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Chapter 6 · Application of Derivatives

Exercise 6.2 · Q16: log⁡sin⁡x\log\sin x rises on (0,π2)\left(0,\frac\pi2\right) and falls on (π2,π)\left(\frac\pi2,\pi\right)

Exercise 6.2Increasing and decreasing functions · Section 6.3
The question

16. Prove that the function ff given by f(x)=log⁡sin⁡xf(x) = \log\sin x is increasing on (0,π2)\left(0, \frac{\pi}{2}\right) and decreasing on (π2,π)\left(\frac{\pi}{2}, \pi\right)⁠.

Exam answer

f(x)=log⁡sin⁡xf(x) = \log\sin x is defined for 0<x<π0 < x < \pi⁠. By the chain rule,

f′(x)=1sin⁡x⋅cos⁡x=cot⁡xf'(x) = \frac{1}{\sin x}\cdot\cos x = \cot x

In (0,π2)\left(0, \frac{\pi}{2}\right)⁠, cot⁡x>0\cot x > 0⁠, so f′(x)>0f'(x) > 0⁠.

In (π2,π)\left(\frac{\pi}{2}, \pi\right)⁠, cot⁡x<0\cot x < 0⁠, so f′(x)<0f'(x) < 0⁠.

Hence ff is increasing on (0,π2)\left(0, \dfrac{\pi}{2}\right) and decreasing on (π2,π)\left(\dfrac{\pi}{2}, \pi\right)⁠.

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

The one idea this problem is built on

The logarithm is itself increasing, so log⁡sin⁡x\log\sin x goes up exactly where sin⁡x\sin x goes up and down where sin⁡x\sin x goes down. The derivative makes this precise: by the chain rule f′(x)=cos⁡xsin⁡x=cot⁡xf'(x) = \frac{\cos x}{\sin x} = \cot x⁠, and on (0,π)(0, \pi) the denominator sin⁡x\sin x is positive, so f′f' has the sign of cos⁡x\cos x⁠: positive before π2\frac{\pi}{2}⁠, negative after.

Given, to show, and the plan
Given
f(x)=log⁡sin⁡xf(x) = \log\sin x⁠, where log⁡\log is the natural logarithm (base ee⁠), the convention of Section 5.4. (Any base greater than 1 would give the same answer, since log⁡bu=log⁡ulog⁡b\log_b u = \frac{\log u}{\log b} with log⁡b>0\log b \gt 0⁠.)
To show
ff is increasing on (0,π2)\left(0, \frac{\pi}{2}\right) and decreasing on (π2,π)\left(\frac{\pi}{2}, \pi\right)⁠.
Plan
(1) Check ff is defined on both intervals. (2) Differentiate with the chain rule. (3) Sign of f′f' on each interval. (4) Theorem 1 on every closed subinterval. (5) Conclude.

Basics you need

1Increasing and decreasing (Definition 1, Section 6.3)

Increasing on I: x1<x2x_1 \lt x_2 in I gives f(x1)≤f(x2)f(x_1) \le f(x_2)⁠. Decreasing: it gives f(x1)≥f(x2)f(x_1) \ge f(x_2)⁠. With <\lt and >\gt these become strictly increasing and strictly decreasing. Lesson: increasing and decreasing functions

2Theorem 1 on an open interval

Apply Theorem 1 on [x1,x2][x_1, x_2] for any x1<x2x_1 \lt x_2 in the open interval: f′>0f' \gt 0 inside gives f(x1)<f(x2)f(x_1) \lt f(x_2)⁠, and f′<0f' \lt 0 inside gives f(x1)>f(x2)f(x_1) \gt f(x_2) (Mean Value Theorem step of the proof). Lesson: increasing and decreasing functions

3Derivative of log⁡\log⁠, and the chain rule

ddulog⁡u=1u\frac{d}{du}\log u = \frac1u for u>0u \gt 0 (Theorem 5, Section 5.4). For log⁡(g(x))\log(g(x))⁠, the chain rule gives g′(x)g(x)\frac{g'(x)}{g(x)}⁠. Basics: the chain rule

4Signs of sine and cosine

On (0,π)(0, \pi)⁠, sin⁡x>0\sin x \gt 0⁠. cos⁡x>0\cos x \gt 0 on (0,π2)\left(0, \frac{\pi}{2}\right) and cos⁡x<0\cos x \lt 0 on (π2,π)\left(\frac{\pi}{2}, \pi\right)⁠. Basics: signs of sin and cos on the unit circle

Solution

  1. ff is defined on both intervals

    log⁡u\log u is defined only for u>0u \gt 0⁠, so f(x)=log⁡sin⁡xf(x) = \log\sin x needs sin⁡x>0\sin x \gt 0⁠. For 0<x<π0 \lt x \lt \pi the angle is in the first or second quadrant, where sin⁡x>0\sin x \gt 0⁠. Both (0,π2)\left(0, \frac{\pi}{2}\right) and (π2,π)\left(\frac{\pi}{2}, \pi\right) lie inside (0,π)(0, \pi)⁠, so ff is defined (and differentiable) on both.

    The book defines log⁡\log only on positive numbers (Section 5.4), so the domain has to be checked first. Outside (0,π)(0, \pi)⁠, for example at x=3π2x = \frac{3\pi}{2} where sin⁡x=−1\sin x = -1⁠, log⁡sin⁡x\log\sin x does not exist.
    Domain of log⁡\log is the positive reals (Section 5.4). Basics: signs of sin and cos

  2. Differentiate with the chain rule

    Inner function u=sin⁡xu = \sin x (with dudx=cos⁡x\frac{du}{dx} = \cos x⁠), outer function log⁡u\log u (with derivative 1u\frac1u⁠):

    f′(x)=1sin⁡x⋅cos⁡x=cos⁡xsin⁡x=cot⁡x,0<x<π.\begin{aligned} f'(x) &= \frac{1}{\sin x}\cdot\cos x\\ &= \frac{\cos x}{\sin x} = \cot x, \quad 0 \lt x \lt \pi. \end{aligned}
    The chain rule multiplies the derivative of the outer function, evaluated at the inner one, by the derivative of the inner function. The formula is valid wherever sin⁡x>0\sin x \gt 0⁠, so on all of (0,π)(0, \pi)⁠.
    Chain rule (Theorem 4, Section 5.3.1); ddxlog⁡x=1x\frac{d}{dx}\log x = \frac1x (Theorem 5, Section 5.4). Basics: the chain rule

  3. Sign of f′f' on each interval

    On (0,π)(0, \pi) the denominator sin⁡x\sin x is positive, so f′(x)=cos⁡xsin⁡xf'(x) = \frac{\cos x}{\sin x} has the same sign as cos⁡x\cos x⁠:

    • 0<x<π20 \lt x \lt \frac{\pi}{2}⁠: cos⁡x>0\cos x \gt 0⁠, so f′(x)>0f'(x) \gt 0⁠.
    • π2<x<π\frac{\pi}{2} \lt x \lt \pi⁠: cos⁡x<0\cos x \lt 0⁠, so f′(x)<0f'(x) \lt 0⁠.
    0 π/2 π cos x + − 0 sin x + + f′(x) + − 0 f increasing decreasing
    Sign chart of f′(x)=cos⁡xsin⁡xf'(x) = \frac{\cos x}{\sin x} on (0,π)(0, \pi)⁠, drawn to scale. f′f' is 00 at π2\frac{\pi}{2}⁠, where cos⁡x=0\cos x = 0⁠.
    Splitting the quotient into its top and bottom and reading each sign is more reliable than remembering where cot⁡x\cot x is positive.
    Quadrant signs: first quadrant all positive; second quadrant sin⁡>0\sin \gt 0⁠, cos⁡<0\cos \lt 0⁠. Basics: signs of sin and cos

  4. Increasing on (0,π2)\left(0, \frac{\pi}{2}\right)

    Take any x1<x2x_1 \lt x_2 in (0,π2)\left(0, \frac{\pi}{2}\right)⁠. ff is continuous on [x1,x2][x_1, x_2] and differentiable on (x1,x2)(x_1, x_2)⁠, with f′>0f' \gt 0 there (Step 3). By Theorem 1 (a), f(x1)≤f(x2)f(x_1) \le f(x_2)⁠, and the Mean Value Theorem step of the proof gives f(x1)<f(x2)f(x_1) \lt f(x_2)⁠. So ff is increasing (strictly) on (0,π2)\left(0, \frac{\pi}{2}\right).

    Theorem 1 is stated on a closed interval; the open interval (0,π2)\left(0, \frac{\pi}{2}\right) is handled pair by pair. We cannot use [0,π2][0, \frac{\pi}{2}] itself, because ff is not defined at 00⁠.
    Theorem 1 (a), Section 6.3. Lesson: increasing and decreasing functions

  5. Decreasing on (π2,π)\left(\frac{\pi}{2}, \pi\right)

    Take any x1<x2x_1 \lt x_2 in (π2,π)\left(\frac{\pi}{2}, \pi\right)⁠. Now f′<0f' \lt 0 on (x1,x2)(x_1, x_2)⁠, so Theorem 1 (b) gives f(x1)≥f(x2)f(x_1) \ge f(x_2)⁠, and the proof gives f(x1)>f(x2)f(x_1) \gt f(x_2)⁠. So ff is decreasing (strictly) on (π2,π)\left(\frac{\pi}{2}, \pi\right).

    x y π/4 3π/4 π −1 −2 −3 (π/2, 0) increasing decreasing
    y=log⁡sin⁡xy = \log\sin x on (0,π)(0, \pi) (cut off at height −3-3⁠; the axes use different scales). It rises to its highest value 00 at (π2,0)\left(\frac{\pi}{2}, 0\right) and then falls; it goes down without bound near 00 and near π\pi⁠, where sin⁡x→0\sin x \to 0⁠.
    Because ff is continuous at π2\frac{\pi}{2}⁠, the same argument also gives increasing on (0,π2]\left(0, \frac{\pi}{2}\right] and decreasing on [π2,π)\left[\frac{\pi}{2}, \pi\right)⁠, as in Example 12 of the book. The question only asks about the open intervals.
    Theorem 1 (b), Section 6.3. Lesson: increasing and decreasing functions
Result

f′(x)=cot⁡xf'(x) = \cot x is positive on (0,π2)\left(0, \frac{\pi}{2}\right) and negative on (π2,π)\left(\frac{\pi}{2}, \pi\right)⁠, so log⁡sin⁡x\log\sin x is increasing on (0,π2)\left(0, \frac{\pi}{2}\right) and decreasing on (π2,π)\left(\frac{\pi}{2}, \pi\right), in fact strictly in both cases. (The answer key has no entry for this proof question.)

Check it with numbers

f(π6)=log⁡12≈−0.693147f\left(\frac{\pi}{6}\right) = \log\frac12 \approx -0.693147⁠, f(π3)=log⁡32≈−0.143841f\left(\frac{\pi}{3}\right) = \log\frac{\sqrt3}{2} \approx -0.143841⁠, f(π2)=log⁡1=0f\left(\frac{\pi}{2}\right) = \log 1 = 0⁠: rising. ✓
f(2π3)≈−0.143841f\left(\frac{2\pi}{3}\right) \approx -0.143841⁠, f(5π6)≈−0.693147f\left(\frac{5\pi}{6}\right) \approx -0.693147⁠: falling after π2\frac{\pi}{2}⁠. ✓
Derivative: cot⁡π4=1>0\cot\frac{\pi}{4} = 1 \gt 0 and cot⁡3π4=−1<0\cot\frac{3\pi}{4} = -1 \lt 0⁠. ✓

Common mistakes
  1. Writing f′(x)=1sin⁡xf'(x) = \frac{1}{\sin x} and forgetting the inner derivative cos⁡x\cos x⁠. The chain rule needs both factors.
  2. Skipping the domain. log⁡sin⁡x\log\sin x exists only where sin⁡x>0\sin x \gt 0⁠; without Step 1 the derivative has nothing to apply to.
  3. Reading the sign of cot⁡x\cot x from memory and getting the quadrant wrong. Split it into cos⁡x\cos x over sin⁡x\sin x and read each sign.
What to take away
  1. ddxlog⁡g(x)=g′(x)g(x)\frac{d}{dx}\log g(x) = \frac{g'(x)}{g(x)} where g(x)>0g(x) \gt 0⁠; with g>0g \gt 0⁠, its sign is the sign of g′g'⁠.
  2. So log⁡g\log g increases exactly where gg increases: here, where sin⁡x\sin x does.
  3. Check the domain first whenever a logarithm (or a square root, or a quotient) appears.

Every fact used, and where it comes from

FactUsed inSource
Increasing and decreasing on an intervalSteps 4, 5Definition 1, Section 6.3 (Lesson: increasing and decreasing functions)
Sign of f′f' decides increasing or decreasingSteps 4, 5Theorem 1, Section 6.3, and the Mean Value Theorem step of its proof
log⁡\log means log⁡e\log_e⁠; its domain is the positive realsStep 1Section 5.4
ddxlog⁡x=1x\frac{d}{dx}\log x = \frac1xStep 2Theorem 5 (2), Section 5.4
Chain ruleStep 2Theorem 4, Section 5.3.1
ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠; signs of sin⁡\sin⁠, cos⁡\cos by quadrantSteps 1–3Class XI, Limits and Derivatives; Trigonometric Functions
Including the turning point by continuityStep 5Example 12, Section 6.3
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