Exercise 5.2 · Q4: Differentiate $\sec(\tan\sqrt x)$

Basics you need

Solution

1. Where is the function defined?

   Working from the inside out:

   • $\sqrt{x}$ needs $x \ge 0$⁠.
   • $\tan(\sqrt{x})$ needs $\cos\sqrt{x} \ne 0$⁠, that is $\sqrt{x}$ not an odd multiple of $\frac{\pi}{2}$⁠.
   • $\sec(\tan\sqrt{x})$ needs $\cos(\tan\sqrt{x}) \ne 0$⁠, that is $\tan\sqrt{x}$ not an odd multiple of $\frac{\pi}{2}$⁠.

   For example, the function is defined on $0 \le x < 1.0077\ldots$⁠; the first excluded point is where $\tan\sqrt x = \frac{\pi}{2}$⁠, that is $\sqrt{x} = \tan^{-1}\frac{\pi}{2} \approx 1.00388$⁠, $x \approx 1.00778$⁠.

   A derivative can only exist where the function itself is defined. Listing the conditions now tells us where the final formula is meaningful, and the same conditions will reappear as "denominator $\ne 0$" conditions in the derivative.
   $\sec$ and $\tan$ are undefined exactly where cosine is zero, at odd multiples of $\frac{\pi}{2}$⁠. Basics: trigonometric identities

2. Name the three links

   $$u = \sqrt{x}, \qquad v = \tan u, \qquad y = \sec v.$$
   On a calculator you would press $\sqrt{\ }$ first, then tan, then sec (that is, $1/\cos$⁠). The order in which the functions are applied gives the order of the links, innermost first.
   Composite of three functions. Basics: composite functions

3. Derive $\frac{d}{dv}(\sec v)$ with the quotient rule

   Write $\sec v = \dfrac{1}{\cos v}$⁠, with numerator $p = 1$ ($p' = 0$⁠) and denominator $q = \cos v$ ($q' = -\sin v$⁠). Where $\cos v \ne 0$⁠:

   $$\begin{aligned} \frac{d}{dv}(\sec v) &= \frac{0\cdot\cos v - 1\cdot(-\sin v)}{\cos^2 v} \\ &= \frac{\sin v}{\cos^2 v} = \frac{1}{\cos v}\cdot\frac{\sin v}{\cos v} \\ &= \sec v\,\tan v. \end{aligned}$$

   So $\dfrac{dy}{dv} = \sec v\tan v$⁠.

   Table 5.3 lists only $x^n$⁠, $\sin$⁠, $\cos$ and $\tan$⁠. Every other trigonometric derivative follows from these by the quotient rule, so we derive the one we need instead of quoting it from memory. Splitting $\frac{\sin v}{\cos^2 v}$ into two factors shows the familiar form $\sec v\tan v$⁠.
   Quotient rule (3), Section 5.3, valid where the denominator $\cos v \ne 0$⁠. Basics: derivative rules

4. The other two link rates

   $$\begin{aligned} \frac{dv}{du} &= \frac{d}{du}(\tan u) = \sec^2 u, \\ \frac{du}{dx} &= \frac{d}{dx}\,x^{1/2} = \frac{1}{2\sqrt{x}}. \end{aligned}$$

   The first holds wherever $\cos u \ne 0$⁠; the second needs $x > 0$⁠.

   $(\tan u)' = \sec^2 u$ is in Table 5.3. For the square root, the power rule with $n = \frac12$ gives $\frac12 x^{-1/2}$⁠, and $x^{-1/2} = \frac{1}{\sqrt x}$⁠. At $x = 0$ this is not a real number, so $\sqrt x$ is not differentiable at $0$ and the chain rule cannot be applied there.
   Power rule $(x^n)' = nx^{n-1}$⁠, here with $n = \frac12$ and $x > 0$ (the textbook uses it this way in Example 39). Basics: derivative rules

5. Multiply the three rates and substitute back

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dv}\cdot\frac{dv}{du}\cdot\frac{du}{dx} \\ &= \sec v\tan v\cdot\sec^2 u\cdot\frac{1}{2\sqrt{x}}. \end{aligned}$$

   Now put back $u = \sqrt{x}$ and $v = \tan\sqrt{x}$⁠:

   $$\frac{dy}{dx} = \frac{\sec(\tan\sqrt{x})\,\tan(\tan\sqrt{x})\,\sec^2\!\sqrt{x}}{2\sqrt{x}}.$$
   Each rate is evaluated at the value entering its link: the secant link sees $v = \tan\sqrt x$⁠, so its rate is $\sec(\tan\sqrt x)\tan(\tan\sqrt x)$⁠; the tangent link sees $u = \sqrt x$⁠, so its rate is $\sec^2\sqrt x$⁠. That is why $\tan$ appears applied twice in the answer.
   Chain rule for three functions, Section 5.3.1 (after Theorem 4). Lesson: the chain rule

6. State where the formula holds

   The formula is valid at every $x > 0$ at which the function is defined (Step 1). At those points every link is differentiable, and every denominator in the answer is non-zero: $\sqrt x \ne 0$⁠, $\cos\sqrt x \ne 0$⁠, $\cos(\tan\sqrt x) \ne 0$⁠.

   At $x = 0$ the chain rule cannot be used, because the first link $\sqrt x$ has no derivative there. (Going deeper: the function still has a right-hand derivative at $0$⁠, equal to $\frac12$⁠, because near $0$ we have $\tan(\tan\sqrt x)\approx\sqrt x$ and the formula tends to $\frac12$⁠. Proving that needs a separate limit argument, and the textbook's answer, like ours, is meant for $x > 0$⁠.)

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