log x$ needs $x \gt 0$ ⁠; $x^3$ is defined everywhere. So the domain is $x \gt 0$ ⁠, and every derivative below is taken there.

A product exists only where both factors exist, so the domain is set by the more restrictive factor, the logarithm.

Section 5.4: $\log x$ is defined for $x \gt 0$ ⁠. Basics: exponentials and logarithms
First derivative: product rule

Take $u = x^3$ ⁠, $v = \log x$ ⁠, so $u' = 3x^2$ and $v' = \frac{1}{x}$ ⁠:
$\begin{aligned} \frac{dy}{dx} &= u'v + uv' = 3x^2\log x + x^3\cdot\frac{1}{x} \\ &= 3x^2\log x + x^2. \end{aligned}$
$y$ is a product of a power and a logarithm, so the product rule applies. In the second term $x^3\cdot\frac1x = x^2$ ⁠: cancelling one $x$ is allowed because $x \gt 0$ ⁠, so $x \ne 0$ ⁠.

Product rule (Section 5.3, rule (2)); $\frac{d}{dx}\log x = \frac1x$ (Theorem 5).
Second derivative: sum rule, then each term

Differentiate $\frac{dy}{dx} = 3x^2\log x + x^2$ term by term (sum rule):
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(3x^2\log x\right) + \frac{d}{dx}\left(x^2\right).$
First term, product rule with $u = 3x^2$ ( $u' = 6x$ ⁠) and $v = \log x$ ( $v' = \frac1x$ ⁠):
$\begin{aligned} \frac{d}{dx}\left(3x^2\log x\right) &= 6x\log x + 3x^2\cdot\frac{1}{x} \\ &= 6x\log x + 3x. \end{aligned}$
Second term, power rule: $\frac{d}{dx}x^2 = 2x$ ⁠.

The second derivative is the derivative of the expression from Step 2. Its first term is again a product of a power and a logarithm, so the product rule is needed once more; the second term is a plain power.

Sum rule (Section 5.3, rule (1)); product rule (rule (2)); power rule (Table 5.3).
Collect and factor
$\begin{aligned} \frac{d^2y}{dx^2} &= 6x\log x + 3x + 2x \\ &= 6x\log x + 5x \\ &= x(5 + 6\log x). \end{aligned}$
So $\dfrac{d^2y}{dx^2} = x(5 + 6\log x)$ for $x \gt 0$ ⁠.

$3x$ and $2x$ are like terms (sum $5x$ ⁠). Taking out the common factor $x$ gives exactly the textbook's form $x(5 + 6\log x)$ ⁠; expanding it back gives $5x + 6x\log x$ ⁠, the same expression.

Common factor: $6x\log x + 5x = x(6\log x + 5)$ ⁠.

Result

For $y = x^3\log x$ ⁠, $x \gt 0$ ⁠:

\begin{aligned} \frac{dy}{dx} &= 3x^2\log x + x^2, \\ \frac{d^2y}{dx^2} &= x(5 + 6\log x). \end{aligned}

(Textbook answer: $x(5 + 6\log x)$ ⁠.)

Check it with numbers

At $x = 2$ ⁠, with $\log 2 \approx 0.6931472$ ⁠:

Formula: $y'' = 2(5 + 6 \times 0.6931472)$ ⁠, which is about $18.3177662$ ⁠.
Second difference with $h = 0.001$ ⁠: $\frac{y(2.001) - 2y(2) + y(1.999)}{0.000001}$ is about $18.3177664$ ⁠.
First derivative: $12\log 2 + 4 \approx 12.3177662$ ⁠; central difference $\frac{y(2.001) - y(1.999)}{0.002}$ is about $12.3177687$ ⁠.

The values agree to about six significant figures. ✓

Common mistakes

Skipping the product rule in the second round. $\frac{d}{dx}(3x^2\log x)$ is not $6x\cdot\frac1x$ ⁠; both terms of the product rule are needed.
Losing the $x^2$ term. The first derivative has two terms, and both must be differentiated; the $2x$ from $x^2$ is part of the $5x$ ⁠.
Ignoring the domain. Everything holds only for $x \gt 0$ ⁠.

What to take away

Simplify the first derivative (here $x^3\cdot\frac1x = x^2$ ⁠) before differentiating again; it makes the second round shorter.
$\frac{d}{dx}(x^n\log x) = nx^{n-1}\log x + x^{n-1}$ for $x \gt 0$ ⁠, a pattern used twice here.

Every fact used, and where it comes from

Fact	Used in	Source
$\log x$ is defined for $x \gt 0$	Step 1	Section 5.4
$(uv)' = u'v + uv'$	Steps 2, 3	Section 5.3, rule (2)
$(u + v)' = u' + v'$	Step 3	Section 5.3, rule (1)
$\frac{d}{dx}\log x = \frac1x$	Steps 2, 3	Theorem 5(2), Section 5.4 (Lesson: exponential and logarithmic functions)
$\frac{d}{dx}x^n = nx^{n-1}$	Steps 2, 3	Table 5.3 (Basics: derivative rules)
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 3	Section 5.7 (Lesson: second order derivative)

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Exercise 5.7 · Q5: Second derivative of x3log⁡xx^3\log xx3logx

Basics you need

Solution

Name the function and its domain

First derivative: product rule

Second derivative: sum rule, then each term

Collect and factor

Every fact used, and where it comes from

Exercise 5.7 · Q5: Second derivative of $x^3\log x$