Exercise 5.4 · Q4: Differentiate $\sin(\tan^{-1}e^{-x})$

Basics you need

Solution

1. Where is the function defined?

   $e^{-x}$ is defined for every real $x$ (and is positive); $\tan^{-1}$ accepts every real number; $\sin$ accepts every real number. So $y$ is defined for all $x \in \R$⁠. In fact, since $e^{-x} > 0$⁠, the angle $\tan^{-1}e^{-x}$ lies in $\left(0, \frac{\pi}{2}\right)$⁠.

   Unlike $\sin^{-1}$⁠, the inverse tangent has no restriction on its input, so nothing is excluded. Knowing that the angle is between $0$ and $\frac{\pi}{2}$ will let us draw it in a right triangle in Step 5.
   Domain $\R$ and principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ of $\tan^{-1}$⁠, Chapter 2. Lesson: derivatives of inverse trig functions

2. Name the three links and differentiate each

   $u = e^{-x}$⁠, $\ v = \tan^{-1}u$⁠, $\ y = \sin v$⁠.

   $$\begin{aligned} \frac{du}{dx} &= -e^{-x}, \\ \frac{dv}{du} &= \frac{1}{1+u^2}, \\ \frac{dy}{dv} &= \cos v. \end{aligned}$$
   The order of the links is the order of calculation: first $e^{-x}$⁠, then the inverse tangent of that number, then the sine of the angle. Each derivative is a standard one: $e^{-x}$ from Example 26 (i), $\tan^{-1}$ from the table after Example 24, $\sin$ from Table 5.3. All three hold for every real input, so the chain rule applies at every $x$⁠.
   Standard derivatives: Section 5.3.3 for $\tan^{-1}$⁠, Section 5.4 for $e^{-x}$⁠. Basics: derivative rules

3. Multiply and substitute back

   $$\begin{aligned} \frac{dy}{dx} &= \cos v\cdot\frac{1}{1+u^2}\cdot\left(-e^{-x}\right) \\ &= -\frac{e^{-x}\cos\left(\tan^{-1}e^{-x}\right)}{1 + e^{-2x}}. \end{aligned}$$

   Here $u^2 = (e^{-x})^2 = e^{-2x}$⁠, and $v = \tan^{-1}e^{-x}$⁠. The denominator $1 + e^{-2x}$ is always positive, so the formula holds for every real $x$⁠.

   Each rate is evaluated at the value entering its link: the sine link sees the angle $v = \tan^{-1}e^{-x}$⁠, so its rate is $\cos(\tan^{-1}e^{-x})$⁠; the inverse-tangent link sees $u = e^{-x}$⁠, so its rate is $\frac{1}{1+e^{-2x}}$⁠. The index law $(e^{a})^2 = e^{2a}$ turns $(e^{-x})^2$ into $e^{-2x}$⁠.
   Chain rule for three functions, Section 5.3.1. Lesson: the chain rule

4. Simplify the cosine with a right triangle

   Let $v = \tan^{-1}u$ with $u = e^{-x} > 0$⁠. Then $v$ is an acute angle with $\tan v = u$⁠, so it sits in a right triangle with adjacent side $1$⁠, opposite side $u$ and hypotenuse $\sqrt{1+u^2}$⁠:

   $$\begin{aligned} \cos v &= \frac{1}{\sqrt{1+e^{-2x}}}, \\ \sin v &= \frac{e^{-x}}{\sqrt{1+e^{-2x}}}. \end{aligned}$$

   So the answer of Step 3 can also be written

   $$\begin{aligned} \frac{dy}{dx} &= -\frac{e^{-x}}{\left(1+e^{-2x}\right)^{3/2}} \\ &= -\frac{e^{2x}}{\left(1+e^{2x}\right)^{3/2}}. \end{aligned}$$

   (For the last form, multiply top and bottom by $e^{3x} = \left(e^{2x}\right)^{3/2}$⁠; then $e^{3x}e^{-x} = e^{2x}$ and $e^{2x}\left(1 + e^{-2x}\right) = e^{2x} + 1$⁠.)

   The triangle picture is legitimate because $v$ is genuinely between $0$ and $\frac{\pi}{2}$ (Step 1), so $\cos v > 0$ and the positive square root is the right one. Multiplying $\frac{1}{1+e^{-2x}}$ by $\frac{1}{\sqrt{1+e^{-2x}}}$ gives the power $\frac{3}{2}$⁠.
   Right-triangle ratios and Pythagoras. Basics: trigonometric identities

5. Cross-check: simplify first, then differentiate

   By Step 4, $y = \sin v = \dfrac{e^{-x}}{\sqrt{1+e^{-2x}}}$⁠. Multiplying top and bottom by $e^{x}$ (and using $e^x = \sqrt{e^{2x}}$⁠) gives

   $$y = \frac{1}{\sqrt{e^{2x} + 1}} = \left(1 + e^{2x}\right)^{-1/2}.$$

   Now the chain rule with inner $w = 1 + e^{2x}$ ($w' = 2e^{2x}$⁠) and outer $w^{-1/2}$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= -\tfrac12\left(1+e^{2x}\right)^{-3/2}\cdot 2e^{2x} \\ &= -\frac{e^{2x}}{\left(1+e^{2x}\right)^{3/2}}. \end{aligned}$$

   This is the same as Step 4, so the two routes agree.

   Getting the same result by two independent methods is strong evidence that neither contains a slip. The power rule with $n = -\frac12$ is used the same way the textbook uses fractional powers (Example 39), and $\frac{d}{dx}e^{2x} = 2e^{2x}$ is the chain rule with Theorem 5.
   Power rule and chain rule. Basics: derivative rules

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