Exercise 5.5 · Q10: Differentiate $x^{x\cos x}+\frac{x^2+1}{x^2-1}$

Basics you need

Solution

1. Where is the function defined? Split it into two terms

   Put

   $$u = x^{x\cos x}, \qquad v = \frac{x^2+1}{x^2-1},$$

   so $y = u + v$⁠. The power $u$ has the variable exponent $x\cos x$⁠, so its base must be positive: $x > 0$⁠. The fraction $v$ needs $x^2 - 1 \ne 0$⁠, that is $x \ne 1$ and $x \ne -1$⁠. Together:

   $$x > 0, \quad x \ne 1.$$

   On this set $u > 0$⁠. By the sum rule, $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$⁠.

   A power such as $x^{x\cos x}$ with a non-integer, varying exponent is only defined (in this chapter) for a positive base; Section 5.5 says $u(x)$ "must always be positive". We split the sum because $\log(u + v)$ cannot be broken into simpler pieces: the log laws handle products, quotients and powers, never sums.
   Sum rule $(u + v)' = u' + v'$ (Section 5.3); positivity requirement (Section 5.5). Lesson: logarithmic differentiation

2. Take logarithms of $u$

   For $x > 0$⁠, $u > 0$⁠, and the law $\log p^n = n\log p$ gives

   $$\log u = x\cos x\cdot\log x.$$
   The log brings the awkward exponent $x\cos x$ down as an ordinary factor. The right side is now a product of three familiar functions, $x$⁠, $\cos x$ and $\log x$⁠, which the product rule can handle.
   $\log_b p^n = n\log_b p$ (Section 5.4). Basics: exponentials and logarithms

3. Differentiate both sides

   Left side: by the chain rule (outer $\log$⁠, inner $u$⁠), $\frac{d}{dx}\log u = \frac1u\frac{du}{dx}$⁠.

   Right side: treat it as $(x\cos x)\cdot\log x$⁠. First, by the product rule,

   $$\begin{aligned} \frac{d}{dx}(x\cos x) &= 1\cdot\cos x + x(-\sin x) \\ &= \cos x - x\sin x. \end{aligned}$$

   Then, by the product rule again,

   $$\begin{aligned} &\frac{d}{dx}\big[(x\cos x)\log x\big] \\ &= (\cos x - x\sin x)\log x + x\cos x\cdot\frac1x \\ &= \cos x\log x - x\sin x\log x + \cos x \\ &= \cos x\,(1 + \log x) - x\sin x\log x. \end{aligned}$$

   So

   $$\frac1u\cdot\frac{du}{dx} = \cos x\,(1 + \log x) - x\sin x\log x.$$
   A product of three factors is handled by grouping two of them. In the last line, $x\cos x\cdot\frac1x = \cos x$ (allowed because $x \ne 0$⁠), and the two $\cos x$ terms are collected as $\cos x\,(1 + \log x)$⁠, which is the form the book's answer uses.
   Product rule; $(\cos x)' = -\sin x$ (Table 5.3); $(\log x)' = \frac1x$ (Theorem 5). Basics: derivative rules

4. Multiply by $u$

   $$\begin{aligned} \frac{du}{dx} = \ &x^{x\cos x}\big[\cos x\,(1 + \log x) \\ &\quad - x\sin x\log x\big]. \end{aligned}$$
   The left side of Step 3 was $\frac1u\frac{du}{dx}$⁠; multiplying by $u = x^{x\cos x}$ isolates $\frac{du}{dx}$⁠. This holds for every $x > 0$⁠.
   Logarithmic differentiation: $u' = u\cdot(\log u)'$⁠. Lesson: logarithmic differentiation

5. Differentiate $v$ by the quotient rule

   Numerator $x^2 + 1$ has derivative $2x$⁠; denominator $x^2 - 1$ has derivative $2x$⁠. For $x \ne \pm 1$⁠:

   $$\begin{aligned} \frac{dv}{dx} &= \frac{2x(x^2-1) - (x^2+1)\cdot 2x}{(x^2-1)^2} \\ &= \frac{2x\big[(x^2-1) - (x^2+1)\big]}{(x^2-1)^2} \\ &= \frac{2x\cdot(-2)}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}. \end{aligned}$$

   A quick second route: $v = 1 + \frac{2}{x^2-1}$⁠, so $\frac{dv}{dx} = -\frac{2\cdot 2x}{(x^2-1)^2}$⁠, the same.

   No logs are needed for $v$⁠: it is a quotient of polynomials. Taking out the common factor $2x$ makes the cancellation $x^2 - 1 - x^2 - 1 = -2$ easy to see. The second route rewrites $v$ as $\frac{(x^2 - 1) + 2}{x^2-1}$ and uses the chain rule on $2(x^2-1)^{-1}$⁠.
   Quotient rule, valid where the denominator is not zero (Section 5.3). Basics: derivative rules

6. Add the two derivatives

   $$\begin{aligned} \frac{dy}{dx} = \ &x^{x\cos x}\big[\cos x\,(1 + \log x) \\ &\quad - x\sin x\log x\big] - \frac{4x}{(x^2-1)^2}, \end{aligned}$$

   for $x > 0$⁠, $x \ne 1$⁠.

   This is the sum rule from Step 1 with Steps 4 and 5 put in. The condition combines $x > 0$ (needed for $u$⁠) with $x \ne 1$ (needed for $v$⁠); $x = -1$ is already excluded by $x > 0$⁠.
   Sum rule (Section 5.3). Basics: derivative rules

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