Exercise 5.7 · Q12: $\frac{d^2y}{dx^2}$ in terms of $y$ for $y=\cos^{-1}x$

Basics you need

Solution

1. Where $y$ lies, and the sign of $\sin y$

   We need $-1 \lt x \lt 1$ for the derivative to exist (Basic 2). By Basic 1, $y = \cos^{-1}x$ lies in $[0, \pi]$⁠. The end values are taken only at the ends: $y = 0$ only for $x = 1$⁠, and $y = \pi$ only for $x = -1$⁠. So for $-1 \lt x \lt 1$⁠,

   $$0 \lt y \lt \pi, \qquad\text{hence}\qquad \sin y \gt 0.$$
   The next step takes a square root, and $\sqrt{\sin^2 y}$ equals $\sin y$ only if $\sin y$ is not negative. The principal range is what guarantees this. Without it the answer could have the wrong sign.
   $\sin y \gt 0$ for $0 \lt y \lt \pi$ (sine is positive in the first and second quadrants). Basics: trigonometric identities

2. Write $\sqrt{1-x^2}$ in terms of $y$

   By Basic 1, $x = \cos(\cos^{-1}x) = \cos y$⁠. So

   $$\begin{aligned} 1 - x^2 &= 1 - \cos^2 y = \sin^2 y, \\ \sqrt{1 - x^2} &= |\sin y| = \sin y, \end{aligned}$$

   the last equality because $\sin y \gt 0$ (Step 1).

   The question wants the answer in $y$ alone, so every $x$ must be replaced. The relation $x = \cos y$ does that, and the Pythagorean identity turns $1 - \cos^2 y$ into $\sin^2 y$⁠.
   Basic 3. Basics: trigonometric identities

3. First derivative in terms of $y$

   By Basic 2 and Step 2, for $-1 \lt x \lt 1$⁠,

   $$\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} = -\frac{1}{\sin y} = -\operatorname{cosec} y.$$
   This is the derivative of $\cos^{-1}x$ rewritten in $y$⁠. You can also get it directly as in Example 24: from $x = \cos y$⁠, $\frac{dx}{dy} = -\sin y$⁠, so $\frac{dy}{dx} = -\frac{1}{\sin y}$⁠.
   Table of derivatives of inverse trigonometric functions, Section 5.3.3. Lesson: derivatives of inverse trig functions

4. Differentiate again, with the chain rule

   $-\operatorname{cosec} y$ is a function of $y$⁠, and $y$ is a function of $x$⁠. By Basics 5 and 4,

   $$\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(-\operatorname{cosec} y) \\ &= -(-\operatorname{cosec} y\cot y)\cdot\frac{dy}{dx} \\ &= \operatorname{cosec} y\cot y\cdot\frac{dy}{dx}. \end{aligned}$$
   We differentiate with respect to $x$⁠, not $y$⁠. So we differentiate $-\operatorname{cosec} y$ as if $y$ were the variable, and then multiply by $\frac{dy}{dx}$⁠, just as in implicit differentiation.
   Chain rule, Theorem 4, Section 5.3.1. Lesson: chain rule

5. Substitute $\frac{dy}{dx} = -\operatorname{cosec} y$

   $$\begin{aligned} \frac{d^2y}{dx^2} &= \operatorname{cosec} y\cot y\cdot(-\operatorname{cosec} y) \\ &= -\cot y\operatorname{cosec}^2 y, \end{aligned}$$

   valid for $0 \lt y \lt \pi$⁠, that is, for $-1 \lt x \lt 1$⁠.

   Replacing $\frac{dy}{dx}$ by its value from Step 3 removes the last trace of $x$⁠, so the answer is in $y$ alone, as asked. This is exactly the answer key's form.

6. Check: differentiate in $x$⁠, convert at the end

   Write $\frac{dy}{dx} = -(1-x^2)^{-\frac12}$⁠. By the chain rule (outer $u^{-\frac12}$⁠, inner $u = 1-x^2$⁠, $\frac{du}{dx} = -2x$⁠):

   $$\begin{aligned} \frac{d^2y}{dx^2} &= -\left(-\tfrac12\right)(1-x^2)^{-\frac32}\cdot(-2x) \\ &= -\frac{x}{(1-x^2)^{3/2}}. \end{aligned}$$

   Now put $x = \cos y$ and $1 - x^2 = \sin^2 y$⁠, so $(1-x^2)^{3/2} = (\sin^2 y)^{3/2} = \sin^3 y$ (again because $\sin y \gt 0$⁠):

   $$\begin{aligned} \frac{d^2y}{dx^2} &= -\frac{\cos y}{\sin^3 y} \\ &= -\frac{\cos y}{\sin y}\cdot\frac{1}{\sin^2 y} \\ &= -\cot y\operatorname{cosec}^2 y. \end{aligned}$$
   An independent route to the same answer. The three minus signs in the first line multiply to a minus. Converting to $y$ only at the end needs the same fact as before, $\sin y \gt 0$⁠.
   Power rule (Table 5.3) with the chain rule; $(p^2)^{3/2} = p^3$ for $p \gt 0$⁠.

Every fact used, and where it comes from