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Chapter 5 · Continuity and Differentiability

Exercise 5.4 · Q9: Differentiate cos⁡xlog⁡x\dfrac{\cos x}{\log x}⁠, x>0x>0

Exercise 5.4Quotient rule with log x · Section 5.4
The question

Differentiate the following w.r.t. xx⁠:

cos⁡xlog⁡x,x>0\frac{\cos x}{\log x}, \qquad x > 0
Exam answer

Let y=cos⁡xlog⁡xy = \frac{\cos x}{\log x}⁠. By the quotient rule,

dydx=log⁡x⋅(−sin⁡x)−cos⁡x⋅1x(log⁡x)2\frac{dy}{dx} = \frac{\log x\cdot(-\sin x) - \cos x\cdot\frac1x}{(\log x)^2}

Multiplying numerator and denominator by xx⁠,

∴ dydx=−xsin⁡xlog⁡x+cos⁡xx(log⁡x)2\therefore\ \dfrac{dy}{dx} = -\dfrac{x\sin x\log x + \cos x}{x(\log x)^2} (x>0, x≠1)(x > 0,\ x \ne 1)⁠.

The answer key, like the question, states only x>0x > 0⁠; at x=1x = 1 neither the function nor the formula is defined, so x≠1x \ne 1 is needed as well.

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

The one idea this problem is built on

One function divided by another: the quotient rule, with ddxlog⁡x=1x\frac{d}{dx}\log x = \frac1x from Theorem 5. The domain needs two conditions, not one: log⁡x\log x needs x>0x > 0⁠, and since it is a denominator it must also be non-zero, which excludes x=1x = 1⁠.

Given, to find, and the plan
Given
y=cos⁡xlog⁡xy = \dfrac{\cos x}{\log x}⁠, x>0x > 0 (natural logarithm).
To find
dydx\dfrac{dy}{dx}⁠, and exactly where it is valid.
Plan
(1) Domain, including x≠1x \ne 1⁠. (2) Name u=cos⁡xu = \cos x⁠, v=log⁡xv = \log x and differentiate each. (3) Apply the quotient rule. (4) Clear the fraction inside the numerator.

Basics you need

1The logarithm function

log⁡x\log x is defined for x>0x > 0 only, and log⁡x=0\log x = 0 exactly at x=1x = 1 (it is increasing and passes through (1,0)(1, 0)⁠, Section 5.4). Basics: exponentials and logarithms

2Quotient rule

(uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}⁠, wherever v≠0v \ne 0 (rule (3), Section 5.3). Basics: derivative rules

3Two standard derivatives

ddxcos⁡x=−sin⁡x\dfrac{d}{dx}\cos x = -\sin x (Table 5.3) and ddxlog⁡x=1x\dfrac{d}{dx}\log x = \dfrac1x⁠, x>0x > 0 (Theorem 5 (2)). Lesson: exponential and logarithmic functions

Solution

  1. Where is the function defined?

    cos⁡x\cos x is defined everywhere. The denominator log⁡x\log x needs x>0x > 0⁠, and we cannot divide by zero, so also log⁡x≠0\log x \ne 0⁠, that is x≠1x \ne 1⁠. The domain is

    0<x<1orx>1.0 < x < 1 \quad\text{or}\quad x > 1.
    The question writes only "x>0x > 0", which makes log⁡x\log x exist; but at x=1x = 1 the fraction is cos⁡10\frac{\cos 1}{0}⁠, which is undefined. So the condition x≠1x \ne 1 must be added, both for the function and for its derivative. The graph below shows what happens near x=1x = 1⁠.
    log⁡1=0\log 1 = 0⁠: the point (1,0)(1, 0) is on the graph of log⁡\log (Section 5.4, observation (3)). Basics: exponentials and logarithms
    x y 2 3 4 5 3 2 1 −1 −2 −3 1 x = 1 excluded: log 1 = 0 y = cos x / log x
    y=cos⁡xlog⁡xy = \frac{\cos x}{\log x} for 0<x≤50 < x \le 5 (values beyond ±3.5\pm 3.5 not drawn). Near x=1x = 1 the denominator is close to 00 and the graph shoots off to −∞-\infty on the left and +∞+\infty on the right, so x=1x = 1 is not in the domain. The open dot at the origin marks the excluded point x=0x = 0⁠.

  2. Name the top and bottom, and differentiate each

    u=cos⁡x,u′=−sin⁡x,v=log⁡x,v′=1x.\begin{aligned} u &= \cos x, & u' &= -\sin x, \\ v &= \log x, & v' &= \frac{1}{x}. \end{aligned}
    Listing the four ingredients of the quotient rule first prevents mixing them up. u′u' is from Table 5.3; v′v' is Theorem 5 (2), valid since x>0x > 0⁠.
    Table 5.3; Theorem 5 (2), Section 5.4. Basics: derivative rules

  3. Apply the quotient rule

    For x>0x > 0⁠, x≠1x \ne 1 (so that v≠0v \ne 0⁠):

    dydx=u′v−uv′v2=(−sin⁡x)log⁡x−cos⁡x⋅1x(log⁡x)2.\begin{aligned} \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ &= \frac{(-\sin x)\log x - \cos x\cdot\frac{1}{x}}{(\log x)^2}. \end{aligned}
    "Bottom times derivative of top, minus top times derivative of bottom, over bottom squared." The bottom squared, (log⁡x)2(\log x)^2⁠, is positive at every point of the domain.
    Quotient rule, rule (3), Section 5.3. Basics: derivative rules

  4. Clear the small fraction

    Multiply the top and the bottom by xx (which is positive, so nothing changes):

    dydx=−xsin⁡x log⁡x+cos⁡xx(log⁡x)2.\frac{dy}{dx} = -\frac{x\sin x\,\log x + \cos x}{x(\log x)^2}.
    The numerator contained cos⁡xx\frac{\cos x}{x}⁠, a fraction inside a fraction. Multiplying top and bottom by the same non-zero number xx removes it. Both terms of the numerator were negative, so we take the minus sign out in front.
    Multiplying numerator and denominator by the same non-zero quantity leaves a fraction unchanged.
Result ddx(cos⁡xlog⁡x)=−xsin⁡x log⁡x+cos⁡xx(log⁡x)2,\frac{d}{dx}\left(\frac{\cos x}{\log x}\right) = -\frac{x\sin x\,\log x + \cos x}{x(\log x)^2},

valid for x>0x > 0⁠, x≠1x \ne 1⁠. (Textbook answer: −xsin⁡x⋅log⁡x+cos⁡xx(log⁡x)2-\dfrac{x\sin x\cdot\log x + \cos x}{x(\log x)^2}⁠, x>0x > 0⁠. The same formula. The key states x>0x > 0⁠; at x=1x = 1 neither the function nor the formula is defined, so x≠1x \ne 1 is needed as well.)

Check it with numbers

Take x=2x = 2⁠: sin⁡2≈0.9092974\sin 2 \approx 0.9092974⁠, cos⁡2≈−0.4161468\cos 2 \approx -0.4161468⁠, log⁡2≈0.6931472\log 2 \approx 0.6931472⁠. Then

xsin⁡x log⁡x=2(0.9092974)(0.6931472)≈1.2605539,numerator≈1.2605539−0.4161468=0.8444071,denominator=2(0.6931472)2≈0.9609060.\begin{aligned} x\sin x\,\log x &= 2(0.9092974)(0.6931472) \\ &\approx 1.2605539, \\ \text{numerator} &\approx 1.2605539 - 0.4161468 \\ &= 0.8444071, \\ \text{denominator} &= 2(0.6931472)^2 \\ &\approx 0.9609060. \end{aligned}

The formula gives −0.84440710.9609060≈−0.8787613-\frac{0.8444071}{0.9609060} \approx -0.8787613⁠. A difference quotient of cos⁡xlog⁡x\frac{\cos x}{\log x} with h=0.0001h = 0.0001 on each side of 22 gives −0.87876-0.87876⁠. They agree.

Common mistakes
  1. Forgetting x≠1x \ne 1⁠. "x>0x > 0" makes log⁡x\log x exist, but division by log⁡1=0\log 1 = 0 is still impossible.
  2. A sign slip in the numerator. u′v=−sin⁡xlog⁡xu'v = -\sin x\log x is already negative; subtracting uv′=cos⁡xxuv' = \frac{\cos x}{x} makes the second term negative too.
  3. Writing log⁡x2\log x^2 for (log⁡x)2(\log x)^2⁠. log⁡x2=2log⁡x\log x^2 = 2\log x⁠, a different quantity.
What to take away
  1. When log⁡x\log x is a denominator, the domain is x>0x > 0 and x≠1x \ne 1⁠.
  2. After the quotient rule, clear fractions inside the numerator by multiplying top and bottom by the same non-zero factor.

Every fact used, and where it comes from

FactUsed inSource
log⁡\log defined on R+\R^+⁠; log⁡x=0  ⟺  x=1\log x = 0 \iff x = 1Step 1Observations (1), (3), (4) on the log function, Section 5.4
ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin xStep 2Table 5.3, Section 5.3
ddxlog⁡x=1x\frac{d}{dx}\log x = \frac1x⁠, x>0x > 0Step 2Theorem 5 (2), Section 5.4 (Lesson: exponential and logarithmic functions)
Quotient ruleStep 3Rule (3), Section 5.3
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