Exercise 5.4 · Q5: Differentiate $\log(\cos e^x)$

Basics you need

Solution

1. Where is the function defined?

   $e^x$ and $\cos$ accept every real number, so the only condition comes from the logarithm:

   $$\cos\left(e^x\right) > 0.$$

   Put $u = e^x$⁠. As $x$ runs over $\R$⁠, $u$ runs over all positive numbers. The positive $u$ with $\cos u > 0$ are

   $$\begin{gathered} 0 < u < \tfrac{\pi}{2}, \quad \tfrac{3\pi}{2} < u < \tfrac{5\pi}{2}, \\ \tfrac{7\pi}{2} < u < \tfrac{9\pi}{2}, \ \dots \end{gathered}$$

   Taking logarithms (which keeps inequalities, because $\log$ is increasing), the domain of $y$ is

   $$\begin{aligned} &x < \log\tfrac{\pi}{2}, \ \text{or} \\ &\log\tfrac{(4n-1)\pi}{2} < x < \log\tfrac{(4n+1)\pi}{2},\ n \in \N. \end{aligned}$$

   Numerically: $x < 0.4516$⁠, or $1.5502 < x < 2.0610$⁠, or $2.3975 < x < 2.6488$⁠, and so on. For example $x = 0$ is in the domain ($\cos 1 > 0$⁠), but $x = 1$ is not: $\cos e \approx -0.9117$⁠, and $\log$ of a negative number is not defined.

   Solving for $u = e^x$ first is easier than solving for $x$ directly. The first interval for $u$ is $\left(0, \frac{\pi}{2}\right)$⁠, not $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, because $e^x$ is never $0$ or negative. Then $x = \log u$ converts each $u$-interval into an $x$-interval; $u \to 0^+$ corresponds to $x \to -\infty$⁠.
   Domain of $\log$ is $\R^+$⁠; $\log$ is increasing (Section 5.4, observations (1) and (4)). Basics: exponentials and logarithms

2. Name the three links and differentiate each

   $u = e^x$⁠, $\ v = \cos u$⁠, $\ y = \log v$⁠.

   $$\begin{aligned} \frac{du}{dx} &= e^x, \\ \frac{dv}{du} &= -\sin u, \\ \frac{dy}{dv} &= \frac{1}{v} \quad (v > 0). \end{aligned}$$
   The links are in calculation order: exponentiate, take the cosine, take the log. The first and last derivatives are Theorem 5; the middle one is in Table 5.3. On the domain found in Step 1, $v = \cos e^x > 0$⁠, so the logarithm link is differentiable there.
   Theorem 5, Section 5.4; Table 5.3. Basics: derivative rules

3. Multiply and simplify

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{v}\cdot(-\sin u)\cdot e^x \\ &= \frac{-\sin(e^x)}{\cos(e^x)}\cdot e^x \\ &= -e^x\tan e^x. \end{aligned}$$

   This holds at every $x$ in the domain of Step 1: there $\cos e^x > 0$⁠, so $\tan e^x$ is defined.

   Each rate is evaluated at the value entering its link: the log link sees $v = \cos e^x$⁠, the cosine link sees $u = e^x$⁠. Then $\frac{\sin}{\cos} = \tan$ tidies the answer.
   Chain rule for three functions, Section 5.3.1; $\tan = \frac{\sin}{\cos}$⁠. Lesson: the chain rule

4. Compare with the condition in the answer key

   The answer key prints $-e^x\tan e^x$ with the condition $e^x \ne (2n+1)\frac{\pi}{2}$⁠, $n \in \N$⁠. That condition only removes the points where $\cos e^x = 0$ (where $\tan e^x$ is undefined). It does not ensure $\cos e^x > 0$⁠: at $x = 1$⁠, $e^1 \approx 2.718$ is not an odd multiple of $\frac{\pi}{2}$⁠, yet $\cos e < 0$ and $y$ is not defined. Also, $n \in \N$ leaves out $n = 0$⁠, the point $e^x = \frac{\pi}{2}$⁠. The correct set is the domain of Step 1.

   Going deeper: with $n = 0$ included, the key's condition describes exactly where $\log\left|\cos e^x\right|$ is defined, and the derivative of $\log|v|$ is also $\frac{v'}{v}$ (where $v < 0$⁠, $\log|v| = \log(-v)$⁠, and the chain rule gives $\frac{-v'}{-v}$⁠). So the formula $-e^x\tan e^x$ is the derivative of $\log|\cos e^x|$ at every such point; for $\log(\cos e^x)$ itself, only the points with $\cos e^x > 0$ count.
   A derivative exists only where the function does. Lesson: exponential and logarithmic functions

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