Chapter 5 · Continuity and Differentiability

Exercise 5.5 · Q4: Differentiate $x^x-2^{\sin x}$ ⁠, $x>0$

Exercise 5.5Logarithmic differentiation · Section 5.5

The question

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

x^x - 2^{\sin x}

Exam answer

Let $u = x^x$ and $v = 2^{\sin x}$ ⁠, so $y = u - v$ ⁠, where $x > 0$ ⁠.

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

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

\therefore\ \frac{du}{dx} = x^x(1 + \log x)

Taking log on both sides of $v = 2^{\sin x}$ ⁠, $\log v = \sin x\log 2$ ⁠. Differentiating w.r.t. $x$ ⁠,

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

\therefore\ \frac{dv}{dx} = 2^{\sin x}\cos x\log 2

$\therefore\ \dfrac{dy}{dx} = \dfrac{du}{dx} - \dfrac{dv}{dx}$ $= x^x(1 + \log x)$ $-\ 2^{\sin x}\cos x\log 2$ $(x > 0)$ ⁠.

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

The one idea this problem is built on

The function is a difference of two terms, and the logarithm of a difference does not split into anything simpler (and the difference can be negative). So we do not take the log of the whole function. Instead we use the difference rule, $(u - v)' = u' - v'$ ⁠, and find each derivative separately by logarithmic differentiation: $u = x^x$ has a variable base and exponent; $v = 2^{\sin x}$ has a constant base and a variable exponent.

Given, to find, and the plan

Given: $y = x^x - 2^{\sin x}$ ⁠.
To find: $\frac{dy}{dx}$ ⁠, and where it holds.
Plan: (1) Find the domain. (2) Write $y = u - v$ ⁠. (3) Find $u'$ by taking $\log u$ ⁠. (4) Find $v'$ by taking $\log v$ ⁠. (5) Subtract.

Basics you need

1Variable powers need a positive base

For $u > 0$ ⁠, $u^v = e^{v\log u} > 0$ and $\log(u^v) = v\log u$ (Example 25, Section 5.4). Basics: exponentials and logarithms

2Sum and difference rule; product rule

$(u \pm v)' = u' \pm v'$ and $(uv)' = u'v + uv'$ (Section 5.3). Basics: derivative rules

3 $\frac{d}{dx}\log y = \frac1y\frac{dy}{dx}$

The chain rule with outer $\log$ and inner $y$ ⁠, for $y > 0$ (Theorems 4 and 5). Basics: composite functions

Solution

Domain: $x > 0$

Let $y = x^x - 2^{\sin x}$ ⁠. The term $x^x$ has a variable exponent, so its base must be positive: $x > 0$ ⁠. The term $2^{\sin x}$ has the positive base $2$ ⁠, so it is defined for every real $x$ ⁠. Together, the function is defined for $x > 0$ ⁠.

$y$ itself can be negative: at $x = 1$ ⁠, $y = 1 - 2^{\sin 1}$ $= 1 - 1.7918762 = -0.7918762$ ⁠. So we could not take $\log y$ here even if it helped, and it would not help, because $\log(u - v)$ has no rule for splitting. This is why each term is handled on its own.

$x^x = e^{x\log x}$ needs $x > 0$ (Basic 1). Basics: exponentials and logarithms
Split with the difference rule

Put $u = x^x$ and $v = 2^{\sin x}$ ⁠, so $y = u - v$ and
$\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}.$
The derivative of a difference is the difference of the derivatives (rule (1) of the algebra of derivatives). Now each piece is a single power, and a single positive power is exactly what logarithmic differentiation handles. Both $u$ and $v$ are positive for $x > 0$ ⁠.

$(u - v)' = u' - v'$ (Section 5.3). Basics: derivative rules
$u = x^x$ ⁠: take logs and differentiate

For $x > 0$ ⁠, $u > 0$ and $\log u = x\log x$ ⁠. Differentiate; the right side is a product ( $x$ times $\log x$ ⁠):
$\begin{aligned} \frac1u\cdot\frac{du}{dx} &= 1\cdot\log x + x\cdot\frac1x \\ &= \log x + 1. \end{aligned}$
So
$\frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x).$
Neither standard rule fits $x^x$ ⁠: the power rule $nx^{n-1}$ needs a fixed exponent and the $a^x$ rule needs a fixed base. Taking logs turns the power into the product $x\log x$ ⁠, and the product rule finishes it.

Product rule; $\frac{d}{dx}\log x = \frac1x$ (Theorem 5). Lesson: logarithmic differentiation
$v = 2^{\sin x}$ ⁠: take logs and differentiate

$v > 0$ and $\log v = \sin x\cdot\log 2$ ⁠, where $\log 2$ is a constant. Differentiate:
$\frac1v\cdot\frac{dv}{dx} = \cos x\cdot\log 2,$
so
$\frac{dv}{dx} = 2^{\sin x}\cos x\log 2.$
The same result comes from Example 28 ( $\frac{d}{dt}2^t = 2^t\log 2$ ⁠) with the chain rule, outer $2^t$ ⁠, inner $t = \sin x$ whose derivative is $\cos x$ ⁠. The factor $\log 2$ is a constant (about $0.693$ ⁠), not something to differentiate.

$\frac{d}{dx}\sin x = \cos x$ (Table 5.3); $\frac{d}{dx}a^x = a^x\log a$ (Example 28). Basics: exponentials and logarithms
Subtract

For $x > 0$ ⁠,
$\frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x}\cos x\log 2.$
Step 2 says the answer is $u' - v'$ ⁠; Steps 3 and 4 supplied them. This is exactly the textbook's answer.

Difference rule (Section 5.3). Basics: derivative rules

Result

For $x > 0$ ⁠,

\frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x}\cos x\log 2.

(Textbook answer: $x^x(1 + \log x)$ $-\ 2^{\sin x}\cos x\log 2$ ⁠; the key does not state $x > 0$ ⁠.)

Check it with numbers

At $x = 1$ ⁠: $1^1(1 + \log 1) = 1$ ⁠, and $2^{\sin 1}\cos 1\log 2$ $= 1.7918762\times 0.5403023\times 0.6931472$ $= 0.6710738$ ⁠. So $\frac{dy}{dx} = 1 - 0.6710738 = 0.3289262$ ⁠. The central difference quotient with $h = 0.0001$ gives $0.3289262$ ⁠.

At $x = 2$ the formula gives $7.3143374$ and the difference quotient $7.3143374$ ⁠.

Common mistakes

Taking $\log$ of the whole function and writing $\log(x^x - 2^{\sin x})$ as $x\log x - \sin x\log 2$ ⁠. The log of a difference is not the difference of the logs.
$\frac{d}{dx}x^x = x\cdot x^{x-1}$ (power rule) or $x^x\log x$ (exponential rule). Each misses one part; the correct $x^x(1 + \log x)$ is their sum.
Forgetting the chain-rule factor $\cos x$ in $\frac{d}{dx}2^{\sin x}$ ⁠.

What to take away

For sums and differences, split first and log-differentiate each power separately.
$\frac{d}{dx}x^x = x^x(1 + \log x)$ for $x > 0$ ⁠: worth remembering.
$\frac{d}{dx}a^{g(x)} = a^{g(x)}\log a\cdot g'(x)$ for a positive constant $a$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Positive base for logarithmic differentiation	Steps 1, 3, 4	Section 5.5 (Lesson: logarithmic differentiation)
$x = e^{\log x}$ for $x > 0$	Step 1	Example 25
Sum, difference and product rules	Steps 2, 3, 5	Algebra of derivatives (1), (2), Section 5.3
$\log p^n = n\log p$	Steps 3, 4	Section 5.4
$\frac{d}{dx}\log x = \frac1x$ ⁠; chain rule	Steps 3, 4	Theorem 5, Section 5.4; Theorem 4, Section 5.3.1
$\frac{d}{dx}\sin x = \cos x$	Step 4	Table 5.3
$\frac{d}{dx}a^x = a^x\log a$	Step 4	Example 28

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Exercise 5.5 · Q4: Differentiate xx−2sin⁡xx^x-2^{\sin x}xx−2sinx⁠, x>0x>0x>0

Basics you need

Solution

Domain: x>0x > 0x>0

Split with the difference rule

u=xxu = x^xu=xx⁠: take logs and differentiate

v=2sin⁡xv = 2^{\sin x}v=2sinx⁠: take logs and differentiate

Subtract

Every fact used, and where it comes from

Exercise 5.5 · Q4: Differentiate $x^x-2^{\sin x}$ ⁠, $x>0$

Domain: $x > 0$

$u = x^x$ ⁠: take logs and differentiate

$v = 2^{\sin x}$ ⁠: take logs and differentiate