Example 41: $f'(x)$ for $f(x) = (\sin x)^{\sin x}$⁠, $0 < x < \pi$

Basics you need

Solution

1. Check that the base is positive

   For $0 < x < \pi$⁠, $0 < \sin x \le 1$⁠. So the base $\sin x$ is positive, $y = (\sin x)^{\sin x}$ is a well-defined positive number, and $\log y$ and $\log(\sin x)$ both exist.

   A real power $p^q$ with a non-integer exponent is defined only for $p > 0$⁠, and the logarithm needs a positive input. The interval $0 < x < \pi$ in the question is exactly what makes the method legal. (The textbook says the function "is defined for all positive real numbers". What matters is that the base $\sin x$ is positive; that holds on $(0,\pi)$ but not for every positive $x$⁠: on $(\pi, 2\pi)$⁠, for example, $\sin x < 0$ and the logarithm step fails.)
   $\sin x > 0$ for $0 < x < \pi$ (the upper half of the unit circle). Basics: trigonometric identities

2. Take logarithms

   $$\begin{aligned} \log y &= \log\left[(\sin x)^{\sin x}\right] \\ &= \sin x\,\log(\sin x). \end{aligned}$$
   The exponent $\sin x$ moves down in front of the logarithm. The right side is now a product of two functions of $x$⁠, which the product rule can handle.
   $\log(p^q) = q\log p$ for $p > 0$⁠, here with $p = q = \sin x$⁠. Basics: exponentials and logarithms

3. Differentiate the left side

   $$\frac{d}{dx}(\log y) = \frac{1}{y}\,\frac{dy}{dx}.$$
   $\log y$ is a composite: $y$ is a function of $x$ and $\log$ is applied to it. Chain rule: (derivative of $\log$ at $y$⁠) times $\frac{dy}{dx}$⁠.
   Theorem 5 (2), Section 5.4: $(\log y)' = \frac1y$ with respect to $y$⁠; chain rule, Theorem 4. Lesson: the chain rule

4. Differentiate the right side (product rule)

   Let $u = \sin x$ and $w = \log(\sin x)$⁠. Then $u' = \cos x$⁠, and by the chain rule $w' = \frac{1}{\sin x}\cdot\cos x$⁠. So

   $$\begin{aligned} \frac{d}{dx}\big(\sin x\log(\sin x)\big) &= u'w + uw' \\ &= \cos x\log(\sin x) + \sin x\cdot\frac{\cos x}{\sin x} \\ &= \cos x\log(\sin x) + \cos x \\ &= (1 + \log(\sin x))\cos x. \end{aligned}$$
   The product rule differentiates one factor at a time. In the second term $\sin x$ cancels (allowed, since $\sin x \ne 0$ on $(0,\pi)$⁠). Finally $\cos x$ is a common factor of both terms.
   Product rule (2), Section 5.3; $(\log s)' = \frac1s$ with $s = \sin x$ (Theorem 5 and chain rule). Basics: derivative rules

5. Solve for $\frac{dy}{dx}$

   Equating the two sides, $\frac1y\frac{dy}{dx} = (1 + \log(\sin x))\cos x$⁠. Multiply by $y = (\sin x)^{\sin x}$⁠:

   $$\frac{dy}{dx} = (1 + \log(\sin x))\,(\sin x)^{\sin x}\cos x.$$
   Logarithmic differentiation always ends this way: the left side was $\frac1y\frac{dy}{dx}$⁠, so we multiply by $y$ and replace $y$ by its formula so that the answer is in terms of $x$ alone.
   Since both sides of $\log y = \sin x\log(\sin x)$ are equal for every $x$ in $(0,\pi)$⁠, their derivatives are equal too. Lesson: logarithmic differentiation

6. An alternative route, and what the answer shows

   Since $\sin x > 0$⁠, $y = e^{\sin x\log(\sin x)}$⁠. Differentiating with the chain rule (outer $e^w$⁠, inner $w = \sin x\log(\sin x)$⁠, whose derivative is Step 4) gives

   $$\begin{aligned} y' &= e^{\sin x\log(\sin x)} \\ &\quad\cdot(1+\log(\sin x))\cos x \\ &= (\sin x)^{\sin x}(1+\log(\sin x))\cos x, \end{aligned}$$

   the same answer.

   The derivative is zero where $\cos x = 0$ ($x = \frac{\pi}{2}$⁠) and where $1 + \log(\sin x) = 0$⁠, that is $\sin x = \frac1e$ ($x \approx 0.3767$ and $x \approx \pi - 0.3767$⁠). The graph has flat tangents exactly there:

   The second route avoids writing $\log y$ but rests on the same fact, $p^q = e^{q\log p}$ for $p > 0$⁠. Checking where the derivative is zero against the picture is a quick test that the formula is right: the flat points of the graph are exactly the zeros of $f'$⁠.
   $e^{\log p} = p$ for $p > 0$ (Example 25); $(e^x)' = e^x$ (Theorem 5). Basics: exponentials and logarithms

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