Exercise 5.7 · Q9: Second derivative of $\log(\log x)$⁠, $x \gt 1$

Basics you need

Solution

1. Find the domain: $x \gt 1$

   Let $y = \log(\log x)$⁠. Two conditions are needed:

   • the inner $\log x$ exists: $x \gt 0$⁠;
   • the outer logarithm gets a positive input: $\log x \gt 0$⁠, that is, $x \gt 1$⁠.

   Together: $x \gt 1$⁠. For example, at $x = \frac12$⁠, $\log x \approx -0.693$ and $\log(-0.693)$ does not exist; at $x = 1$⁠, $\log x = 0$ and $\log 0$ does not exist. All the work below is on $x \gt 1$⁠.

   The formula for a derivative can look fine at points where the function itself does not exist; stating the domain first prevents that. Here it also guarantees $x \gt 0$ and $\log x \gt 0$⁠, so $x\log x \gt 0$ and none of the denominators below can be zero.
   $\log t \gt 0 \iff t \gt 1$⁠, since $\log 1 = 0$ and $\log$ is increasing. Basics: exponentials and logarithms

2. First derivative: chain rule

   Outer function $\log u$⁠, with derivative $\frac1u$⁠; inner function $u = \log x$⁠, with derivative $\frac1x$⁠. So

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{\log x}\cdot\frac{d}{dx}(\log x) \\ &= \frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x\log x}. \end{aligned}$$
   The derivative of the outer function is evaluated at the inner one (giving $\frac{1}{\log x}$⁠) and then multiplied by the derivative of the inner function. This is the same pattern as the book's Example 26(ii), $\sin(\log x)$⁠.
   Chain rule (Theorem 4, Section 5.3.1); $\frac{d}{dx}\log x = \frac1x$ (Theorem 5). Lesson: chain rule

3. Derivative of the inner $x\log x$⁠: product rule

   We will need $\frac{d}{dx}(x\log x)$⁠. With $u = x$ ($u' = 1$⁠) and $v = \log x$ ($v' = \frac1x$⁠):

   $$\begin{aligned} \frac{d}{dx}(x\log x) &= 1\cdot\log x + x\cdot\frac1x \\ &= \log x + 1. \end{aligned}$$
   The first derivative is $1$ over the product $x\log x$⁠, so differentiating it will need the derivative of that product. $x\cdot\frac1x = 1$ because $x \ne 0$⁠.
   Product rule (Section 5.3, rule (2)).

4. Second derivative: chain rule on $(x\log x)^{-1}$

   Write $\frac{dy}{dx} = (x\log x)^{-1}$⁠. Outer function $w^{-1}$⁠, with derivative $-w^{-2}$ (power rule, $n = -1$⁠); inner function $w = x\log x$⁠, with derivative $\log x + 1$ (Step 3). So

   $$\begin{aligned} \frac{d^2y}{dx^2} &= -(x\log x)^{-2}\cdot(\log x + 1) \\ &= -\frac{1 + \log x}{(x\log x)^2}. \end{aligned}$$

   So $\dfrac{d^2y}{dx^2} = -\dfrac{1 + \log x}{(x\log x)^2}$ for $x \gt 1$⁠.

   The second derivative is the derivative of the first derivative from Step 2. Writing "one over something" as "something to the power $-1$" lets the power rule and the chain rule do the work. (The quotient rule gives the same: $\frac{0 - 1\cdot(\log x + 1)}{(x\log x)^2}$⁠.) The denominator is positive for $x \gt 1$ (Step 1).
   Chain rule (Theorem 4); power rule (Table 5.3). Lesson: second order derivative

5. Compare with the textbook and read the sign

   The textbook's answer is $-\frac{(1 + \log x)}{(x\log x)^2}$⁠, exactly our expression. An equivalent expanded form is $-\frac{1 + \log x}{x^2(\log x)^2}$⁠, since $(x\log x)^2 = x^2(\log x)^2$⁠.

   For $x \gt 1$⁠, $\log x \gt 0$⁠, so $1 + \log x \gt 0$ and the second derivative is negative: the slope $\frac{1}{x\log x}$ decreases as $x$ grows, and the graph of $\log(\log x)$ bends downward everywhere on its domain.

   The book's answer does not state the domain; it is understood to be the domain $x \gt 1$ of the function, which we found in Step 1.

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