Exercise 5.5 · Q3: Differentiate $(\log x)^{\cos x}$⁠, $x>1$

Basics you need

Solution

1. Where is the function defined? ($x > 1$⁠)

   Let $y = (\log x)^{\cos x}$⁠. First, $\log x$ needs $x > 0$⁠. For a power whose exponent varies, the base must be positive, so we also need $\log x > 0$⁠, that is, $x > 1$⁠. Then $y = e^{\cos x\,\log(\log x)} > 0$⁠.

   For $0 < x < 1$ the base is negative: at $x = \frac12$⁠, $\log\frac12 \approx -0.693$ and $\cos\frac12 \approx 0.878$⁠, and $(-0.693)^{0.878}$ is not a real number. At $x = 1$ the base is $0$ and $\log 0$ does not exist. The book's condition for logarithmic differentiation, that $u(x)$ and $f(x)$ be positive, holds exactly for $x > 1$⁠; the answer key does not state it.
   $\log x > 0 \iff x > 1$ (Basic 2). Basics: exponentials and logarithms

2. Take logarithms

   For $x > 1$⁠, by the power law for logs,

   $$\log y = \cos x\cdot\log(\log x).$$
   $\log(u^v) = v\log u$ with $u = \log x > 0$ and $v = \cos x$⁠. The exponent has come down as a factor, so the right side is a product of two functions of $x$⁠.
   $\log p^n = n\log p$ for $p > 0$ and real $n$ (Section 5.4). Basics: exponentials and logarithms

3. The derivative of $\log(\log x)$

   Chain rule with outer function $\log t$ and inner function $t = \log x$ (positive for $x > 1$⁠):

   $$\frac{d}{dx}\log(\log x) = \frac{1}{\log x}\cdot\frac{d}{dx}\log x = \frac{1}{x\log x}.$$
   Outer derivative: $\frac{d}{dt}\log t = \frac1t$⁠, evaluated at $t = \log x$⁠. Inner derivative: $\frac{d}{dx}\log x = \frac1x$⁠. Their product is the chain rule.
   Chain rule (Theorem 4); $\frac{d}{dx}\log x = \frac1x$ (Theorem 5). Lesson: the chain rule

4. Differentiate both sides

   Left (chain rule, outer $\log$⁠, inner $y$⁠): $\frac1y\frac{dy}{dx}$⁠. Right: product rule with $u = \cos x$ ($u' = -\sin x$⁠) and $v = \log(\log x)$ ($v' = \frac{1}{x\log x}$⁠, Step 3):

   $$\begin{aligned} \frac1y\cdot\frac{dy}{dx} &= -\sin x\cdot\log(\log x) \\ &\quad + \cos x\cdot\frac{1}{x\log x}. \end{aligned}$$
   Both factors vary with $x$⁠, so the product rule is needed: differentiate one factor at a time, keeping the other, and add.
   Product rule (Section 5.3); $\frac{d}{dx}\cos x = -\sin x$ (Table 5.3). Lesson: logarithmic differentiation

5. Multiply by $y$

   For $x > 1$⁠,

   $$\begin{aligned} \frac{dy}{dx} &= y\left[\frac{\cos x}{x\log x} - \sin x\log(\log x)\right] \\ &= (\log x)^{\cos x} \\ &\quad\times\left[\frac{\cos x}{x\log x} - \sin x\log(\log x)\right]. \end{aligned}$$
   Multiplying by $y$ (positive) isolates $\frac{dy}{dx}$⁠; then $y$ is replaced by its formula. The terms are reordered to match the textbook's answer.
   Rearranging: $\frac1y Y = R$ with $y \ne 0$ gives $Y = yR$⁠. Basics: solving for dy/dx

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