Example 26: Derivatives of $e^{-x}$⁠, $\sin(\log x)$⁠, $\cos^{-1}(e^x)$⁠, $e^{\cos x}$

Basics you need

Part (i): $y = e^{-x}$

1. Name the pieces

   Inner: $t = -x$⁠. Outer: $y = e^t$⁠. Both are defined and differentiable for every real number, so $y = e^{-x}$ is differentiable on all of $\R$⁠.

   To evaluate $e^{-x}$ you first change the sign of $x$⁠, then raise $e$ to that power. The job done first is the inner function.
   Composite functions. Basics: composite functions

2. Differentiate each piece

   $$\begin{gathered} \frac{dt}{dx} = \frac{d}{dx}(-x) = -1, \\ \frac{dy}{dt} = \frac{d}{dt}e^t = e^t. \end{gathered}$$
   $-x = (-1)\cdot x$⁠, and a constant multiple differentiates to the same constant times the derivative of $x$⁠, which is $1$⁠. The outer derivative is Theorem 5 (1), written in the variable $t$⁠.
   Theorem 5 (1), Section 5.4. Lesson: exponential and logarithmic functions

3. Multiply

   $$\frac{dy}{dx} = e^t\cdot(-1) = -e^{-x}, \qquad x \in \R.$$
   The answer is negative for every $x$⁠, because $e^{-x} > 0$⁠: the graph of $e^{-x}$ falls everywhere. It is the mirror image of $e^x$ in the $y$-axis, and mirroring turns every slope into its negative.
   Chain rule (Theorem 4, Section 5.3.1). Lesson: chain rule

Part (ii): $y = \sin(\log x)$⁠, $x > 0$

1. Name the pieces, and the domain

   Inner: $t = \log x$⁠, defined and differentiable for $x > 0$⁠. Outer: $y = \sin t$⁠, differentiable for all real $t$⁠. So $y$ is differentiable for every $x > 0$⁠, which is exactly the condition in the question.

   The restriction $x > 0$ in the question is not decoration: $\log x$ does not exist otherwise (Example 25).
   Domain of $\log$ is $\R^+$ (Section 5.4). Basics: exponentials and logarithms

2. Differentiate each piece

   $$\frac{dt}{dx} = \frac{d}{dx}\log x = \frac1x, \qquad \frac{dy}{dt} = \cos t.$$
   The first is Theorem 5 (2); the second is from Table 5.3.
   Theorem 5 (2), Section 5.4; Table 5.3. Basics: derivative rules

3. Multiply

   $$\frac{dy}{dx} = \cos t\cdot\frac1x = \frac{\cos(\log x)}{x}, \qquad x > 0.$$
   We substitute $t = \log x$ back so the answer is in $x$ alone. Keep the brackets: $\cos(\log x)$ is the cosine of the number $\log x$⁠, not $\cos$ times $\log x$⁠.
   Chain rule (Theorem 4). Lesson: chain rule

Part (iii): $y = \cos^{-1}(e^x)$

1. Where is this function defined?

   $\cos^{-1}$ accepts only inputs in $[-1, 1]$⁠. The input here is $e^x$⁠, which is always positive, so we need

   $$0 < e^x \le 1 \iff x \le 0.$$

   So $y = \cos^{-1}(e^x)$ is defined exactly for $x \le 0$⁠. The derivative of $\cos^{-1}t$ exists only for $-1 < t < 1$⁠, so we will also need $e^x < 1$⁠, that is, $x < 0$⁠.

   The book does not state a domain for this part, but the formula only makes sense where $e^x$ is a legal input for $\cos^{-1}$⁠. For $x > 0$⁠, $e^x > 1$ and $\cos^{-1}(e^x)$ does not exist (for example $e^1 \approx 2.718$⁠, and no angle has cosine $2.718$⁠).
   Domain of $\cos^{-1}$ is $[-1,1]$ (Chapter 2, Section 2.2); $e^x \le 1 \iff x \le 0$ because $e^x$ is increasing with $e^0 = 1$⁠. Basics: exponentials and logarithms

2. Name the pieces and differentiate each

   Inner: $t = e^x$⁠, with $\frac{dt}{dx} = e^x$⁠. Outer: $y = \cos^{-1}t$⁠, with

   $$\frac{dy}{dt} = \frac{-1}{\sqrt{1 - t^2}}, \qquad -1 < t < 1.$$
   For $x < 0$ we have $0 < t = e^x < 1$⁠, inside the interval where the derivative of $\cos^{-1}$ exists, so the chain rule applies.
   Derivative of $\cos^{-1}$⁠, table after Example 24 (Section 5.3.3). Lesson: derivatives of inverse trig functions

3. Multiply and simplify

   $$\begin{aligned} \frac{dy}{dx} &= \frac{-1}{\sqrt{1 - (e^x)^2}}\cdot e^x \\ &= \frac{-e^x}{\sqrt{1 - e^{2x}}}, \qquad x < 0. \end{aligned}$$
   $(e^x)^2 = e^{x}\cdot e^{x} = e^{2x}$ by the index law $a^m a^n = a^{m+n}$⁠. Note $e^{2x}$ is not $e^{x^2}$⁠. At $x = 0$ the denominator is $\sqrt{1-1} = 0$⁠: there the graph ends with a vertical tangent, so $y$ is not differentiable at $0$⁠.
   Index law $(e^x)^2 = e^{2x}$⁠. Basics: exponentials and logarithms

4. Picture

   The graph confirms the sign: $\frac{dy}{dx} < 0$ for all $x < 0$⁠, so $y$ decreases. This makes sense, as $e^x$ increases and $\cos^{-1}$ is a decreasing function.
   A negative derivative means the graph falls. Basics: slope at a point

Part (iv): $y = e^{\cos x}$

1. Name the pieces

   Inner: $t = \cos x$⁠. Outer: $y = e^t$⁠. Both are differentiable everywhere, so $y$ is differentiable for every real $x$⁠.

   To evaluate $e^{\cos x}$⁠: first find $\cos x$⁠, then raise $e$ to that power.
   Composite functions. Basics: composite functions

2. Differentiate each piece

   $$\frac{dt}{dx} = -\sin x, \qquad \frac{dy}{dt} = e^t.$$
   Table 5.3 gives $\frac{d}{dx}\cos x = -\sin x$⁠; Theorem 5 (1) gives the outer derivative.
   Table 5.3; Theorem 5 (1). Basics: derivative rules

3. Multiply

   $$\begin{aligned} \frac{dy}{dx} &= e^{\cos x}\cdot(-\sin x) \\ &= -(\sin x)\,e^{\cos x}, \qquad x\in\R. \end{aligned}$$
   Since $e^{\cos x} > 0$⁠, the sign of the derivative is the sign of $-\sin x$⁠: the function falls on $(0, \pi)$ and rises on $(\pi, 2\pi)$⁠. It is largest ($e$⁠) where $\cos x = 1$ and smallest ($e^{-1}$⁠) where $\cos x = -1$⁠.
   Chain rule (Theorem 4). Lesson: chain rule

Every fact used, and where it comes from