Chapter 5 · Continuity and Differentiability

Miscellaneous · Q9: Differentiate $(\sin x-\cos x)^{\sin x-\cos x}$ on $\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$

Miscellaneous ExerciseLogarithmic differentiation · Miscellaneous Exercise

The question

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

(\sin x - \cos x)^{(\sin x - \cos x)},

\frac{\pi}{4} < x < \frac{3\pi}{4}.

Exam answer

Let $u = \sin x - \cos x$ ⁠, so $y = u^u$ ⁠. For $\frac{\pi}{4} \lt x \lt \frac{3\pi}{4}$ ⁠, $u = \sqrt2\sin\big(x - \frac{\pi}{4}\big) \gt 0$ ⁠.

Taking log on both sides, $\log y = u\log u$ ⁠.

Differentiating both sides w.r.t. $x$ ⁠, with $\frac{du}{dx} = \cos x + \sin x$ ⁠,

\begin{aligned} \frac1y\frac{dy}{dx} &= \frac{du}{dx}\log u + u\cdot\frac1u\frac{du}{dx} \\ &= (\cos x + \sin x)(1 + \log u) \end{aligned}

$\therefore\ \dfrac{dy}{dx} = (\sin x - \cos x)^{\sin x - \cos x}$ $\times(\cos x + \sin x)$ $\times\big(1 + \log(\sin x - \cos x)\big)$ $\big(\dfrac{\pi}{4} \lt x \lt \dfrac{3\pi}{4}\big)$

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

The one idea this problem is built on

Write $u = \sin x - \cos x$ ⁠. The function is $y = u^u$ ⁠: the same quantity is base and exponent, so we use logarithmic differentiation, $\log y = u\log u$ ⁠. This needs $u > 0$ ⁠, and the interval $\frac{\pi}{4} < x < \frac{3\pi}{4}$ guarantees it, because $u = \sqrt2\sin\left(x - \frac{\pi}{4}\right)$ ⁠.

Given, to find, and the plan

Given: $y = (\sin x - \cos x)^{\sin x - \cos x}$ for $\frac{\pi}{4} < x < \frac{3\pi}{4}$ ⁠.
To find: $\dfrac{dy}{dx}$ ⁠.
Plan: (1) Show the base $u = \sin x - \cos x$ is positive on the interval. (2) Take logarithms. (3) Differentiate both sides, using $u' = \cos x + \sin x$ ⁠. (4) Multiply by $y$ ⁠.

Basics you need

1Compound-angle formula

$\sin(A - B) = \sin A\cos B - \cos A\sin B$ ⁠, and $\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt2}$ (Class XI). Basics: trigonometric identities

2Logarithms

$\log$ is defined for positive numbers only, and $\log p^n = n\log p$ (Section 5.4). Basics: exponentials and logarithms

3Product and chain rules

$(uv)' = u'v + uv'$ ⁠; $\frac{d}{dx}\log u = \frac{1}{u}\cdot\frac{du}{dx}$ ⁠. Basics: derivative rules

4Logarithmic differentiation

Take logs, differentiate, multiply by $y$ (Section 5.5; Example 41 does $(\sin x)^{\sin x}$ the same way). Lesson: logarithmic differentiation

Solution

The base is positive on $\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$

Let $u = \sin x - \cos x$ ⁠. Using $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠:
$\begin{aligned} \sqrt2\sin\left(x - \tfrac{\pi}{4}\right) &= \sqrt2\left(\tfrac{\sin x}{\sqrt2} - \tfrac{\cos x}{\sqrt2}\right) \\ &= \sin x - \cos x = u. \end{aligned}$
For $\frac{\pi}{4} < x < \frac{3\pi}{4}$ ⁠, the angle $x - \frac{\pi}{4}$ lies in $\left(0, \frac{\pi}{2}\right)$ ⁠, where sine is positive. Hence
$0 < u < \sqrt2 \quad\text{on the whole interval.}$
Logarithmic differentiation takes $\log u$ ⁠, so $u$ must be positive. Rewriting $u$ as a single sine shows at a glance where it is positive. The key states this condition as " $\sin x > \cos x$ ", which is the same thing as $u > 0$ ⁠.

$\sin(A - B) = \sin A\cos B - \cos A\sin B$ (Class XI). Basics: trigonometric identities
Take logarithms

$y = u^u$ with $u > 0$ ⁠, so $y > 0$ and
$\begin{aligned} \log y &= u\log u \\ &= (\sin x - \cos x)\log(\sin x - \cos x). \end{aligned}$
$\log p^n = n\log p$ brings the exponent $u$ down, turning the power into a product of $u$ and $\log u$ ⁠.

$\log_b p^n = n\log_b p$ ⁠, Section 5.4. Basics: exponentials and logarithms
Differentiate both sides

Left side, by the chain rule: $\frac{d}{dx}\log y = \frac1y\frac{dy}{dx}$ ⁠.

Right side, by the product rule with factors $u$ and $\log u$ ⁠, using $\frac{d}{dx}\log u = \frac{u'}{u}$ ⁠:
$\begin{aligned} \frac{d}{dx}(u\log u) &= u'\log u + u\cdot\frac{u'}{u} \\ &= u'(\log u + 1). \end{aligned}$
The inner derivative is
$u' = \frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x.$
The $u$ cancels in the second term (allowed since $u > 0$ ⁠), and $u'$ is then a common factor. The derivative of $-\cos x$ is $+\sin x$ ⁠, which is why $u'$ is a sum.

Product rule (2), Section 5.3; chain rule, Theorem 4; $\frac{d}{dx}\log x = \frac1x$ ⁠, Theorem 5; Table 5.3. Lesson: chain rule
Multiply by $y$ and substitute back

From Step 3, $\frac1y\frac{dy}{dx} = (\cos x + \sin x)\left(1 + \log u\right)$ ⁠. Multiplying by $y = u^u$ ⁠:
$\begin{aligned} \frac{dy}{dx} = {}&(\sin x - \cos x)^{\sin x - \cos x} \\ &\times(\cos x + \sin x) \\ &\times\left(1 + \log(\sin x - \cos x)\right). \end{aligned}$
The final answer must contain only $x$ ⁠, so every $u$ is replaced by $\sin x - \cos x$ ⁠.

Logarithmic differentiation, Section 5.5. Lesson: logarithmic differentiation

The base

u = \sin x - \cos x = \sqrt2\sin\left(x - \frac{\pi}{4}\right)

for

0 \le x \le \pi

(horizontal and vertical scales differ). On the shaded interval

\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)

it is positive (blue), so

\log u

exists and logarithmic differentiation is allowed. The open circles mark the excluded end points, where

u = 0

and

u = \sqrt2

⁠.

Result

For $\frac{\pi}{4} < x < \frac{3\pi}{4}$ ⁠,

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

(Textbook answer: $(\sin x - \cos x)^{\sin x - \cos x}(\cos x + \sin x)$ $(1 + \log(\sin x - \cos x))$ ⁠, $\sin x > \cos x$ ⁠; the same.)

Check it with numbers

At $x = \frac{\pi}{2}$ ⁠: $u = 1 - 0 = 1$ ⁠, $y = 1^1 = 1$ ⁠, $u' = 0 + 1 = 1$ and $\log 1 = 0$ ⁠, so the formula gives $1\cdot 1\cdot(1 + 0) = 1$ ⁠. With $h = 0.0001$ ⁠: $y\left(\frac{\pi}{2} + h\right) \approx 1.000100005$ and $y\left(\frac{\pi}{2} - h\right) \approx 0.999900005$ ⁠, so the quotient is about $1.0000000$ ⁠. ✓

At $x = 1.2$ ⁠: $u \approx 0.5696813$ ⁠, $u' \approx 1.2943968$ ⁠, $\log u \approx -0.5626781$ ⁠, $y = u^u \approx 0.7257518$ ⁠. Formula:

\begin{aligned} y' &\approx 0.7257518 \times 1.2943968 \\ &\quad\times 0.4373219 \approx 0.4108249. \end{aligned}

Difference quotient with $h = 0.0001$ ⁠: $y(1.2001) \approx 0.72579287$ and $y(1.1999) \approx 0.72571070$ ⁠, so the quotient $\frac{0.72579287 - 0.72571070}{0.0002}$ is about $0.4108249$ ⁠. ✓

Common mistakes

Writing $u' = \cos x - \sin x$ ⁠. The derivative of $-\cos x$ is $+\sin x$ ⁠, so $u' = \cos x + \sin x$ ⁠.
Skipping the sign check. For $0 < x < \frac{\pi}{4}$ the base is negative and $\log u$ does not exist; the interval is part of the question for that reason.
Using the power rule $u\cdot u^{u-1}\cdot u'$ ⁠: this misses the $\log u$ term, because the exponent is not constant.

What to take away

$\frac{d}{dx}\left(u^u\right) = u^u(1 + \log u)\,u'$ wherever $u > 0$ ⁠; this matches Example 41 with $u = \sin x$ ⁠.
$\sin x - \cos x = \sqrt2\sin\left(x - \frac{\pi}{4}\right)$ makes its sign easy to read.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin(A - B)$ formula; $\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt2}$	Step 1	Class XI, Trigonometric Functions
$\log$ defined for positive numbers; $\log p^n = n\log p$	Steps 1, 2	Section 5.4 ( $\log p^n = n\log p$ is stated there for every real $n$ ⁠, without proof)
Product rule; chain rule; Table 5.3	Step 3	Rule (2), Section 5.3; Theorem 4, Section 5.3.1
$\frac{d}{dx}\log x = \frac1x$	Step 3	Theorem 5, Section 5.4
Logarithmic differentiation; $(\sin x)^{\sin x}$	Steps 2–4	Section 5.5; Example 41 (Lesson: logarithmic differentiation)
Answer	Result	Answers, Miscellaneous Exercise on Chapter 5, Q9
Numerical values, figure coordinates	Check, figure	Recomputed in the site's verification script

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Miscellaneous · Q9: Differentiate (sin⁡x−cos⁡x)sin⁡x−cos⁡x(\sin x-\cos x)^{\sin x-\cos x}(sinx−cosx)sinx−cosx on (π4,3π4)\left(\frac{\pi}{4},\frac{3\pi}{4}\right)(4π​,43π​)

Basics you need

Solution

The base is positive on (π4,3π4)\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)(4π​,43π​)

Take logarithms

Differentiate both sides

Multiply by yyy and substitute back

Every fact used, and where it comes from

Miscellaneous · Q9: Differentiate $(\sin x-\cos x)^{\sin x-\cos x}$ on $\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$

The base is positive on $\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$

Multiply by $y$ and substitute back