Miscellaneous · Q5: Differentiate $\cos^{-1}\frac{x}{2}$ over $\sqrt{2x+7}$

Basics you need

Solution

1. Both parts are defined and differentiable on $(-2,2)$

   Dividing $-2 < x < 2$ by 2 gives $-1 < \frac{x}{2} < 1$⁠. So $\cos^{-1}\frac x2$ is defined, and its inner value is strictly inside $(-1,1)$⁠, where the derivative formula holds.

   For the denominator, $-2 < x < 2$ gives $3 < 2x + 7 < 11$⁠. So $2x + 7 > 0$⁠, the root $v = \sqrt{2x+7}$ is defined, differentiable and never zero.

   The quotient rule needs $u$ and $v$ differentiable and $v \ne 0$⁠. The given interval delivers all three. At $x = \pm 2$ the inner value $\frac x2 = \pm1$ would make $\sqrt{1 - w^2} = 0$⁠, which is why the end points are excluded.
   $\cos^{-1}: [-1,1] \to [0,\pi]$ (Chapter 2, Section 2.2); the formula for its derivative needs $w \in (-1,1)$⁠.

2. Differentiate the numerator $u = \cos^{-1}\frac{x}{2}$

   Outer function $\cos^{-1}$⁠, inner function $w = \frac x2$ with $\frac{dw}{dx} = \frac12$⁠. By the chain rule:

   $$\frac{du}{dx} = -\frac{1}{\sqrt{1 - \frac{x^2}{4}}}\cdot\frac12.$$

   Simplify the root: $1 - \frac{x^2}{4} = \frac{4 - x^2}{4}$⁠, so $\sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4-x^2}}{2}$⁠. Therefore

   $$\frac{du}{dx} = -\frac{2}{\sqrt{4-x^2}}\cdot\frac12 = -\frac{1}{\sqrt{4-x^2}}.$$
   $\sqrt 4 = 2$ comes out of the root, and it cancels the $\frac12$ from the inner function. Here $4 - x^2 > 0$ on the interval, so the root is a positive real number.
   Chain rule, Theorem 4, Section 5.3.1; $\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}}$⁠, table after Example 24, Section 5.3.3. Lesson: derivatives of inverse trig functions

3. Differentiate the denominator $v = \sqrt{2x+7}$

   Outer function $\sqrt{\ }$ (derivative $\frac{1}{2\sqrt{s}}$⁠), inner function $s = 2x+7$ with $\frac{ds}{dx} = 2$⁠:

   $$\frac{dv}{dx} = \frac{1}{2\sqrt{2x+7}}\cdot 2 = \frac{1}{\sqrt{2x+7}}.$$
   Again the inner derivative cancels the 2 in $\frac{1}{2\sqrt s}$⁠. The formula $\frac{d}{ds}\sqrt s = \frac{1}{2\sqrt s}$ needs $s > 0$⁠, which Step 1 gave.
   $\frac{d}{ds}s^{1/2} = \frac12 s^{-1/2}$⁠; chain rule. Lesson: chain rule

4. Apply the quotient rule

   Here $v^2 = 2x + 7$⁠. With $u'$ from Step 2 and $v'$ from Step 3:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{u'v - uv'}{v^2} \\ &= \frac{1}{2x+7}\Bigl[-\frac{\sqrt{2x+7}}{\sqrt{4-x^2}} \\ &\qquad - \frac{\cos^{-1}\frac x2}{\sqrt{2x+7}}\Bigr]. \end{aligned}$$
   The first term of the numerator is $u'v = -\frac{1}{\sqrt{4-x^2}}\cdot\sqrt{2x+7}$⁠; the second is $uv' = \cos^{-1}\frac{x}{2}\cdot\frac{1}{\sqrt{2x+7}}$⁠, subtracted. The order matters: it is $u'v - uv'$⁠, not the other way round.
   Quotient rule (3), Section 5.3. Basics: derivative rules

5. Divide each term by $2x+7$

   Split the fraction and divide each term separately. Using $\frac{\sqrt{2x+7}}{2x+7} = \frac{1}{\sqrt{2x+7}}$ and $\sqrt{2x+7}\,(2x+7) = (2x+7)^{3/2}$⁠:

   $$\begin{aligned} \frac{dy}{dx} = {}&-\frac{1}{\sqrt{4-x^2}\,\sqrt{2x+7}} \\ &- \frac{\cos^{-1}\frac x2}{(2x+7)^{3/2}}. \end{aligned}$$

   Taking out the common minus sign gives the key's form:

   $$\begin{aligned} \frac{dy}{dx} = {}&-\Bigl[\frac{1}{\sqrt{4-x^2}\,\sqrt{2x+7}} \\ &\quad + \frac{\cos^{-1}\frac x2}{(2x+7)^{3/2}}\Bigr]. \end{aligned}$$
   Both simplifications use $a = \sqrt a\cdot\sqrt a$ for $a = 2x + 7 > 0$⁠: one $\sqrt a$ cancels in the first term, and in the second $\sqrt a\cdot a = a^{1/2}a^1 = a^{3/2}$⁠.
   Laws of indices: $a^{1/2}\cdot a = a^{3/2}$ for $a > 0$⁠. Basics: solving for dy/dx

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