Chapter 5 · Continuity and Differentiability

Exercise 5.5 · Q7: Differentiate $(\log x)^x+x^{\log x}$ ⁠, $x>1$

Exercise 5.5Logarithmic differentiation · Section 5.5

The question

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

(\log x)^x + x^{\log x}

Exam answer

Let $u = (\log x)^x$ and $v = x^{\log x}$ ⁠, so $y = u + v$ ⁠, where $x > 1$ ⁠.

Taking log on both sides, $\log u = x\log(\log x)$ ⁠. Differentiating w.r.t. $x$ ⁠,

\begin{aligned} \frac1u\cdot\frac{du}{dx} &= 1\cdot\log(\log x) \\ &\quad + x\cdot\frac{1}{\log x}\cdot\frac1x \\ &= \log(\log x) + \frac{1}{\log x} \end{aligned}

\begin{aligned} \therefore\ \frac{du}{dx} &= (\log x)^x \\ &\quad\times\bigg[\log(\log x) + \frac{1}{\log x}\bigg] \\ &= (\log x)^{x - 1} \\ &\quad\times\big[1 + \log x\cdot\log(\log x)\big] \end{aligned}

Also $\log v = \log x\cdot\log x = (\log x)^2$ ⁠. Differentiating w.r.t. $x$ ⁠,

\frac1v\cdot\frac{dv}{dx} = 2\log x\cdot\frac1x

\begin{aligned} \therefore\ \frac{dv}{dx} &= x^{\log x}\cdot\frac{2\log x}{x} \\ &= 2x^{\log x - 1}\log x \end{aligned}

$\begin{aligned} \therefore\ \dfrac{dy}{dx} &= (\log x)^{x - 1} \\ &\quad\times\big[1 + \log x\cdot\log(\log x)\big] \\ &\quad + 2x^{\log x - 1}\log x \end{aligned}$

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

The one idea this problem is built on

Two variable powers are added, so we differentiate them separately (sum rule) and take logarithms of each one. For $(\log x)^x$ ⁠, the log brings down the exponent $x$ ⁠; for $x^{\log x}$ ⁠, the log gives $\log x\cdot\log x = (\log x)^2$ ⁠, which is easy. The base $\log x$ must be positive, so everything happens for $x > 1$ ⁠.

Given, to find, and the plan

Given: $y = u + v$ ⁠, with $u = (\log x)^x$ ⁠, $v = x^{\log x}$ ⁠.
To find: $\frac{dy}{dx}$ ⁠, and where it holds.
Plan: (1) Domain. (2) $u'$ from $\log u = x\log(\log x)$ ⁠. (3) $v'$ from $\log v = (\log x)^2$ ⁠. (4) Add and match the textbook's form.

Basics you need

1Variable powers need a positive base

For $p > 0$ ⁠: $p^q = e^{q\log p} > 0$ and $\log p^q = q\log p$ (Example 25, Section 5.4). Also $\log x > 0 \iff x > 1$ ⁠. Basics: exponentials and logarithms

2Sum rule and product rule

$(u + v)' = u' + v'$ ⁠, $(fg)' = f'g + fg'$ (Section 5.3). Basics: derivative rules

3Chain rule with logs

$\frac{d}{dx}\log(\log x) = \frac{1}{x\log x}$ for $x > 1$ ⁠; $\frac{d}{dx}(\log x)^2 = \frac{2\log x}{x}$ (Theorems 4 and 5). Basics: composite functions

Solution

Domain: $x > 1$

$v = x^{\log x}$ needs $x > 0$ (for $\log x$ and for the positive base $x$ ⁠). $u = (\log x)^x$ has a variable exponent, so its base must be positive: $\log x > 0$ ⁠, that is, $x > 1$ ⁠. So $y$ is defined, and both terms are positive, for $x > 1$ ⁠.

For $0 < x < 1$ ⁠, $\log x < 0$ and, for example, $\left(\log\frac12\right)^{1/2}$ is the square root of a negative number. At $x = 1$ ⁠, $u = 0^1 = 0$ ⁠, and $\log u$ does not exist. So logarithmic differentiation of $u$ works for $x > 1$ only; the answer key does not say so.

$\log$ is increasing with $\log 1 = 0$ (Section 5.4). Basics: exponentials and logarithms
Split with the sum rule

$\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ⁠.

The log of a sum does not split, so we never take $\log y$ here. Each term separately is a positive power, ready for logarithmic differentiation.

$(u + v)' = u' + v'$ (Section 5.3). Basics: derivative rules
$u = (\log x)^x$

$\log u = x\log(\log x)$ ⁠. Product rule, with $\frac{d}{dx}\log(\log x) = \frac{1}{\log x}\cdot\frac1x$ (chain rule, inner $\log x$ ⁠):
$\begin{aligned} \frac1u\cdot\frac{du}{dx} &= 1\cdot\log(\log x) + x\cdot\frac{1}{x\log x} \\ &= \log(\log x) + \frac{1}{\log x}. \end{aligned}$
So
$\frac{du}{dx} = (\log x)^x\left[\log(\log x) + \frac{1}{\log x}\right].$
The $x$ in front cancels the $x$ in $\frac{1}{x\log x}$ ⁠. Both terms are defined because $\log x > 0$ ⁠.

Product rule; chain rule with $\frac{d}{dx}\log x = \frac1x$ (Theorem 5). Lesson: logarithmic differentiation
Rewrite $u'$ in the textbook's form

Take $\frac{1}{\log x}$ out of the bracket:
$\begin{aligned} \frac{du}{dx} &= \frac{(\log x)^x}{\log x} \\ &\quad\times\left[\log x\cdot\log(\log x) + 1\right] \\ &= (\log x)^{x-1} \\ &\quad\times\left[1 + \log x\cdot\log(\log x)\right]. \end{aligned}$
$\log(\log x) = \frac{1}{\log x}\cdot\log x\log(\log x)$ ⁠, so the bracket is unchanged; and $(\log x)^x\cdot(\log x)^{-1} = (\log x)^{x-1}$ by the index law (the base is positive).

Index law $p^m\cdot p^{-1} = p^{m-1}$ for $p > 0$ ⁠. Basics: exponentials and logarithms
$v = x^{\log x}$

For $x > 0$ ⁠, $\log v = \log x\cdot\log x = (\log x)^2$ ⁠. Chain rule, outer $t^2$ ⁠, inner $t = \log x$ ⁠:
$\frac1v\cdot\frac{dv}{dx} = 2\log x\cdot\frac1x.$
So
$\begin{aligned} \frac{dv}{dx} &= x^{\log x}\cdot\frac{2\log x}{x} \\ &= 2x^{\log x - 1}\log x. \end{aligned}$
The exponent $\log x$ comes down as a factor in front of $\log x$ ⁠, giving $(\log x)^2$ ⁠. Dividing $x^{\log x}$ by $x$ lowers the exponent by $1$ ⁠. This term alone is valid for all $x > 0$ ⁠.

$\log p^q = q\log p$ ⁠; chain rule (Theorem 4). Lesson: the chain rule
Add

For $x > 1$ ⁠,
$\begin{aligned} \frac{dy}{dx} = \ &(\log x)^{x-1}\left[1 + \log x\cdot\log(\log x)\right] \\ &+ 2x^{\log x - 1}\cdot\log x. \end{aligned}$
This is $u' + v'$ from Steps 4 and 5. It matches the textbook's answer, where " $2x^{\log x - 1}\,.\,\log x$ " means $2\cdot x^{\log x - 1}\cdot\log x$ ⁠.

Sum rule (Section 5.3). Basics: derivative rules

Result

For $x > 1$ ⁠,

\begin{aligned} \frac{dy}{dx} = \ &(\log x)^{x-1}\left[1 + \log x\cdot\log(\log x)\right] \\ &+ 2x^{\log x - 1}\cdot\log x. \end{aligned}

(Textbook answer: the same; the key does not state $x > 1$ ⁠.)

Check it with numbers

At $x = e$ ⁠: $\log e = 1$ and $\log(\log e) = 0$ ⁠, so $u' = 1^{e-1}[1 + 0] = 1$ and $v' = 2e^{0}\cdot 1 = 2$ ⁠. Total $3$ ⁠; the central difference quotient with $h = 0.0001$ gives $3.0000000$ ⁠.

At $x = 3$ ⁠: $\log 3 = 1.0986123$ ⁠, $\log(\log 3) = 0.0940478$ ⁠. Then $u' = 1.2069490\times 1.1033221$ $= 1.3316535$ and $v' = 2\times 1.1144229\times 1.0986123$ $= 2.4486373$ ⁠. Total $3.7802908$ ⁠; the difference quotient gives $3.7802908$ ⁠.

Common mistakes

Taking $\log$ of the sum. Split first.
Writing $\log v = \log x\cdot x$ for $v = x^{\log x}$ ⁠: the exponent is $\log x$ ⁠, so $\log v = \log x\cdot\log x$ ⁠.
Missing the inner derivative $\frac1x$ in $\frac{d}{dx}\log(\log x)$ ⁠.
Domain: $x > 0$ is not enough; the base $\log x$ needs $x > 1$ ⁠.

What to take away

$\frac{d}{dx}x^{\log x} = 2x^{\log x - 1}\log x$ ⁠, because $\log(x^{\log x}) = (\log x)^2$ ⁠.
Pulling a factor out of a bracket and using index laws turns one correct form into another; check equal forms numerically.
The domain of a power with variable exponent is where its base is positive.

Every fact used, and where it comes from

Fact	Used in	Source
Positive base for logarithmic differentiation	Step 1	Section 5.5 (Lesson: logarithmic differentiation)
$\log$ increasing, $\log 1 = 0$	Step 1	Section 5.4
Sum rule, product rule	Steps 2, 3, 6	Algebra of derivatives (1), (2), Section 5.3
$\log p^n = n\log p$	Steps 3, 5	Section 5.4
$\frac{d}{dx}\log x = \frac1x$ ⁠; chain rule	Steps 3, 5	Theorem 5, Section 5.4; Theorem 4, Section 5.3.1
Index laws for a positive base	Steps 4, 5	Exponents (earlier classes)

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Exercise 5.5 · Q7: Differentiate (log⁡x)x+xlog⁡x(\log x)^x+x^{\log x}(logx)x+xlogx⁠, x>1x>1x>1

Basics you need

Solution

Domain: x>1x > 1x>1

Split with the sum rule

u=(log⁡x)xu = (\log x)^xu=(logx)x

Rewrite u′u'u′ in the textbook's form

v=xlog⁡xv = x^{\log x}v=xlogx

Add

Every fact used, and where it comes from

Exercise 5.5 · Q7: Differentiate $(\log x)^x+x^{\log x}$ ⁠, $x>1$

Domain: $x > 1$

$u = (\log x)^x$

Rewrite $u'$ in the textbook's form

$v = x^{\log x}$