Exercise 5.2 · Q8: Differentiate $\cos\sqrt x$

Basics you need

Solution

1. Domain, and name the inner and outer functions

   $\sqrt x$ is real only for $x \ge 0$⁠, and cosine accepts every real number, so $y$ is defined for $x \ge 0$⁠. Put

   $$t = \sqrt{x}, \qquad y = \cos t.$$
   The inner function is the one applied first: you take the square root of $x$⁠, then the cosine of the result. Naming it $t$ turns one complicated function into two simple ones.
   Composite function $y = (v\circ u)(x)$ with $u(x) = \sqrt x$⁠, $v(t) = \cos t$⁠. Basics: composite functions

2. Differentiate each link

   $$\frac{dy}{dt} = -\sin t, \qquad \frac{dt}{dx} = \frac{1}{2\sqrt{x}} \quad (x > 0).$$
   The first is straight from Table 5.3. The second is the power rule with $n = \frac12$⁠: $\frac12 x^{-1/2}$⁠, and $x^{-1/2} = \frac{1}{\sqrt x}$⁠. At $x = 0$ this would need division by $0$⁠, so the inner link has no derivative there and the chain rule (which needs $\frac{dt}{dx}$ to exist) applies only for $x > 0$⁠.
   Power rule $(x^n)' = nx^{n-1}$⁠, here $n = \frac12$⁠. Basics: derivative rules

3. Multiply and substitute back

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\cdot\frac{dt}{dx} = (-\sin t)\cdot\frac{1}{2\sqrt x} \\ &= -\frac{\sin\sqrt{x}}{2\sqrt{x}}. \end{aligned}$$
   The book's practice is to give the final answer in terms of $x$ only (as in Example 21), so we replace $t$ by $\sqrt x$⁠. The sine is evaluated at $\sqrt x$⁠, the value that entered the cosine.
   Chain rule, Theorem 4, Section 5.3.1. Lesson: the chain rule

4. What about $x = 0$⁠?

   The formula is meaningless at $x = 0$⁠, but the function has a right-hand derivative there. Using $1 - \cos\theta = 2\sin^2\frac{\theta}{2}$ with $\theta = \sqrt h$⁠, for $h > 0$⁠:

   $$\begin{aligned} \frac{\cos\sqrt h - \cos 0}{h} &= -\frac{2\sin^2(\sqrt h/2)}{h} \\ &= -\frac12\left(\frac{\sin(\sqrt h/2)}{\sqrt h/2}\right)^2, \end{aligned}$$

   which tends to $-\frac12\cdot 1^2 = -\frac12$ as $h \to 0^+$⁠.

   This step goes beyond the question, which (like the textbook's answer) concerns $x > 0$⁠. It shows that "the chain rule does not apply" is not the same as "there is no derivative": the chain rule is a sufficient method, not the definition. Only the right-hand derivative makes sense at $0$⁠, because $y$ is not defined for $x < 0$⁠.
   $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$ (Class XI, Limits and Derivatives). Basics: left-hand and right-hand limits

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