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Chapter 5 · Continuity and Differentiability

Exercise 5.4 · Q8: Differentiate log⁡(log⁡x)\log(\log x)⁠, x>1x>1

Exercise 5.4Chain rule with log x · Section 5.4
The question

Differentiate the following w.r.t. xx⁠:

log⁡(log⁡x),x>1\log(\log x), \qquad x > 1
Exam answer

Let y=log⁡(log⁡x)y = \log(\log x)⁠. By the chain rule,

dydx=1log⁡x⋅ddx(log⁡x)=1log⁡x⋅1x\begin{aligned} \frac{dy}{dx} &= \frac{1}{\log x}\cdot\frac{d}{dx}(\log x) \\ &= \frac{1}{\log x}\cdot\frac1x \end{aligned}

∴ dydx=1xlog⁡x\therefore\ \dfrac{dy}{dx} = \dfrac{1}{x\log x} (x>1)(x > 1)⁠.

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

The one idea this problem is built on

A logarithm of a logarithm: inner function log⁡x\log x⁠, outer function log⁡t\log t⁠. Each has derivative "one over its input" (Theorem 5), and the chain rule multiplies them: 1log⁡x⋅1x\frac{1}{\log x}\cdot\frac1x⁠. The condition x>1x > 1 is exactly what makes the outer logarithm's input positive.

Given, to find, and the plan
Given
y=log⁡(log⁡x)y = \log(\log x)⁠, x>1x > 1 (natural logarithms).
To find
dydx\dfrac{dy}{dx}⁠.
Plan
(1) Show that the domain is exactly x>1x > 1⁠. (2) Name inner t=log⁡xt = \log x and outer y=log⁡ty = \log t⁠; differentiate each. (3) Multiply.

Basics you need

1The logarithm function

log⁡\log is defined only for positive inputs, is increasing, and log⁡1=0\log 1 = 0 (Section 5.4). So log⁡x>0\log x > 0 exactly when x>1x > 1⁠. Basics: exponentials and logarithms

2Theorem 5 (2)

ddxlog⁡x=1x\dfrac{d}{dx}\log x = \dfrac{1}{x} for x>0x > 0⁠. Lesson: exponential and logarithmic functions

3Chain rule

If y=f(t)y = f(t) and t=g(x)t = g(x)⁠, then dydx=dydt⋅dtdx\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} (Theorem 4). Basics: composite functions

Solution

  1. The domain is exactly x>1x > 1

    Two logarithms, two conditions:

    • inner log⁡x\log x needs x>0x > 0⁠;
    • outer log⁡(log⁡x)\log(\log x) needs log⁡x>0\log x > 0⁠.

    Because log⁡\log is increasing and log⁡1=0\log 1 = 0⁠, log⁡x>0\log x > 0 holds exactly when x>1x > 1⁠. That also gives x>0x > 0⁠. So the domain is x>1x > 1⁠, precisely the condition in the question.

    For 0<x≤10 < x \le 1⁠, log⁡x≤0\log x \le 0⁠, and the outer logarithm would be asked for the log of zero or a negative number, which is not defined. For example log⁡(log⁡12)\log(\log\frac12) would be log⁡\log of about −0.69-0.69⁠.
    Observations (1), (3), (4) on the log function, Section 5.4. Basics: exponentials and logarithms

  2. Name the inner and outer functions, and differentiate each

    Inner: t=log⁡xt = \log x⁠. Outer: y=log⁡ty = \log t⁠.

    dtdx=1x,dydt=1t.\frac{dt}{dx} = \frac{1}{x}, \qquad \frac{dy}{dt} = \frac{1}{t}.

    The first needs x>0x > 0⁠; the second needs t>0t > 0⁠, which holds since x>1x > 1⁠.

    Both links are logarithms, so both derivatives come from Theorem 5 (2); the outer one is written in the letter tt because its input is tt⁠.
    Theorem 5 (2), Section 5.4. Lesson: exponential and logarithmic functions

  3. Multiply (chain rule) and substitute back

    dydx=1t⋅1x=1xlog⁡x,x>1.\frac{dy}{dx} = \frac{1}{t}\cdot\frac{1}{x} = \frac{1}{x\log x}, \qquad x > 1.
    The outer rate 1t\frac1t is evaluated at the inner value t=log⁡xt = \log x⁠. For x>1x > 1 both xx and log⁡x\log x are positive, so the derivative is positive: log⁡(log⁡x)\log(\log x) increases, which makes sense because it is an increasing function of an increasing function.
    Chain rule, Theorem 4, Section 5.3.1. Lesson: the chain rule
Result ddxlog⁡(log⁡x)=1xlog⁡x,x>1.\frac{d}{dx}\log(\log x) = \frac{1}{x\log x}, \qquad x > 1.

(Textbook answer: 1xlog⁡x\dfrac{1}{x\log x}⁠, x>1x > 1⁠. The same.)

Check it with numbers

Take x=ex = e⁠: log⁡e=1\log e = 1⁠, so the formula gives 1e⋅1=1e≈0.3678794\dfrac{1}{e\cdot 1} = \dfrac1e \approx 0.3678794⁠. A difference quotient of log⁡(log⁡x)\log(\log x) with h=0.0001h = 0.0001 on each side of ee gives 0.367880.36788⁠. They agree.

Common mistakes
  1. Answering 1log⁡x\frac{1}{\log x}⁠. That is only the outer rate; the chain rule also needs the inner rate 1x\frac1x⁠.
  2. Writing 1log⁡x2\frac{1}{\log x^2} or 1(log⁡x)2\frac{1}{(\log x)^2}⁠. The product is 1x⋅1log⁡x=1xlog⁡x\frac{1}{x}\cdot\frac{1}{\log x} = \frac{1}{x\log x}⁠: xx times log⁡x\log x⁠.
  3. Taking the domain to be x>0x > 0⁠. For 0<x≤10 < x \le 1 the outer logarithm is not defined.
What to take away
  1. ddxlog⁡g(x)=g′(x)g(x)\dfrac{d}{dx}\log g(x) = \dfrac{g'(x)}{g(x)}⁠, valid where g(x)>0g(x) > 0⁠.
  2. Each logarithm in a chain adds its own "input must be positive" condition.

Every fact used, and where it comes from

FactUsed inSource
log⁡\log defined on R+\R^+⁠, increasing, log⁡1=0\log 1 = 0Step 1Observations (1), (3), (4) on the log function, Section 5.4
ddxlog⁡x=1x\frac{d}{dx}\log x = \frac1x⁠, x>0x > 0Step 2Theorem 5 (2), Section 5.4 (Lesson: exponential and logarithmic functions)
Chain ruleStep 3Theorem 4, Section 5.3.1
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