Chapter 5 · Continuity and Differentiability

Exercise 5.5 · Q9: Differentiate $x^{\sin x}+(\sin x)^{\cos x}$

Exercise 5.5Logarithmic differentiation · Section 5.5

The question

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

x^{\sin x} + (\sin x)^{\cos x}

Exam answer

Let $y = u + v$ ⁠, where $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$ ⁠. Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ⁠.

Taking log on both sides of $u = x^{\sin x}$ ⁠,

\log u = \sin x\log x

Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} \frac1u\frac{du}{dx} &= \cos x\log x + \frac{\sin x}{x} \\ \therefore\ \frac{du}{dx} &= x^{\sin x}\bigg[\frac{\sin x}{x} \\ &\qquad + \cos x\log x\bigg] \end{aligned}

Taking log on both sides of $v = (\sin x)^{\cos x}$ ⁠,

\log v = \cos x\log\sin x

Differentiating both sides w.r.t. $x$ ⁠,

\begin{aligned} \frac1v\frac{dv}{dx} &= -\sin x\log\sin x \\ &\qquad + \cos x\cdot\frac{\cos x}{\sin x} \\ &= \cos x\cot x \\ &\qquad - \sin x\log\sin x \\ \therefore\ \frac{dv}{dx} &= (\sin x)^{\cos x} \\ &\quad\times\big[\cos x\cot x \\ &\qquad - \sin x\log\sin x\big] \end{aligned}

Hence, for $x > 0$ ⁠, $\sin x > 0$ ⁠, $\begin{aligned} \dfrac{dy}{dx} &= x^{\sin x}\bigg[\dfrac{\sin x}{x} + \cos x\log x\bigg] \\ &\quad + (\sin x)^{\cos x}\big[\cos x\cot x \\ &\qquad - \sin x\log\sin x\big] \end{aligned}$

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

The one idea this problem is built on

Both terms are powers with a variable base and a variable exponent, added together. The log of a sum does not split, so we use the sum rule and log-differentiate each term on its own. The first term is Example 29. Each base must be positive: $x > 0$ for the first and $\sin x > 0$ for the second, so the function lives on the intervals $(2n\pi, (2n+1)\pi)$ ⁠, $n = 0, 1, 2, \ldots$

Given, to find, and the plan

Given: $y = u + v$ ⁠, with $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$ ⁠.
To find: $\frac{dy}{dx}$ ⁠, and where it holds.
Plan: (1) Domain: both bases positive. (2) $u'$ via $\log u = \sin x\log x$ ⁠. (3) $v'$ via $\log v = \cos x\log\sin x$ ⁠. (4) Add.

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). Basics: exponentials and logarithms

2Where $\sin x > 0$

$\sin x > 0$ exactly when $2n\pi < x < (2n+1)\pi$ for an integer $n$ (the upper half of the unit circle, repeated every $2\pi$ ⁠). Basics: trigonometric identities

3Sum rule, product rule, chain rule

$(u + v)' = u' + v'$ ⁠, $(fg)' = f'g + fg'$ (Section 5.3); $\frac{d}{dx}\log\sin x = \frac{\cos x}{\sin x} = \cot x$ for $\sin x > 0$ (Theorems 4 and 5). Basics: derivative rules

Solution

Domain: $x > 0$ and $\sin x > 0$

$u = x^{\sin x}$ needs its base positive: $x > 0$ ⁠. $v = (\sin x)^{\cos x}$ needs $\sin x > 0$ ⁠. Both together:
$x \in (2n\pi,\ (2n+1)\pi), \quad n = 0, 1, 2, \ldots$

The domain (green) for $0 < x < 5\pi$ ⁠: the intervals $(0, \pi)$ ⁠, $(2\pi, 3\pi)$ ⁠, $(4\pi, 5\pi)$ ⁠, and so on to the right (34 px per unit). The endpoints are excluded: $0$ because the base $x$ must be positive, the multiples of $\pi$ because $\sin x = 0$ there.

The book requires the base of each power to be positive for logarithmic differentiation (Section 5.5). Negative $x$ are excluded by the first term even where $\sin x > 0$ ⁠; the gaps $[(2n-1)\pi, 2n\pi]$ are excluded by the second. For example, at $x = \frac{3\pi}{2}$ ⁠, $v = (-1)^0 = 1$ happens to be defined, but at $x = \frac{5\pi}{3}$ ⁠, where $\sin x = -\frac{\sqrt3}{2}$ and $\cos x = \frac12$ ⁠, $v$ would be the square root of a negative number, which is not real. The answer key does not state the domain.

Sign of $\sin x$ on each quarter-turn. Basics: trigonometric identities
Split with the sum rule

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

We cannot take $\log(u + v)$ apart, so each term is handled separately; each is a positive power on the domain.

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

$\log u = \sin x\cdot\log x$ ⁠. Product rule:
$\frac1u\cdot\frac{du}{dx} = \cos x\cdot\log x + \sin x\cdot\frac1x,$
so
$\frac{du}{dx} = x^{\sin x}\left[\frac{\sin x}{x} + \cos x\log x\right].$
This is exactly Example 29 (valid for all $x > 0$ ⁠). The two terms in the bracket come from differentiating $\log x$ (giving $\frac1x$ ⁠) and $\sin x$ (giving $\cos x$ ⁠) in turn.

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

$\log v = \cos x\cdot\log\sin x$ ⁠. Product rule, with $\frac{d}{dx}\cos x = -\sin x$ and $\frac{d}{dx}\log\sin x = \frac{\cos x}{\sin x}$ (chain rule, inner $\sin x > 0$ ⁠):
$\begin{aligned} \frac1v\cdot\frac{dv}{dx} &= -\sin x\cdot\log\sin x + \cos x\cdot\frac{\cos x}{\sin x} \\ &= \cos x\cot x - \sin x\log\sin x. \end{aligned}$
So
$\begin{aligned} \frac{dv}{dx} = \ &(\sin x)^{\cos x} \\ &\times\left[\cos x\cot x - \sin x\log\sin x\right]. \end{aligned}$
$\cos x\cdot\frac{\cos x}{\sin x} = \cos x\cot x$ (it could also be written $\frac{\cos^2 x}{\sin x}$ ⁠). The terms are ordered as in the textbook's answer. Multiplying by $v$ (positive) isolates $\frac{dv}{dx}$ ⁠.

Chain rule (Theorem 4); $\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$ (Table 5.3). Lesson: the chain rule
Add

For $x > 0$ with $\sin x > 0$ ⁠:
$\begin{aligned} \frac{dy}{dx} = \ &x^{\sin x}\left[\frac{\sin x}{x} + \cos x\log x\right] \\ &+ (\sin x)^{\cos x} \\ &\quad\times\left[\cos x\cot x - \sin x\log\sin x\right]. \end{aligned}$
This is $u' + v'$ ⁠, and it is exactly the textbook's answer.

Sum rule (Section 5.3). Basics: derivative rules

Result

For $x \in (2n\pi, (2n+1)\pi)$ ⁠, $n = 0, 1, 2, \ldots$ ⁠,

\begin{aligned} \frac{dy}{dx} = \ &x^{\sin x}\left[\frac{\sin x}{x} + \cos x\log x\right] \\ &+ (\sin x)^{\cos x} \\ &\quad\times\left[\cos x\cot x - \sin x\log\sin x\right]. \end{aligned}

(Textbook answer: the same expression; the key does not state the domain.)

Check it with numbers

At $x = \frac{\pi}{2}$ ⁠: $\sin x = 1$ ⁠, $\cos x = 0$ ⁠. Then $u' = \frac{\pi}{2}\left[\frac{1}{\pi/2} + 0\right] = 1$ and $v' = 1^0\,[0 - 1\cdot\log 1] = 0$ ⁠. Total $1$ ⁠; the central difference quotient with $h = 0.0001$ gives $1.0000000$ ⁠.

At $x = 1$ ⁠: $\sin 1 = 0.8414710$ ⁠, $\cos 1 = 0.5403023$ ⁠. $u' = 1\cdot[0.8414710 + 0] = 0.8414710$ ⁠; $v = 0.9109583$ and its bracket is $0.4921652$ ⁠, so $v' = 0.4483419$ ⁠. Total $1.2898129$ ⁠; the difference quotient gives $1.2898129$ ⁠. In the second piece of the domain, at $x = 7$ (between $2\pi \approx 6.283$ and $3\pi \approx 9.425$ ⁠), the formula gives $6.4365068$ and the difference quotient $6.4365068$ ⁠.

Common mistakes

Taking $\log$ of the sum instead of splitting.
$\frac{d}{dx}\log\sin x = \frac{1}{\sin x}$ : the inner derivative $\cos x$ is missing; it is $\cot x$ ⁠.
Sign of $\sin x\log\sin x$ : it comes from $\frac{d}{dx}\cos x = -\sin x$ ⁠.
Domain " $x > 0$ " only: the second base also needs $\sin x > 0$ ⁠.

What to take away

For $u^v$ with $u > 0$ ⁠: $\frac{d}{dx}u^v = u^v\left[v'\log u + \frac{v}{u}u'\right]$ ⁠.
Second term: $u = \sin x$ ⁠, $v = \cos x$ ⁠: $v'\log u = -\sin x\log\sin x$ and $\frac{v}{u}u' = \frac{\cos x}{\sin x}\cos x$ ⁠.
The domain of a sum is the intersection of the terms' domains.

Every fact used, and where it comes from

Fact	Used in	Source
Positive bases for logarithmic differentiation	Step 1	Section 5.5 (Lesson: logarithmic differentiation)
$\sin x > 0$ on $(2n\pi, (2n+1)\pi)$	Step 1	Trigonometric functions (Class XI)
Sum rule, product rule	Steps 2, 3, 4, 5	Algebra of derivatives (1), (2), Section 5.3
$\log p^n = n\log p$	Steps 3, 4	Section 5.4
$\frac{d}{dx}x^{\sin x}$	Step 3	Example 29
$\frac{d}{dx}\log x = \frac1x$ ⁠; chain rule	Steps 3, 4	Theorem 5, Section 5.4; Theorem 4, Section 5.3.1
$\frac{d}{dx}\sin x = \cos x$ ⁠, $\frac{d}{dx}\cos x = -\sin x$	Step 4	Table 5.3

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Exercise 5.5 · Q9: Differentiate xsin⁡x+(sin⁡x)cos⁡xx^{\sin x}+(\sin x)^{\cos x}xsinx+(sinx)cosx

Basics you need

Solution

Domain: x>0x > 0x>0 and sin⁡x>0\sin x > 0sinx>0

Split with the sum rule

u=xsin⁡xu = x^{\sin x}u=xsinx

v=(sin⁡x)cos⁡xv = (\sin x)^{\cos x}v=(sinx)cosx

Add

Every fact used, and where it comes from

Exercise 5.5 · Q9: Differentiate $x^{\sin x}+(\sin x)^{\cos x}$

Domain: $x > 0$ and $\sin x > 0$

$u = x^{\sin x}$

$v = (\sin x)^{\cos x}$