Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q4: Second derivative of $\log x$

Exercise 5.7Second order derivative · Section 5.7

The question

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

4. $\log x$

Exam answer

Let $y = \log x$ ( $x > 0$ ⁠). Then

\frac{dy}{dx} = \frac1x = x^{-1}

$\therefore\ \dfrac{d^2y}{dx^2} = -x^{-2} = -\dfrac{1}{x^2}$

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

The one idea this problem is built on

The first derivative of $\log x$ is $\frac{1}{x}$ ⁠, which is the power $x^{-1}$ ⁠. Writing it as a power lets the ordinary power rule finish the job:

\frac{d}{dx}x^{-1} = -1\cdot x^{-2} = -\frac{1}{x^2}.

Everything happens on the domain of $\log x$ ⁠, which is $x \gt 0$ ⁠.

Given, to find, and the plan

Given: $y = \log x$ (natural logarithm, base $e$ ⁠, as everywhere in this book), defined for $x \gt 0$ ⁠.
To find: $\dfrac{d^2y}{dx^2}$ ⁠.
Plan: (1) State the domain. (2) $\frac{dy}{dx} = \frac1x$ from Theorem 5. (3) Rewrite $\frac1x = x^{-1}$ and use the power rule. (4) Read the sign of the answer on the graph.

Basics you need

1Logarithm: domain and derivative

$\log x$ is defined only for $x \gt 0$ (Section 5.4), and $\frac{d}{dx}\log x = \frac{1}{x}$ for $x \gt 0$ (Theorem 5(2), Section 5.4). Basics: exponentials and logarithms

2Power rule for negative powers

$\frac{d}{dx}x^n = nx^{n-1}$ (Table 5.3) also holds for negative whole numbers $n$ ⁠, wherever $x \ne 0$ ⁠. With $n = -1$ ⁠: $\frac{d}{dx}x^{-1} = -x^{-2}$ ⁠. Basics: derivative rules

3Second 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 $y = \log x$ ⁠. The logarithm is defined only for positive numbers, so we work on $x \gt 0$ ⁠. Both derivatives below exist at every $x \gt 0$ ⁠.

$\log x$ is the power to which $e$ must be raised to give $x$ ⁠. Since $e^t \gt 0$ for every $t$ ⁠, only positive $x$ can be reached, so $\log 0$ and $\log(-2)$ do not exist. A derivative can only be taken where the function exists.

Section 5.4: $\log x$ is defined for $x \gt 0$ ⁠. Basics: exponentials and logarithms
First derivative: Theorem 5
$\frac{dy}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}, \qquad x \gt 0.$
This is one of the two standard results of Section 5.4 (Theorem 5(2)); the book states it without proof.

$\frac{d}{dx}\log x = \frac1x$ ⁠. Lesson: exponential and logarithmic functions
Second derivative: power rule with $n = -1$

Write $\frac{1}{x}$ as a power, then differentiate:
$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(x^{-1}\right) = (-1)\,x^{-1-1} \\ &= -x^{-2} = -\frac{1}{x^2}. \end{aligned}$
So $\dfrac{d^2y}{dx^2} = -\dfrac{1}{x^2}$ for $x \gt 0$ ⁠.

The power rule is stated for powers $x^n$ ⁠, so rewriting the fraction as $x^{-1}$ puts it in that form. The power $-1$ comes down as a factor and the new power is $-1 - 1 = -2$ ⁠. (The quotient rule gives the same: $\frac{0\cdot x - 1\cdot 1}{x^2} = -\frac{1}{x^2}$ ⁠.)

Power rule (Table 5.3); $x^{-2} = \frac{1}{x^2}$ ⁠. Basics: derivative rules
Read the sign: the slope keeps falling

For every $x \gt 0$ ⁠, $x^2 \gt 0$ ⁠, so $\frac{d^2y}{dx^2} = -\frac{1}{x^2} \lt 0$ ⁠. A negative second derivative means the slope $\frac{dy}{dx} = \frac1x$ decreases as $x$ increases: it is 2 at $x = \frac12$ ⁠, 1 at $x = 1$ and $\frac13$ at $x = 3$ ⁠. The graph of $\log x$ keeps rising but ever more slowly, so it bends downward.

$y = \log x$ (one unit = 90 px on both axes). The tangent slopes $\frac1x$ at $x = \frac12, 1, 3$ are $2, 1, \frac13$ ⁠: the slope falls as $x$ grows, which is exactly what $y'' = -\frac{1}{x^2} \lt 0$ says.

The second derivative is the rate of change of the slope. Seeing it on the graph makes the minus sign in the answer believable rather than just a result of algebra.

Result

For $y = \log x$ ⁠, $x \gt 0$ ⁠:

\frac{dy}{dx} = \frac{1}{x}, \qquad \frac{d^2y}{dx^2} = -\frac{1}{x^2}.

(Textbook answer: $-\dfrac{1}{x^2}$ ⁠.)

Check it with numbers

At $x = 2$ ⁠: the formula gives $y'' = -\frac{1}{4} = -0.25$ ⁠. With $h = 0.001$ ⁠, $\log 2.001 \approx 0.6936471$ ⁠, $\log 2 \approx 0.6931472$ ⁠, $\log 1.999 \approx 0.6926471$ ⁠, and

\begin{aligned} &\frac{\log 2.001 - 2\log 2 + \log 1.999}{0.000001} \\ &\approx -0.2500000. \end{aligned}

The first difference $\frac{\log 2.001 - \log 1.999}{0.002}$ is about $0.5000000 = \frac12$ ⁠. ✓

Common mistakes

$\frac{d}{dx}\frac1x = \frac{1}{x^2}$ ⁠: the sign is lost. The power $-1$ comes down as a factor, so the answer is negative.
$\frac{d}{dx}x^{-1} = -x^{0}$ ⁠: the power drops by one, from $-1$ to $-2$ ⁠, not up to $0$ ⁠.
Forgetting the domain. The answer is only meaningful for $x \gt 0$ ⁠, where $\log x$ exists.

What to take away

Rewrite $\frac1{x^k}$ as $x^{-k}$ and use the power rule.
$\log x$ has $y' = \frac1x \gt 0$ and $y'' = -\frac{1}{x^2} \lt 0$ ⁠: always increasing, always bending downward.

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 (Basics: exponentials and logarithms)
$\frac{d}{dx}\log x = \frac{1}{x}$	Step 2	Theorem 5(2), Section 5.4 (Lesson: exponential and logarithmic functions)
$\frac{d}{dx}x^n = nx^{n-1}$ ⁠, here $n = -1$	Step 3	Table 5.3, Section 5.3
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Step 3	Section 5.7 (Lesson: second order derivative)

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Exercise 5.7 · Q4: Second derivative of log⁡x\log xlogx

Basics you need

Solution

Name the function and its domain

First derivative: Theorem 5

Second derivative: power rule with n=−1n = -1n=−1

Read the sign: the slope keeps falling

Every fact used, and where it comes from

Exercise 5.7 · Q4: Second derivative of $\log x$

Second derivative: power rule with $n = -1$