Example 29: Derivative of $x^{\sin x}$⁠, $x>0$

Basics you need

Solution

1. Take logarithms (allowed because $y > 0$⁠)

   Let $y = x^{\sin x}$⁠. For $x > 0$ the base is positive, so $y = e^{\sin x\log x} > 0$ and $\log y$ exists. By the power law for logs,

   $$\log y = \sin x\cdot\log x.$$
   This is why the question says $x > 0$⁠. For $x \le 0$⁠, $\log x$ is undefined and a power like $(-2)^{\sin(-2)} = (-2)^{-0.909\ldots}$ has no real value. The book's general remark for $y = [u(x)]^{v(x)}$ is that $u(x)$ and $f(x)$ must be positive; here $u(x) = x > 0$⁠.
   $\log p^n = n\log p$ with $p = x > 0$ and $n = \sin x$ (Section 5.4). Basics: exponentials and logarithms

2. Differentiate both sides with respect to $x$

   Left (chain rule, outer $\log$⁠, inner $y$⁠): $\frac1y\cdot\frac{dy}{dx}$⁠. Right: a product of $u = \sin x$ and $v = \log x$⁠, so by the product rule

   $$\begin{aligned} \frac1y\cdot\frac{dy}{dx} &= \sin x\cdot\frac{d}{dx}(\log x) \\ &\quad + \log x\cdot\frac{d}{dx}(\sin x) \\ &= \sin x\cdot\frac1x + \log x\cdot\cos x. \end{aligned}$$
   Both factors $\sin x$ and $\log x$ change with $x$⁠, so the product rule is needed: differentiate one factor at a time, keeping the other, and add. The derivatives are $\frac1x$ (Theorem 5) and $\cos x$ (Table 5.3).
   Product rule (Section 5.3); $\frac{d}{dx}\log x = \frac1x$ (Theorem 5). Lesson: logarithmic differentiation

3. Multiply by $y$

   $$\begin{aligned} \frac{dy}{dx} &= y\left[\frac{\sin x}{x} + \cos x\log x\right] \\ &= x^{\sin x}\left[\frac{\sin x}{x} + \cos x\log x\right]. \end{aligned}$$
   Multiplying both sides by $y$ isolates $\frac{dy}{dx}$⁠, and replacing $y$ by $x^{\sin x}$ puts the answer in $x$ alone.
   Rearranging; $y \ne 0$⁠. Basics: solving for dy/dx

4. Expand, and see the two parts

   Multiply out the bracket, using $\frac{x^{\sin x}}{x} = x^{\sin x - 1}$⁠:

   $$\frac{dy}{dx} = x^{\sin x - 1}\cdot\sin x + x^{\sin x}\cdot\cos x\log x.$$

   The two terms have a clear meaning:

   • $\sin x\cdot x^{\sin x - 1}$ is what the power rule would give if the exponent $\sin x$ were frozen;
   • $x^{\sin x}\log x\cdot\cos x$ is what the $a^x$ rule (Example 28) with the chain rule would give if the base $x$ were frozen.
   When both base and exponent vary, the total rate of change is the sum of the two "one at a time" rates. This is the product rule at work behind the scenes, and it is a quick way to check any answer of this type.
   Index law $\frac{x^m}{x} = x^{m-1}$ for $x > 0$⁠. Basics: exponentials and logarithms

5. See it on the graph

   These two points are good tests because one term of the bracket vanishes at each: at $\frac{\pi}{2}$⁠, $\cos x = 0$⁠; at $\pi$⁠, $\sin x = 0$⁠. Near $0$ the function tends to $1$ (open dot), since $\sin x\log x \to 0$⁠, but $x = 0$ itself is not in the domain.
   The derivative at a point is the slope of the tangent there. Basics: slope at a point

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