Chapter 5 · Continuity and Differentiability

Miscellaneous · Q7: Differentiate $(\log x)^{\log x}$ for $x>1$

Miscellaneous ExerciseLogarithmic differentiation · Miscellaneous Exercise

The question

Differentiate w.r.t. $x$ the function

(\log x)^{\log x}, \quad x > 1.

Exam answer

Let $y = (\log x)^{\log x}$ ⁠, $x \gt 1$ ⁠, so $\log x \gt 0$ ⁠.

Taking log on both sides,

\log y = \log x\cdot\log(\log x)

Differentiating both sides w.r.t. $x$ (product rule),

\begin{aligned} \frac1y\frac{dy}{dx} &= \frac1x\log(\log x) \\ &\quad + \log x\cdot\frac{1}{\log x}\cdot\frac1x \\ &= \frac1x + \frac{\log(\log x)}{x} \end{aligned}

$\therefore\ \dfrac{dy}{dx} = (\log x)^{\log x}\Big[\dfrac1x + \dfrac{\log(\log x)}{x}\Big]$ $(x \gt 1)$

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

The one idea this problem is built on

The same expression $\log x$ is both the base and the exponent, so neither the power rule nor the rule for $a^x$ applies. Take logarithms: $\log y = \log x\cdot\log(\log x)$ ⁠, a product. The condition $x > 1$ is exactly what makes the base $\log x$ positive, so that its logarithm exists.

Given, to find, and the plan

Given: $y = (\log x)^{\log x}$ for $x > 1$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: (1) Check the base is positive. (2) Take logarithms. (3) Differentiate both sides (chain rule on the left, product and chain rules on the right). (4) Multiply by $y$ ⁠.

Basics you need

1The log function

$\log$ (base $e$ ⁠) is defined for positive numbers, is increasing, and $\log 1 = 0$ (Section 5.4). So $\log x > 0$ exactly when $x > 1$ ⁠. Basics: exponentials and logarithms

2 $\log p^n = n\log p$

For $p > 0$ and any real $n$ (Section 5.4).

3Derivatives of $\log$

$\frac{d}{dx}\log x = \frac1x$ (Theorem 5). By the chain rule, $\frac{d}{dx}\log(\log x) = \frac{1}{\log x}\cdot\frac1x$ (Exercise 5.4, Q8). Basics: derivative rules

4Logarithmic differentiation

For $y = [u(x)]^{v(x)}$ with $u(x) > 0$ ⁠: take logs, differentiate, multiply by $y$ (Section 5.5). Lesson: logarithmic differentiation

Solution

The base is positive

Let $y = (\log x)^{\log x}$ ⁠. Since $\log$ is increasing and $\log 1 = 0$ ⁠,
$x > 1 \implies \log x > 0.$
So the base is a positive number, $y$ is a positive real number, and $\log y$ and $\log(\log x)$ are both defined.

A variable real power needs a positive base, and logarithmic differentiation takes the logarithm of the base. For $0 < x \le 1$ the base $\log x$ would be zero or negative, which is why the question insists on $x > 1$ ⁠.

Properties (1), (3), (4) of the log function, Section 5.4. Basics: exponentials and logarithms
Take logarithms
$\begin{aligned} \log y &= \log\left[(\log x)^{\log x}\right] \\ &= \log x\cdot\log(\log x). \end{aligned}$
The rule $\log p^n = n\log p$ with $p = \log x$ and $n = \log x$ brings the exponent down. Now we have a product of $\log x$ and $\log(\log x)$ ⁠.

$\log_b p^n = n\log_b p$ ⁠, Section 5.4.
Differentiate the left side
$\frac{d}{dx}(\log y) = \frac1y\cdot\frac{dy}{dx}.$
Outer function $\log$ ⁠, inner function $y$ ⁠. The chain rule gives $\frac1y$ times the inner derivative $\frac{dy}{dx}$ ⁠.

Chain rule, Theorem 4, Section 5.3.1; Theorem 5. Lesson: chain rule
Differentiate the right side by the product rule

First factor $\log x$ ⁠, with derivative $\frac1x$ ⁠. Second factor $\log(\log x)$ ⁠: outer $\log$ ⁠, inner $\log x$ ⁠, so its derivative is $\frac{1}{\log x}\cdot\frac1x$ ⁠. The product rule gives
$\begin{aligned} &\frac{d}{dx}\left[\log x\cdot\log(\log x)\right] \\ &= \frac1x\log(\log x) + \log x\cdot\frac{1}{x\log x} \\ &= \frac{\log(\log x)}{x} + \frac1x. \end{aligned}$
In the second term $\log x$ cancels: it is not zero because $x > 1$ ⁠. This cancellation is what produces the neat " $+\frac1x$ ".

Product rule (2), Section 5.3; $\frac{d}{dx}\log x = \frac1x$ ⁠, Theorem 5. Basics: derivative rules
Multiply by $y$

Equating Steps 3 and 4 and multiplying by $y = (\log x)^{\log x}$ ⁠:
$\begin{aligned} \frac{dy}{dx} = {}&(\log x)^{\log x} \\ &\times\left[\frac{1}{x} + \frac{\log(\log x)}{x}\right]. \end{aligned}$
Equivalently, $\dfrac{dy}{dx} = \dfrac{(\log x)^{\log x}\left(1 + \log(\log x)\right)}{x}$ ⁠.

The answer must be in terms of $x$ ⁠, so we substitute $y$ back. The second form just takes out the common factor $\frac1x$ ⁠.

Logarithmic differentiation, Section 5.5 (compare Example 29). Lesson: logarithmic differentiation

Result

For $x > 1$ ⁠,

\begin{aligned} \frac{dy}{dx} = {}&(\log x)^{\log x} \\ &\times\left[\frac{1}{x} + \frac{\log(\log x)}{x}\right]. \end{aligned}

(Textbook answer: $(\log x)^{\log x}\left[\frac1x + \frac{\log(\log x)}{x}\right]$ ⁠, $x > 1$ ⁠, the same.)

Check it with numbers

Take $x = e^2 \approx 7.3890561$ ⁠, so $\log x = 2$ and $y = 2^2 = 4$ ⁠. With $\log 2 \approx 0.6931472$ ⁠, the formula gives

\frac{4(1 + 0.6931472)}{7.3890561} \approx 0.9165702.

Difference quotient with $h = 0.0001$ ⁠: $y(e^2 + h) \approx 4.00009166$ and $y(e^2 - h) \approx 3.99990834$ ⁠, so the quotient is about $0.9165702$ ⁠. ✓

Common mistakes

Using the power rule: $\log x\cdot(\log x)^{\log x - 1}\cdot\frac1x$ treats the exponent as a constant and misses the $\log(\log x)$ term.
Differentiating $\log(\log x)$ as $\frac{1}{\log x}$ , forgetting the inner derivative $\frac1x$ ⁠.
Ignoring $x > 1$ ⁠. Without it the base $\log x$ can be zero or negative and the logarithm in Step 2 does not exist.

What to take away

For $y = u^v$ with both varying: $\frac{dy}{dx} = u^v\left[v'\log u + \frac{v\,u'}{u}\right]$ ⁠, valid where $u > 0$ ⁠.
The domain condition in the question usually tells you exactly where the base is positive.

Every fact used, and where it comes from

Fact	Used in	Source
$\log$ is increasing, $\log 1 = 0$ ⁠, defined on positive numbers	Step 1	Section 5.4, properties of the log function
$\log p^n = n\log p$	Step 2	Section 5.4 (stated there for every real $n$ ⁠, without proof)
$\frac{d}{dx}\log x = \frac1x$	Steps 3, 4	Theorem 5, Section 5.4
Chain rule; product rule	Steps 3, 4	Theorem 4, Section 5.3.1; rule (2), Section 5.3 (Lesson: chain rule)
Logarithmic differentiation	Steps 2–5	Section 5.5, Example 29 (Lesson: logarithmic differentiation)
Answer	Result	Answers, Miscellaneous Exercise on Chapter 5, Q7
Numerical values	Check	Recomputed in the site's verification script

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Miscellaneous · Q7: Differentiate (log⁡x)log⁡x(\log x)^{\log x}(logx)logx for x>1x>1x>1

Basics you need

Solution

The base is positive

Take logarithms

Differentiate the left side

Differentiate the right side by the product rule

Multiply by yyy

Every fact used, and where it comes from

Miscellaneous · Q7: Differentiate $(\log x)^{\log x}$ for $x>1$

Multiply by $y$