Exercise 5.5 · Q6: Differentiate $\left(x+\frac1x\right)^x$ $+\,x^{1+\frac1x}$

Basics you need

Solution

1. Domain: $x > 0$

   The second term $x^{1 + \frac1x}$ has a variable exponent, so its base $x$ must be positive. For $x > 0$ the first base is positive too:

   $$x + \frac1x = \frac{x^2 + 1}{x} > 0.$$

   So both terms are defined and positive for $x > 0$⁠, and that is where we work.

   For $x < 0$ both bases are negative ($x + \frac1x = \frac{x^2+1}{x} < 0$⁠), and a negative base with a non-integer exponent is in general not a real number: at $x = -\frac12$ the first term is $\left(-\frac52\right)^{-1/2}$ $= \frac{1}{\sqrt{-5/2}}$⁠, which is not real. $x = 0$ is excluded because $\frac1x$ is undefined. Logarithmic differentiation needs each base positive (Section 5.5), which is exactly $x > 0$⁠.
   $x^2 + 1 > 0$ always, so $\frac{x^2+1}{x}$ has the sign of $x$⁠. Basics: exponentials and logarithms

2. Split with the sum rule

   $y = u + v$⁠, so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$⁠.

   We cannot simplify $\log(u + v)$⁠, so we do not take the log of $y$⁠. Each of $u$ and $v$ is a single positive power, which is exactly what logarithmic differentiation handles.
   $(u + v)' = u' + v'$ (Section 5.3). Basics: derivative rules

3. $u = \left(x + \frac1x\right)^x$⁠: take logs and differentiate

   $\log u = x\log\left(x + \frac1x\right)$⁠. The right side is a product of $x$ and $\log\left(x + \frac1x\right)$⁠. The second factor needs the chain rule, inner $g = x + \frac1x$⁠, $g' = 1 - \frac{1}{x^2}$⁠:

   $$\begin{aligned} \frac1u\cdot\frac{du}{dx} &= 1\cdot\log\left(x + \tfrac1x\right) \\ &\quad + x\cdot\frac{1 - \frac{1}{x^2}}{x + \frac1x}. \end{aligned}$$
   Product rule: differentiate $x$ (giving $1$⁠) keeping the log, then differentiate the log (giving $\frac{g'}{g}$⁠) keeping $x$⁠.
   Product rule; $\frac{d}{dx}\log g = \frac{g'}{g}$ (Theorems 4 and 5). Lesson: logarithmic differentiation

4. Simplify the fraction and finish $u'$

   Write $x\left(1 - \frac{1}{x^2}\right) = x - \frac1x = \frac{x^2 - 1}{x}$ and $x + \frac1x = \frac{x^2 + 1}{x}$⁠. Dividing, the $\frac1x$ cancels:

   $$x\cdot\frac{1 - \frac{1}{x^2}}{x + \frac1x} = \frac{x^2 - 1}{x^2 + 1}.$$

   So

   $$\begin{aligned} \frac{du}{dx} = \ &\left(x + \frac1x\right)^x \\ &\times\left[\frac{x^2 - 1}{x^2 + 1} + \log\left(x + \frac1x\right)\right]. \end{aligned}$$
   Clearing the small fractions puts the term in the textbook's form. Multiplying $\frac1u\frac{du}{dx}$ by $u$ (positive) isolates $\frac{du}{dx}$⁠.
   Simplifying a compound fraction by writing both parts over $x$⁠. Basics: solving for dy/dx

5. $v = x^{1 + \frac1x}$⁠: take logs and differentiate

   $\log v = \left(1 + \frac1x\right)\log x$⁠, again a product. With $\frac{d}{dx}\left(1 + \frac1x\right) = -\frac{1}{x^2}$ and $\frac{d}{dx}\log x = \frac1x$⁠:

   $$\begin{aligned} \frac1v\cdot\frac{dv}{dx} &= -\frac{1}{x^2}\log x + \left(1 + \frac1x\right)\frac1x \\ &= \frac{-\log x + x + 1}{x^2}. \end{aligned}$$

   So

   $$\frac{dv}{dx} = x^{1 + \frac1x}\left(\frac{x + 1 - \log x}{x^2}\right).$$
   $\left(1 + \frac1x\right)\frac1x = \frac1x + \frac{1}{x^2} = \frac{x + 1}{x^2}$⁠; putting everything over $x^2$ collects the bracket.
   Product rule; power rule $\frac{d}{dx}x^{-1} = -x^{-2}$ (Table 5.3). Basics: derivative rules

6. Add the two derivatives

   For $x > 0$⁠,

   $$\begin{aligned} \frac{dy}{dx} = \ &\left(x + \frac1x\right)^x \\ &\times\left[\frac{x^2 - 1}{x^2 + 1} + \log\left(x + \frac1x\right)\right] \\ &+ x^{1 + \frac1x}\left(\frac{x + 1 - \log x}{x^2}\right). \end{aligned}$$
   This is $u' + v'$ from Steps 4 and 5, and it is exactly the textbook's answer.
   Sum rule (Section 5.3). Basics: derivative rules

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