Exercise 5.4 · Q7: Differentiate $\sqrt{e^{\sqrt x}}$⁠, $x>0$

Basics you need

Solution

1. Where is the function defined, and why $x > 0$⁠?

   $\sqrt{x}$ needs $x \ge 0$⁠. Then $e^{\sqrt x}$ is always positive, so the outer square root is always defined. So $y$ is defined for $x \ge 0$⁠, with $y(0) = \sqrt{e^0} = 1$⁠. The question restricts to $x > 0$ because $y$ has no derivative at $0$ (proved in Step 5); the source of the trouble is the innermost link $\sqrt x$⁠, which has no derivative at $0$⁠.

   Each square root needs a non-negative input. The outer one is safe automatically, since an exponential is never negative. The restriction $x > 0$ is there for the derivative, not for the function: Step 3 will have $\sqrt{x}$ in a denominator.
   Range of the exponential function is the positive reals (Section 5.4). Basics: exponentials and logarithms

2. Name the three links and differentiate each

   $u = \sqrt{x}$⁠, $\ v = e^u$⁠, $\ y = \sqrt{v}$⁠.

   $$\begin{aligned} \frac{du}{dx} &= \frac{1}{2\sqrt{x}} \quad (x > 0), \\ \frac{dv}{du} &= e^u, \\ \frac{dy}{dv} &= \frac{1}{2\sqrt{v}} \quad (v > 0). \end{aligned}$$
   The links are in calculation order: take $\sqrt{x}$⁠, raise $e$ to it, take the square root of the result. The two square-root links both use the power rule with $n = \frac12$⁠; the middle link is Theorem 5. Since $v = e^{\sqrt x} > 0$⁠, the outer link is always differentiable; only the inner link needs $x > 0$⁠.
   Power rule (Table 5.3) and Theorem 5 (1), Section 5.4. Basics: derivative rules

3. Multiply (chain rule) and substitute back

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{2\sqrt{v}}\cdot e^u\cdot\frac{1}{2\sqrt{x}} \\ &= \frac{e^{\sqrt{x}}}{4\sqrt{x}\,\sqrt{e^{\sqrt{x}}}} \\ &= \frac{e^{\sqrt{x}}}{4\sqrt{x\,e^{\sqrt{x}}}}, \qquad x > 0. \end{aligned}$$
   Each rate is evaluated at the value entering its link: the outer root sees $v = e^{\sqrt x}$⁠, the exponential sees $u = \sqrt x$⁠. In the last line the two square roots are combined using $\sqrt{p}\sqrt{q} = \sqrt{pq}$⁠, allowed because $x > 0$ and $e^{\sqrt x} > 0$⁠. This last line is exactly the textbook's form.
   Chain rule for three functions, Section 5.3.1. Lesson: the chain rule

4. Simplify, and check by a second route

   Since $\sqrt{e^{\sqrt x}} = e^{\sqrt{x}/2}$⁠, the middle line of Step 3 simplifies:

   $$\frac{e^{\sqrt{x}}}{4\sqrt{x}\;e^{\sqrt{x}/2}} = \frac{e^{\sqrt{x}/2}}{4\sqrt{x}}.$$

   Second route: $y = e^{\sqrt{x}/2}$ is a two-link chain with inner $t = \frac{\sqrt x}{2}$⁠, $\frac{dt}{dx} = \frac{1}{4\sqrt x}$⁠. So

   $$\frac{dy}{dx} = e^{\sqrt{x}/2}\cdot\frac{1}{4\sqrt{x}} = \frac{e^{\sqrt{x}/2}}{4\sqrt{x}}.$$

   The two routes agree.

   $e^{\sqrt x}$ divided by $e^{\sqrt x/2}$ is $e^{\sqrt x - \sqrt x/2} = e^{\sqrt x/2}$ (index law for division). Rewriting the function first is usually the quicker way; doing it both ways shows the two forms are the same function and catches slips.
   Index laws $(e^a)^{1/2} = e^{a/2}$ and $e^a/e^b = e^{a-b}$⁠. Basics: exponentials and logarithms

5. Why not at $x = 0$⁠?

   The function is defined at $0$⁠, but it has no derivative there. Only the right-hand side is available, since the domain starts at $0$⁠. The right-hand difference quotient is $\dfrac{e^{\sqrt{h}/2} - 1}{h}$ for $h > 0$⁠. Using $e^s \ge 1 + s$ with $s = \frac{\sqrt h}{2}$⁠, it is at least $\dfrac{\sqrt{h}/2}{h} = \dfrac{1}{2\sqrt{h}}$⁠, which grows without bound as $h \to 0^+$⁠. So there is no finite limit, and no derivative at $0$⁠.

   This is why the question says $x > 0$⁠. The graph of $y$ leaves the point $(0, 1)$ with a vertical tangent, just as $\sqrt{x}$ does. The inequality $e^s \ge 1 + s$ for $s \ge 0$ (the graph of $e^s$ lies on or above its tangent line $1 + s$ at $s = 0$⁠) is not proved in Chapter 5; it follows from Theorem 1 of Section 6.3 in Chapter 6, applied to $g(s) = e^s - 1 - s$⁠, whose derivative $e^s - 1$ is positive for $s > 0$⁠, so $g(s) \ge g(0) = 0$⁠.
   Differentiable at a point means the difference quotient has a finite limit (Section 5.3). Lesson: differentiability

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