Exercise 5.2 · Q7: Differentiate $2\sqrt{\cot(x^2)}$

Basics you need

Solution

1. Where is the function defined and differentiable?

   • $\cot(x^2)$ needs $\sin(x^2) \ne 0$⁠, so $x^2$ is not a multiple of $\pi$⁠; in particular $x \ne 0$⁠.
   • The square root needs $\cot(x^2) \ge 0$⁠, and its derivative $\frac{1}{2\sqrt w}$ needs $\cot(x^2) > 0$⁠.

   Since $x^2 \ge 0$⁠, the condition $\cot(x^2) > 0$ means

   $$\begin{gathered} n\pi < x^2 < n\pi + \tfrac{\pi}{2} \\ \text{for some } n = 0, 1, 2, \ldots \end{gathered}$$

   The first such piece is $0 < x^2 < \frac{\pi}{2}$⁠, that is $0 < |x| < 1.2533\ldots$⁠. We work at points of this kind.

   Cotangent is positive exactly when sine and cosine have the same sign: in the first quarter-turn $(0, \frac{\pi}{2})$ and in $(\pi, \frac{3\pi}{2})$⁠, and every $\pi$ after that. Where $\cot(x^2) = 0$ the function is still defined, but the square-root link has no derivative there (its graph meets the axis vertically), so the chain rule cannot be applied.
   Signs of $\sin$ and $\cos$ in the four quadrants (Class XI). Basics: trigonometric identities

2. Name the three links

   $$t = x^2, \qquad w = \cot t, \qquad y = 2\sqrt{w}.$$
   On a calculator you would square $x$ first, then take the cotangent, then the square root, then double. The innermost operation is the first link. The constant factor $2$ simply stays in front.
   Composite of three functions. Basics: composite functions

3. Derive $\frac{d}{dt}(\cot t)$ with the quotient rule

   Write $\cot t = \dfrac{\cos t}{\sin t}$⁠: numerator $p = \cos t$⁠, $p' = -\sin t$⁠; denominator $q = \sin t$⁠, $q' = \cos t$⁠. Where $\sin t \ne 0$⁠:

   $$\begin{aligned} \frac{d}{dt}(\cot t) &= \frac{(-\sin t)\sin t - \cos t\cos t}{\sin^2 t} \\ &= -\frac{\sin^2 t + \cos^2 t}{\sin^2 t} \\ &= -\frac{1}{\sin^2 t} = -\operatorname{cosec}^2 t. \end{aligned}$$
   Table 5.3 gives only $\sin$⁠, $\cos$⁠, $\tan$ and $x^n$⁠. Writing cotangent as a quotient of sine and cosine lets the quotient rule produce its derivative, and the identity $\sin^2 t + \cos^2 t = 1$ collapses the numerator to $1$⁠.
   Quotient rule (3), Section 5.3; $(\sin t)' = \cos t$⁠, $(\cos t)' = -\sin t$ from Table 5.3. Basics: derivative rules

4. The other two link rates

   $$\begin{aligned} \frac{dy}{dw} &= 2\cdot\frac{1}{2\sqrt w} = \frac{1}{\sqrt w}, \\ \frac{dt}{dx} &= 2x. \end{aligned}$$
   $\sqrt w = w^{1/2}$⁠, so the power rule gives $\frac12 w^{-1/2} = \frac{1}{2\sqrt w}$⁠; the factor $2$ in front of the root cancels the $\frac12$⁠. This needs $w = \cot(x^2) > 0$⁠, which Step 1 arranged.
   Power rule $(w^n)' = nw^{n-1}$ with $n = \frac12$⁠, and with $n = 2$ for $x^2$ (Table 5.3). Basics: derivative rules

5. Multiply the three rates and substitute back

   $$\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dw}\cdot\frac{dw}{dt}\cdot\frac{dt}{dx} \\ &= \frac{1}{\sqrt{\cot t}}\cdot(-\operatorname{cosec}^2 t)\cdot 2x \\ &= -\frac{2x\operatorname{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}. \end{aligned}$$
   Each rate is evaluated at the value entering its link: the root sees $w = \cot(x^2)$⁠, the cotangent sees $t = x^2$⁠. This is already a complete answer, valid wherever $\cot(x^2) > 0$⁠.
   Chain rule for three functions, Section 5.3.1. Lesson: the chain rule

6. Rewrite in the textbook's form, carefully

   Write $s = \sin x^2$ and $c = \cos x^2$⁠. On our domain $\frac{c}{s} > 0$⁠, so $s$ and $c$ have the same sign and $sc > 0$⁠. Since $\frac{c}{s} = \frac{sc}{s^2}$⁠,

   $$\sqrt{\cot x^2} = \sqrt{\frac{sc}{s^2}} = \frac{\sqrt{sc}}{|s|}.$$

   Also $sc = \frac12\sin 2x^2$⁠, so $\sqrt{sc} = \dfrac{\sqrt{\sin 2x^2}}{\sqrt 2}$⁠. Therefore

   $$\begin{aligned} \frac{\operatorname{cosec}^2 x^2}{\sqrt{\cot x^2}} &= \frac{1}{s^2}\cdot\frac{|s|}{\sqrt{sc}} = \frac{1}{|s|\sqrt{sc}} \\ &= \frac{\sqrt 2}{|\sin x^2|\sqrt{\sin 2x^2}}, \end{aligned}$$

   and the derivative becomes

   $$\frac{dy}{dx} = -\frac{2\sqrt2\,x}{|\sin x^2|\,\sqrt{\sin 2x^2}}.$$

   Where $\sin x^2 > 0$ (for instance on the first piece $0 < x^2 < \frac{\pi}{2}$⁠) we have $|\sin x^2| = \sin x^2$⁠, and this is exactly the textbook's answer $-\dfrac{2\sqrt2\,x}{\sin x^2\sqrt{\sin 2x^2}}$⁠.

   The step that needs care is $\sqrt{s^2} = |s|$⁠. If we wrote $\sqrt{s^2} = s$ we would get the textbook's form everywhere, but on the pieces $n\pi < x^2 < n\pi + \frac{\pi}{2}$ with $n$ odd (such as $\pi < x^2 < \frac{3\pi}{2}$⁠) the sine is negative, and the printed form then has the wrong sign. The numbers in the check below show this.
   $\sqrt{a^2} = |a|$ for every real $a$⁠; $\sin 2\theta = 2\sin\theta\cos\theta$⁠. Basics: trigonometric identities

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