Example 40: Derivatives of $\cos^{-1}(\sin x)$⁠, $\tan^{-1}\frac{\sin x}{1+\cos x}$⁠, $\sin^{-1}\frac{2^{x+1}}{1+4^x}$

Basics you need

Part (i): $f(x) = \cos^{-1}(\sin x)$

1. Domain

   For every real $x$⁠, $-1 \le \sin x \le 1$⁠, which is exactly the domain of $\cos^{-1}$⁠. So $f$ is defined for all real $x$⁠, with values in $[0,\pi]$⁠.

   Before simplifying, we check where the function exists, so that we know which $x$ the final answer must cover: here, every real number.
   $\cos^{-1}$ has domain $[-1,1]$ and range $[0,\pi]$ (Chapter 2, Section 2.2).

2. Rewrite sine as a cosine

   $$\begin{aligned} \sin x &= \cos\left(\tfrac{\pi}{2} - x\right), \quad\text{so} \\ f(x) &= \cos^{-1}\left(\cos\left(\tfrac{\pi}{2} - x\right)\right). \end{aligned}$$
   An inverse cosine "undoes" a cosine, not a sine. The complementary-angle identity turns the sine into a cosine so the two can meet.
   Complementary angles: $\cos(\frac{\pi}{2} - x) = \sin x$ for every real $x$⁠. Basics: trigonometric identities

3. Where does the inverse cancel?

   $\cos^{-1}(\cos\theta) = \theta$ holds exactly when $\theta \in [0,\pi]$⁠. With $\theta = \frac{\pi}{2} - x$⁠:

   $$0 \le \tfrac{\pi}{2} - x \le \pi \iff -\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}.$$

   So $f(x) = \frac{\pi}{2} - x$ for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$⁠. This is the textbook's simplification, and this interval is where it is valid.

   $\cos^{-1}$ can only return an angle in $[0,\pi]$⁠. If $\frac{\pi}{2} - x$ were outside that range, it could not be the output. For instance at $x = \pi$ the formula $\frac{\pi}{2} - x$ gives $-\frac{\pi}{2}$⁠, a negative number, while $f(\pi) = \cos^{-1}(0) = \frac{\pi}{2}$⁠.
   Section 2.3: the cancellation results hold only within the principal value branch. Lesson: inverse trig derivatives

4. Differentiate on $(-\frac{\pi}{2}, \frac{\pi}{2})$

   $$f'(x) = \frac{d}{dx}\left(\tfrac{\pi}{2} - x\right) = -1, \qquad -\tfrac{\pi}{2} < x < \tfrac{\pi}{2}.$$
   On this open interval $f$ is the straight line $\frac{\pi}{2} - x$ of slope $-1$⁠. We use the open interval because at its end points $f$ also has neighbours outside the interval, where the formula is different (Part (i), Step 6).
   The derivative of a constant is 0 (Class XI) and of $x$ is 1 (Table 5.3 with $n = 1$⁠).

5. What happens outside that interval

   Take $\frac{\pi}{2} \le x \le \frac{3\pi}{2}$⁠. Then $\theta = \frac{\pi}{2} - x$ lies in $[-\pi, 0]$⁠, outside the range. But cosine is even, so $\cos\theta = \cos(-\theta)$⁠, and $-\theta = x - \frac{\pi}{2}$ lies in $[0,\pi]$⁠. Hence

   $$f(x) = \cos^{-1}\left(\cos\left(x - \tfrac{\pi}{2}\right)\right) = x - \tfrac{\pi}{2},$$

   with $f'(x) = +1$ on $(\frac{\pi}{2}, \frac{3\pi}{2})$⁠. Because $\sin x$ has period $2\pi$⁠, so does $f$⁠, and the graph is a zigzag:

   To use $\cos^{-1}(\cos\theta) = \theta$ we need an angle in $[0,\pi]$ with the same cosine. Replacing $\theta$ by $-\theta$ supplies one. In general, $f'(x) = -1$ wherever $\cos x > 0$ and $+1$ wherever $\cos x < 0$⁠.
   $\cos(-\theta) = \cos\theta$⁠; $\sin(x + 2\pi) = \sin x$⁠. Basics: trigonometric identities

6. Confirm with the inverse-cosine formula

   Where $|\sin x| < 1$⁠, the chain rule with $u = \sin x$ and $(\cos^{-1}u)' = \frac{-1}{\sqrt{1-u^2}}$ gives

   $$f'(x) = \frac{-\cos x}{\sqrt{1 - \sin^2 x}} = \frac{-\cos x}{|\cos x|},$$

   because $\sqrt{\cos^2 x} = |\cos x|$⁠, not $\cos x$⁠. This is $-1$ when $\cos x > 0$ and $+1$ when $\cos x < 0$⁠, exactly as in Part (i), Step 5.

   At $x = \frac{\pi}{2}$ (where $\sin x = 1$⁠) the formula does not apply, so we use difference quotients. Here $f(\frac{\pi}{2}) = \cos^{-1}1 = 0$⁠, and for small $h > 0$⁠, Part (i), Steps 3 and 5 give $f(\frac{\pi}{2} - h) = h$ and $f(\frac{\pi}{2} + h) = h$⁠. So

   $$\begin{aligned} \text{LHD} &= \lim_{h\to0^+}\frac{f(\frac{\pi}{2}-h) - 0}{-h} = -1, \\ \text{RHD} &= \lim_{h\to0^+}\frac{f(\frac{\pi}{2}+h) - 0}{h} = +1. \end{aligned}$$

   They differ, so $f$ is not differentiable at $\frac{\pi}{2}$⁠; the same happens at every odd multiple of $\frac{\pi}{2}$⁠.

   The formula route and the simplification route must agree, and they do. The step $\sqrt{\cos^2 x} = |\cos x|$ is where the sign information lives: dropping the absolute value would wrongly give $-1$ everywhere.
   Table in Section 5.3.3; $\sqrt{a^2} = |a|$⁠; left and right derivatives (Section 5.3). Lesson: differentiability

Part (ii): $f(x) = \tan^{-1}\frac{\sin x}{1+\cos x}$

1. Domain

   The fraction needs $1 + \cos x \ne 0$⁠, that is $\cos x \ne -1$⁠, that is $x$ is not an odd multiple of $\pi$⁠. $\tan^{-1}$ accepts every real number, so $f$ is defined exactly for

   $$x \ne (2n+1)\pi, \qquad n \in \Z.$$
   The textbook says the function is defined for all real numbers where $\cos x \ne -1$⁠, "i.e., at all odd multiplies of $\pi$" (it means multiples). The odd multiples of $\pi$ are the excluded points: $\cos x = -1$ exactly there.
   $\cos x = -1$ exactly when $x = (2n+1)\pi$⁠. The domain of $\tan^{-1}$ is $\R$⁠. Basics: trigonometric identities

2. Half-angle identities

   $$\begin{aligned} \frac{\sin x}{1+\cos x} &= \frac{2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2} \\ &= \frac{\sin\frac x2}{\cos\frac x2} = \tan\frac x2. \end{aligned}$$
   We cancel one factor $\cos\frac x2$⁠. That is allowed because $\cos\frac x2 = 0$ would need $\frac x2$ to be an odd multiple of $\frac{\pi}{2}$⁠, that is $x$ an odd multiple of $\pi$⁠, and those points are not in the domain.
   $\sin 2A = 2\sin A\cos A$ and $1 + \cos 2A = 2\cos^2 A$⁠, with $A = \frac x2$⁠. Basics: trigonometric identities

3. Where does the inverse cancel?

   $\tan^{-1}(\tan\theta) = \theta$ holds exactly when $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$⁠. With $\theta = \frac x2$ this means $-\pi < x < \pi$⁠. There

   $$\begin{aligned} f(x) &= \tan^{-1}\left(\tan\tfrac x2\right) = \tfrac x2, \\ f'(x) &= \tfrac12. \end{aligned}$$
   This is the textbook's computation, valid on $(-\pi, \pi)$⁠. The principal range of $\tan^{-1}$ is open, so no end points need special treatment.
   Principal range of $\tan^{-1}$⁠: $(-\frac{\pi}{2}, \frac{\pi}{2})$ (Section 2.2).

4. The other intervals of the domain

   On $((2n-1)\pi, (2n+1)\pi)$ the angle $\frac x2 - n\pi$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$⁠, and $\tan(\frac x2 - n\pi) = \tan\frac x2$ because tangent has period $\pi$⁠. So

   $$f(x) = \tfrac x2 - n\pi, \qquad f'(x) = \tfrac12.$$
   Subtracting $n\pi$ shifts the angle into the principal range without changing its tangent. Only a constant changes from piece to piece, and constants do not affect derivatives. So $f'(x) = \frac12$ at every point of the domain, and the textbook's answer needs no restriction here.
   $\tan(\theta - n\pi) = \tan\theta$⁠. Basics: trigonometric identities

5. Confirm with the inverse-tangent formula

   With $u = \tan\frac x2$⁠, $u' = \frac12\sec^2\frac x2$ and $(\tan^{-1}u)' = \frac{1}{1+u^2}$⁠:

   $$f'(x) = \frac{\frac12\sec^2\frac x2}{1+\tan^2\frac x2} = \frac{\frac12\sec^2\frac x2}{\sec^2\frac x2} = \frac12.$$
   The chain rule needs no principal-range check, because it never cancels an inverse; it works at every point of the domain at once and agrees with Part (ii), Step 4.
   $1 + \tan^2 A = \sec^2 A$⁠; table in Section 5.3.3. Lesson: inverse trig derivatives

Part (iii): $f(x) = \sin^{-1}\frac{2^{x+1}}{1+4^x}$

1. Domain: the inside is between 0 and 1

   Write $u = \frac{2^{x+1}}{1+4^x} = \frac{2\cdot 2^x}{1+(2^x)^2}$⁠. It is positive. And

   $$\begin{aligned} u \le 1 &\iff 2\cdot 2^x \le 1 + (2^x)^2 \\ &\iff 0 \le (2^x - 1)^2, \end{aligned}$$

   which is true for every $x$⁠, with equality only when $2^x = 1$⁠, that is $x = 0$⁠. So $0 < u \le 1$⁠, $f$ is defined for all real $x$⁠, and $u = 1$ only at $x = 0$⁠.

   $\sin^{-1}$ needs its input in $[-1,1]$⁠. The textbook divides by $2^x$ and uses $2 \le 2^{-x} + 2^x$⁠; moving everything to one side shows the same thing as a perfect square. Knowing that $u$ reaches $1$ at $x = 0$ will matter: that is where $\sin^{-1}$ has its steepest point.
   $4^x = (2^2)^x = (2^x)^2$ and $2^{x+1} = 2\cdot 2^x$ (laws of exponents); a square is never negative. Basics: exponentials and logarithms

2. Substitute $2^x = \tan\theta$ and check it is legitimate

   Since $2^x > 0$⁠, we can take $\theta = \tan^{-1}(2^x)$⁠, which lies in $(0, \frac{\pi}{2})$ and satisfies $\tan\theta = 2^x$⁠. Then

   $$u = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta, \qquad 0 < 2\theta < \pi.$$
   Defining $\theta$ as $\tan^{-1}(2^x)$ makes the substitution exact and tells us precisely which angle we have. Its range $(0,\frac{\pi}{2})$ is what decides the next step.
   $\frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta$⁠. Basics: trigonometric identities

3. Where does the inverse sine cancel?

   $\sin^{-1}(\sin 2\theta) = 2\theta$ needs $2\theta \le \frac{\pi}{2}$⁠, that is $\theta \le \frac{\pi}{4}$⁠, that is $2^x \le 1$⁠, that is $x \le 0$⁠. For $x > 0$ we have $\frac{\pi}{2} < 2\theta < \pi$⁠; then $\pi - 2\theta$ lies in $(0, \frac{\pi}{2})$ and has the same sine. So

   $$f(x) = \begin{cases} 2\tan^{-1}(2^x), & x \le 0, \\ \pi - 2\tan^{-1}(2^x), & x > 0. \end{cases}$$
   The textbook writes $\sin^{-1}(\sin 2\theta) = 2\theta$ for all $x$⁠. That is only true for $x \le 0$⁠. For $x > 0$⁠, $2\theta$ exceeds $\frac{\pi}{2}$⁠, but $\sin^{-1}$ never returns more than $\frac{\pi}{2}$⁠. Example: at $x = 1$⁠, $2\tan^{-1}2 \approx 2.2142974$⁠, yet $f(1) = \sin^{-1}\frac45 \approx 0.9272952$⁠, which is $\pi - 2.2142974$⁠.
   $\sin(\pi - A) = \sin A$⁠; principal range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (Section 2.2). Lesson: inverse trig derivatives

4. Differentiate each piece

   By the chain rule with $(\tan^{-1}v)' = \frac{1}{1+v^2}$ and $(2^x)' = 2^x\log 2$⁠:

   $$\begin{aligned} \frac{d}{dx}\,2\tan^{-1}(2^x) &= \frac{2}{1+(2^x)^2}\cdot 2^x\log 2 \\ &= \frac{2^{x+1}\log 2}{1+4^x}. \end{aligned}$$

   So $f'(x) = \dfrac{2^{x+1}\log 2}{1+4^x}$ for $x < 0$⁠, and $f'(x) = -\dfrac{2^{x+1}\log 2}{1+4^x}$ for $x > 0$ (the constant $\pi$ has derivative $0$⁠).

   For $x < 0$ this is exactly the textbook's answer. For $x > 0$ the minus sign in $\pi - 2\tan^{-1}(2^x)$ flips it.
   Chain rule (Theorem 4); $\tan^{-1}$ derivative (Section 5.3.3); $(a^x)' = a^x\log a$ (Example 28). Lesson: the chain rule

5. The point $x = 0$⁠: one-sided derivatives

   $f(0) = \sin^{-1}1 = \frac{\pi}{2}$⁠. Let $g(x) = 2\tan^{-1}(2^x)$⁠, which is differentiable everywhere with $g(0) = \frac{\pi}{2}$ and $g'(0) = \frac{2\log 2}{2} = \log 2$⁠. Then $f = g$ on $x \le 0$ and $f = \pi - g$ on $x \ge 0$⁠, so

   $$\begin{aligned} \text{LHD} &= \lim_{h\to0^-}\frac{g(h) - g(0)}{h} = \log 2, \\ \text{RHD} &= \lim_{h\to0^+}\frac{-(g(h) - g(0))}{h} = -\log 2. \end{aligned}$$

   Since $\log 2 \ne -\log 2$⁠, $f$ is not differentiable at $x = 0$: the graph has a corner at its peak.

   At $x = 0$ the formula for $f$ changes, so we must not just plug $0$ into a derivative formula. Using $g$ on each side turns both one-sided quotients into the ordinary derivative of $g$ at $0$⁠, with a sign.
   Left- and right-hand derivatives (Section 5.3). Lesson: differentiability

6. Two independent confirmations

   Symmetry. $u(-x) = \frac{2\cdot 2^{-x}}{1+4^{-x}}$⁠; multiplying top and bottom by $4^x$ gives $\frac{2\cdot 2^x}{4^x+1} = u(x)$⁠. So $f$ is even, and the derivative of an even function is odd: $f'(-x) = -f'(x)$⁠. The textbook's expression $\frac{2^{x+1}\log2}{1+4^x}$ is even and never zero, so it cannot be $f'$ on both sides of $0$⁠.

   The inverse-sine formula. $u' = \dfrac{2^{x+1}\log 2\,(1-4^x)}{(1+4^x)^2}$ and $\sqrt{1-u^2} = \dfrac{|1-4^x|}{1+4^x}$⁠, so for $x \ne 0$

   $$\begin{aligned} f'(x) &= \frac{u'}{\sqrt{1-u^2}} \\ &= \frac{2^{x+1}\log 2}{1+4^x}\cdot\frac{1-4^x}{|1-4^x|}. \end{aligned}$$

   The last factor is $+1$ for $x < 0$ and $-1$ for $x > 0$⁠, matching Part (iii), Step 4.

   Two arguments that use no principal-range reasoning at all reach the same conclusion. The second one shows where the sign comes from: $1 - u^2 = \frac{(1-4^x)^2}{(1+4^x)^2}$⁠, and its square root is an absolute value.
   Quotient rule for $u'$⁠; $\sqrt{a^2} = |a|$⁠; $(\sin^{-1}u)' = \frac{u'}{\sqrt{1-u^2}}$ for $|u| < 1$ (Section 5.3.3). Basics: derivative rules

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