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Chapter 5 · Continuity and Differentiability

Exercise 5.4 · Q10: Differentiate cos⁡(log⁡x+ex)\cos(\log x+e^x)⁠, x>0x>0

Exercise 5.4Chain rule with log x and eˣ · Section 5.4
The question

Differentiate the following w.r.t. xx⁠:

cos⁡(log⁡x+ex),x>0\cos\left(\log x + e^x\right), \qquad x > 0
Exam answer

Let y=cos⁡(log⁡x+ex)y = \cos(\log x + e^x)⁠. By the chain rule,

dydx=−sin⁡(log⁡x+ex)×ddx(log⁡x+ex)\begin{aligned} \frac{dy}{dx} &= -\sin(\log x + e^x) \\ &\quad\times\frac{d}{dx}(\log x + e^x) \end{aligned}

∴ dydx\therefore\ \dfrac{dy}{dx} =−(1x+ex)sin⁡(log⁡x+ex)= -\left(\dfrac1x + e^x\right)\sin(\log x + e^x) (x>0)(x > 0)⁠.

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

The one idea this problem is built on

A cosine of a sum. The chain rule handles the cosine (outer function) and the sum inside (inner function); the inner function is differentiated with the sum rule and both parts of Theorem 5. The condition x>0x > 0 is there because log⁡x\log x needs it.

Given, to find, and the plan
Given
y=cos⁡(log⁡x+ex)y = \cos(\log x + e^x)⁠, x>0x > 0 (natural logarithm).
To find
dydx\dfrac{dy}{dx}⁠.
Plan
(1) Domain. (2) Name inner t=log⁡x+ext = \log x + e^x and outer y=cos⁡ty = \cos t⁠; differentiate each. (3) Multiply and substitute back.

Basics you need

1Theorem 5

ddxex=ex\dfrac{d}{dx}e^x = e^x for all xx⁠, and ddxlog⁡x=1x\dfrac{d}{dx}\log x = \dfrac{1}{x} for x>0x > 0⁠. Basics: exponentials and logarithms

2Sum rule

(u+v)′=u′+v′(u + v)' = u' + v' (rule (1), Section 5.3). Basics: derivative rules

3Chain rule

If y=f(t)y = f(t) and t=g(x)t = g(x)⁠, then dydx=dydt⋅dtdx\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} (Theorem 4), with ddtcos⁡t=−sin⁡t\frac{d}{dt}\cos t = -\sin t (Table 5.3). Basics: composite functions

Solution

  1. Where is the function defined?

    log⁡x\log x needs x>0x > 0⁠; exe^x and cos⁡\cos accept every real number. So the domain is exactly x>0x > 0⁠, as in the question.

    The only restriction comes from the logarithm. On x>0x > 0 both log⁡x\log x and exe^x are differentiable, and so is cos⁡\cos⁠, so a derivative will exist at every point of the domain.
    Domain of log⁡\log is R+\R^+ (Section 5.4). Basics: exponentials and logarithms

  2. Name the inner and outer functions, and differentiate each

    Inner: t=log⁡x+ext = \log x + e^x⁠. Outer: y=cos⁡ty = \cos t⁠.

    dtdx=1x+ex(x>0),dydt=−sin⁡t.\begin{aligned} \frac{dt}{dx} &= \frac{1}{x} + e^x \quad (x > 0), \\ \frac{dy}{dt} &= -\sin t. \end{aligned}
    To compute yy you first work out the sum log⁡x+ex\log x + e^x⁠, then take its cosine: the sum is the inner function. Its derivative uses the sum rule (differentiate the two terms separately) and both parts of Theorem 5.
    Sum rule, Section 5.3; Theorem 5, Section 5.4; Table 5.3. Lesson: exponential and logarithmic functions

  3. Multiply (chain rule) and substitute back

    dydx=−sin⁡t⋅(1x+ex)=−(1x+ex)sin⁡(log⁡x+ex).\begin{aligned} \frac{dy}{dx} &= -\sin t\cdot\left(\frac{1}{x} + e^x\right) \\ &= -\left(\frac{1}{x} + e^x\right)\sin\left(\log x + e^x\right). \end{aligned}

    This holds for every x>0x > 0⁠.

    The outer rate −sin⁡t-\sin t is evaluated at the inner value t=log⁡x+ext = \log x + e^x⁠. The bracket 1x+ex\frac1x + e^x must be kept whole: both of its terms multiply the sine. If you prefer one fraction, 1x+ex=1+xexx\frac1x + e^x = \frac{1 + xe^x}{x}⁠.
    Chain rule, Theorem 4, Section 5.3.1. Lesson: the chain rule
Result dydx=−(1x+ex)sin⁡(log⁡x+ex),\frac{dy}{dx} = -\left(\frac{1}{x} + e^x\right)\sin\left(\log x + e^x\right),

valid for every x>0x > 0⁠. (Textbook answer: −(1x+ex)sin⁡(log⁡x+ex)-\left(\dfrac1x + e^x\right)\sin(\log x + e^x)⁠, x>0x > 0⁠. The same.)

Check it with numbers

Take x=1x = 1⁠: log⁡1+e1=e≈2.7182818\log 1 + e^1 = e \approx 2.7182818⁠, sin⁡e≈0.4107813\sin e \approx 0.4107813⁠, and 11+e≈3.7182818\frac11 + e \approx 3.7182818⁠. The formula gives −3.7182818×0.4107813-3.7182818\times 0.4107813⁠, which is about −1.5274006-1.5274006⁠. A difference quotient of cos⁡(log⁡x+ex)\cos(\log x + e^x) with h=0.0001h = 0.0001 on each side of 11 gives −1.52740-1.52740⁠. They agree.

Common mistakes
  1. Losing the minus sign. ddtcos⁡t=−sin⁡t\frac{d}{dt}\cos t = -\sin t⁠.
  2. Multiplying the sine by only one term. −1xsin⁡(log⁡x+ex)+ex-\frac1x\sin(\log x + e^x) + e^x is wrong; the whole bracket 1x+ex\frac1x + e^x multiplies the sine.
  3. Changing the inside of the sine. The sine is evaluated at the original inner value log⁡x+ex\log x + e^x⁠, not at its derivative.
What to take away
  1. When the inner function is itself a sum, differentiate it with the sum rule and keep the result in brackets.
  2. ddxcos⁡g(x)=−g′(x)sin⁡g(x)\dfrac{d}{dx}\cos g(x) = -g'(x)\sin g(x)⁠.

Every fact used, and where it comes from

FactUsed inSource
log⁡\log is defined on R+\R^+Step 1Observation (1) on the log function, Section 5.4
ddxex=ex\frac{d}{dx}e^x = e^x⁠, ddxlog⁡x=1x\frac{d}{dx}\log x = \frac1xStep 2Theorem 5, Section 5.4 (Lesson: exponential and logarithmic functions)
Sum ruleStep 2Rule (1), Section 5.3
ddtcos⁡t=−sin⁡t\frac{d}{dt}\cos t = -\sin tStep 2Table 5.3, Section 5.3
Chain ruleStep 3Theorem 4, Section 5.3.1
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