Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q10: Second derivative of $\sin(\log x)$

Exercise 5.7Second order derivative · Section 5.7

The question

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

10. $\sin(\log x)$

Exam answer

Let $y = \sin(\log x)$ ⁠, $x \gt 0$ ⁠. By the chain rule,

\begin{aligned} \frac{dy}{dx} &= \cos(\log x)\cdot\frac1x \\ &= \frac{\cos(\log x)}{x} \end{aligned}

Differentiating again by the quotient rule,

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{1}{x^2}\bigg[x\cdot\Big(-\frac{\sin(\log x)}{x}\Big) \\ &\qquad - \cos(\log x)\cdot 1\bigg] \end{aligned}

$\therefore\ \dfrac{d^2y}{dx^2} = -\dfrac{\sin(\log x) + \cos(\log x)}{x^2}$ $(x \gt 0)$

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

The one idea this problem is built on

Differentiate twice, one rule at a time. The first derivative needs the chain rule (sine of something), and it comes out as a fraction,

\frac{dy}{dx} = \frac{\cos(\log x)}{x}.

The second derivative is then a derivative of a fraction, so it needs the quotient rule, with the chain rule once more for the top.

Given, to find, and the plan

Given: $y = \sin(\log x)$ ⁠. Here $\log$ is the natural logarithm, defined only for $x \gt 0$ ⁠, so the function lives on $(0, \infty)$ ⁠.
To find: $\dfrac{d^2y}{dx^2}$ for $x \gt 0$ ⁠.
Plan: (1) State the domain. (2) First derivative by the chain rule (this is Example 26(ii) of the book). (3) Second derivative by the quotient rule, with the chain rule for $\cos(\log x)$ ⁠. (4) Tidy up and compare with the answer key.

Basics you need

1Derivative of $\log x$

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

2Chain rule

If $y = f(u)$ and $u = g(x)$ ⁠, then $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ ⁠: derivative of the outer function at the inner one, times the derivative of the inner function (Theorem 4, Section 5.3.1). Lesson: chain rule

3Quotient rule

$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ wherever $v \neq 0$ (Section 5.3, rule (3)). Basics: derivative rules

4Second order derivative (Section 5.7)

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

Solution

The domain

$\log x$ is defined only for $x \gt 0$ ⁠, and $\sin$ accepts every real number. So $y = \sin(\log x)$ is defined exactly for $x \gt 0$ ⁠, and every derivative below is for $x \gt 0$ ⁠.

A derivative can only exist where the function exists. Stating the domain first also tells us that dividing by $x$ later is safe, since $x \neq 0$ throughout.

The natural logarithm is defined on $(0, \infty)$ (Section 5.4). Basics: exponentials and logarithms
First derivative by the chain rule

Outer function $\sin u$ ⁠, inner function $u = \log x$ ⁠. Then $\frac{dy}{du} = \cos u$ and $\frac{du}{dx} = \frac{1}{x}$ ⁠, so
$\frac{dy}{dx} = \cos(\log x)\cdot\frac{1}{x} = \frac{\cos(\log x)}{x}.$
$\sin(\log x)$ is a composite: first take the log, then the sine. The chain rule differentiates the outer layer (sine becomes cosine, with the inside left alone) and multiplies by the derivative of the inner layer. The book works exactly this out in Example 26(ii).

$\frac{d}{du}\sin u = \cos u$ (Table 5.3); Basics 1 and 2. Lesson: chain rule
Set up the quotient rule

Write $\frac{dy}{dx} = \frac{u}{v}$ with
$u = \cos(\log x), \qquad v = x.$
Their derivatives: by the chain rule (outer $\cos$ ⁠, inner $\log x$ ⁠),
$u' = -\sin(\log x)\cdot\frac{1}{x} = -\frac{\sin(\log x)}{x},$
and $v' = 1$ ⁠.

The first derivative is a fraction whose top and bottom both depend on $x$ ⁠, so the quotient rule is the natural tool. Working out $u'$ and $v'$ separately first keeps the next line clean and makes sign errors easy to spot.

$\frac{d}{du}\cos u = -\sin u$ (Table 5.3). Basics: derivative rules
Second derivative

Quotient rule, $\frac{u'v - uv'}{v^2}$ ⁠. First the numerator:
$\begin{aligned} u'v - uv' &= -\frac{\sin(\log x)}{x}\cdot x - \cos(\log x)\cdot 1 \\ &= -\sin(\log x) - \cos(\log x). \end{aligned}$
Dividing by $v^2 = x^2$ ⁠:
$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{-\sin(\log x) - \cos(\log x)}{x^2} \\ &= -\frac{\sin(\log x) + \cos(\log x)}{x^2}. \end{aligned}$
This holds for every $x \gt 0$ (the denominator $x^2$ is never zero there).

In $u'v$ the factor $\frac{1}{x}$ cancels against $v = x$ ⁠, which is allowed because $x \neq 0$ ⁠. Then the two terms on top both carry a minus sign, so we take $-1$ out as a common factor. This is exactly the form in the answer key.

Quotient rule, Section 5.3, rule (3). Basics: derivative rules
Another route, as a check: the product rule

Write $\frac{dy}{dx} = \frac{1}{x}\cdot\cos(\log x)$ ⁠, a product. Using $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$ ⁠:
$\begin{aligned} \frac{d^2y}{dx^2} &= -\frac{1}{x^2}\cos(\log x) \\ &\quad + \frac{1}{x}\cdot\left(-\frac{\sin(\log x)}{x}\right) \\ &= -\frac{\cos(\log x) + \sin(\log x)}{x^2}, \end{aligned}$
the same as Step 4.

Getting the same answer by a second rule is a cheap and reliable check on signs. Either route is fine in an exam.

Product rule, Section 5.3, rule (2); $\frac{d}{dx}x^{-1} = -x^{-2}$ (Table 5.3).

Result

For $x \gt 0$ ⁠,

\frac{d^2y}{dx^2} = -\frac{\sin(\log x) + \cos(\log x)}{x^2}.

(Textbook answer: $-\dfrac{\sin(\log x) + \cos(\log x)}{x^2}$ ⁠, the same.)

Check it with numbers

Take $x = 2$ ⁠. Then $\log 2 \approx 0.6931472$ ⁠, $\sin(\log 2) \approx 0.6389613$ and $\cos(\log 2) \approx 0.7692389$ ⁠, so

\begin{aligned} \frac{dy}{dx} &\approx \frac{0.7692389}{2} \approx 0.3846195, \\ \frac{d^2y}{dx^2} &\approx -\frac{1.4082002}{4} \approx -0.3520500. \end{aligned}

From values of $y = \sin(\log x)$ alone, with $h = 0.001$ ⁠, the second difference $\frac{y(2.001) - 2y(2) + y(1.999)}{h^2}$ is about $-0.352050$ ⁠, matching.

Common mistakes

Forgetting the inner derivative. $\frac{d}{dx}\sin(\log x)$ is $\cos(\log x)\cdot\frac{1}{x}$ ⁠, not just $\cos(\log x)$ ⁠. The same factor $\frac{1}{x}$ appears again when differentiating $\cos(\log x)$ ⁠.
Sign of $\cos$ 's derivative. $\frac{d}{du}\cos u = -\sin u$ ⁠. Losing this minus gives $\frac{\sin(\log x) - \cos(\log x)}{x^2}$ ⁠, which is wrong.
Treating $x$ in the denominator as a constant. $\frac{\cos(\log x)}{x}$ has $x$ on top and bottom, so it needs the quotient (or product) rule.
Leaving out the domain. The answer is only meaningful for $x \gt 0$ ⁠.

What to take away

For a second derivative, first get a clean first derivative, then decide which rule its shape needs (here a quotient).
Every time $\log x$ sits inside another function, the chain rule brings out a factor $\frac{1}{x}$ ⁠.
Functions of $\log x$ lead to second derivatives with $x^2$ underneath; that is why Question 13 multiplies by $x^2$ and $x$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\log x$ is defined for $x \gt 0$ ⁠; $\frac{d}{dx}\log x = \frac{1}{x}$	Steps 1, 2, 3	Section 5.4, Theorem 5 (Lesson: exponential and logarithmic functions)
$\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$ ⁠, $\frac{d}{dx}x^n = nx^{n-1}$	Steps 2, 3, 5	Table 5.3, Section 5.3
Chain rule; $\frac{d}{dx}\sin(\log x) = \frac{\cos(\log x)}{x}$	Steps 2, 3	Theorem 4, Section 5.3.1; Example 26(ii) (Lesson: chain rule)
Product and quotient rules	Steps 4, 5	Section 5.3, rules (2) and (3) (Basics: derivative rules)
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$	Steps 4, 5	Section 5.7 (Lesson: second order derivative)

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Exercise 5.7 · Q10: Second derivative of sin⁡(log⁡x)\sin(\log x)sin(logx)

Basics you need

Solution

The domain

First derivative by the chain rule

Set up the quotient rule

Second derivative

Another route, as a check: the product rule

Every fact used, and where it comes from

Exercise 5.7 · Q10: Second derivative of $\sin(\log x)$