Exercise 1: Restricted ranges: the one angle an inverse function returns

Sine, cosine and tangent take each of their values infinitely often, so none of them has an inverse on its whole domain. Thomas (section 1.6) restricts each one to an interval where it is one-to-one and inverts the restriction. $y = \sin^{-1}x = \arcsin x$ is THE angle $y$ of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin y = x$ , defined for $-1 \le x \le 1$ . $y = \cos^{-1}x = \arccos x$ is the angle of $[0, \pi]$ with $\cos y = x$ . $y = \tan^{-1}x = \arctan x$ is the angle of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan y = x$ , defined for every real $x$ . The $-1$ of $\sin^{-1}$ names an inverse function, it is NOT an exponent.

The figure shows $y = \sin x$ for $-2\pi \le x \le 2\pi$ , the restricted piece on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ in blue, and the dashed line $y = \frac{1}{2}$ . A scientific calculator is allowed, in RADIAN mode.

a) Find exactly, in radians: $\arcsin\left(-\frac{1}{2}\right)$ , $\arccos\left(-\frac{\sqrt 3}{2}\right)$ , $\arctan\sqrt 3$ , $\arccos 0$ and $\arctan(-1)$ .
b) Using the figure, how many solutions does the equation $\sin x = \frac{1}{2}$ have on $[-2\pi, 2\pi]$ ? Give them all exactly, and say which one is $\arcsin\frac{1}{2}$ and why.
c) Find the domain of $f(x) = \arcsin(2x - 3)$ , of $g(x) = \arccos\frac{x - 1}{3}$ and of $h(x) = \arctan\sqrt{4 - x}$ .
d) In radian mode, a calculator gives $\cos^{-1}(0.2) \approx 1.3694$ . Using only this value, the identity $\arccos(-x) = \pi - \arccos x$ and the identity $\arcsin x + \arccos x = \frac{\pi}{2}$ of Thomas 1.6, find $\arccos(-0.2)$ and $\arcsin(0.2)$ to four decimals. Explain the first identity with the unit circle.
e) A student whose calculator is in degree mode types $\tan^{-1}(1)$ and reads $45$ , then types $\sin^{-1}(2)$ and reads an error message. Give $\arctan 1$ in radians, exactly and to four decimals, and explain the error message.

Type your answers, the page tells you right or wrong 0/15

a)

\arcsin\left(-\frac{1}{2}\right)

(in multiples of

\pi

)

\arccos\left(-\frac{\sqrt 3}{2}\right)

(in multiples of

\pi

)

\arctan\sqrt 3

(in multiples of

\pi

)

\arccos 0

(in multiples of

\pi

)

\arctan(-1)

(in multiples of

\pi

)

b)

Number of solutions on the interval

The smallest solution (in multiples of

\pi

)

\arcsin\frac{1}{2}

(in multiples of

\pi

)

c)

Domain of

f

,

Domain of

g

,

Domain of

h

,

d)

\arccos(-0.2)

\arcsin(0.2)

e)

\arctan 1

in radians

\sin^{-1}(2)

is

Show the solution

Answers

a) $-\frac{\pi}{6}$ , $\frac{5\pi}{6}$ , $\frac{\pi}{3}$ , $\frac{\pi}{2}$ , $-\frac{\pi}{4}$
b) Four solutions: $-\frac{11\pi}{6}$ , $-\frac{7\pi}{6}$ , $\frac{\pi}{6}$ , $\frac{5\pi}{6}$ ; $\arcsin\frac{1}{2} = \frac{\pi}{6}$ , the only one on the blue piece
c) $f$ : $[1, 2]$ ; $g$ : $[-2, 4]$ ; $h$ : $(-\infty, 4]$
d) $\arccos(-0.2) \approx 1.7722$ , $\arcsin(0.2) \approx 0.2014$
e) $\arctan 1 = \frac{\pi}{4} \approx 0.7854$ ; $\sin^{-1}(2)$ does not exist, since $2$ is not the sine of any angle

a)

Each answer is found by asking which angle OF THE RANGE has the given sine, cosine or tangent. $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$ and $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , so $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ ; the angle $\frac{7\pi}{6}$ also has sine $-\frac{1}{2}$ , but it is not in the range. $\cos\frac{5\pi}{6} = -\frac{\sqrt 3}{2}$ with $\frac{5\pi}{6} \in [0, \pi]$ , so $\arccos\left(-\frac{\sqrt 3}{2}\right) = \frac{5\pi}{6}$ : the answer $-\frac{\pi}{6}$ , which a student gets by copying the sine case, is outside $[0, \pi]$ and its cosine is $+\frac{\sqrt 3}{2}$ . $\tan\frac{\pi}{3} = \sqrt 3$ , so $\arctan\sqrt 3 = \frac{\pi}{3}$ . $\cos\frac{\pi}{2} = 0$ , so $\arccos 0 = \frac{\pi}{2}$ . $\tan\left(-\frac{\pi}{4}\right) = -1$ , so $\arctan(-1) = -\frac{\pi}{4}$ , and not $\frac{3\pi}{4}$ , which has the same tangent but lies outside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . The rule to memorize: a negative input sends $\arcsin$ and $\arctan$ below zero, and sends $\arccos$ into the second quadrant.

b)

The dashed line $y = \frac{1}{2}$ crosses the grey curve four times on $[-2\pi, 2\pi]$ . On $[0, 2\pi]$ , the solutions are $\frac{\pi}{6}$ and $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$ , since $\sin(\pi - x) = \sin x$ . Subtracting one period $2\pi$ gives the two others: $\frac{\pi}{6} - 2\pi = -\frac{11\pi}{6}$ and $\frac{5\pi}{6} - 2\pi = -\frac{7\pi}{6}$ . Four solutions, and the calculator key $\sin^{-1}$ returns only ONE of them, $\frac{\pi}{6} \approx 0.5236$ : the only one on the blue piece, because $\arcsin$ is by definition the inverse of the RESTRICTED sine. That is the whole chapter in one picture: an inverse trigonometric function answers the question 'which angle of my range', never 'which angles'.

c)

Each inverse function imposes a condition on what is inside it, and the domain is found by solving an inequality: this is the algebra that costs marks. $\arcsin u$ needs $-1 \le u \le 1$ : $-1 \le 2x - 3 \le 1$ , add $3$ to the three members, $2 \le 2x \le 4$ , divide by $2$ : $1 \le x \le 2$ , so the domain of $f$ is $[1, 2]$ . $\arccos u$ needs the same: $-1 \le \frac{x - 1}{3} \le 1$ , multiply by $3$ , $-3 \le x - 1 \le 3$ , so $-2 \le x \le 4$ . $\arctan$ accepts every real number, so the only condition on $h$ comes from the square root: $4 - x \ge 0$ , that is $x \le 4$ , and the domain is $(-\infty, 4]$ . The classic slip is to answer $[-1, 1]$ for $f$ , the domain of $\arcsin x$ itself: the condition is on the INSIDE $2x - 3$ , not on $x$ .

d)

On the unit circle, the angle $\alpha = \arccos x$ of $[0, \pi]$ has the point $(x, \sin\alpha)$ ; the angle $\pi - \alpha$ , also in $[0, \pi]$ , is its mirror image in the vertical axis, with first coordinate $-x$ . So $\cos(\pi - \alpha) = -x$ , and since $\pi - \alpha$ is in the range, $\arccos(-x) = \pi - \arccos x$ . With $\arccos(0.2) \approx 1.369438$ : $\arccos(-0.2) = \pi - 1.369438 \approx 1.772154$ , that is $1.7722$ to four decimals. Then $\arcsin(0.2) = \frac{\pi}{2} - \arccos(0.2) \approx 1.570796 - 1.369438 = 0.201358$ , that is $0.2014$ . Both are checked directly on the calculator. The false symmetry $\arccos(-x) = -\arccos x$ gives a NEGATIVE answer, impossible for $\arccos$ , whose values are in $[0, \pi]$ : that alone refutes it.

e)

In degree mode the calculator returns angles in degrees: $45$ means $45°$ , and $45° = \frac{\pi}{4}$ rad. In calculus every angle is in radians, because the derivative formulas of the chapter, such as $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$ , are only true in radians. So $\arctan 1 = \frac{\pi}{4} \approx 0.7854$ ; a $45$ copied into a tangent line or a limit makes the whole answer wrong by a factor $\frac{180}{\pi}$ . The error message on $\sin^{-1}(2)$ is correct: $\sin^{-1}$ is only defined on $[-1, 1]$ , since no angle has a sine equal to $2$ . It is not the reciprocal $\frac{1}{\sin 2} \approx 1.0998$ , which does exist: the reciprocal of the sine is $\csc$ . Check the mode before the first question of any test: $\sin^{-1}(1)$ must display $1.5708$ .

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Exercise 2: Compositions and the reference triangle: the range fixes the sign

$\sin(\arcsin x) = x$ for every $x$ of $[-1, 1]$ : the inverse acts first and the sine undoes it. The other order is the trap: $\arcsin(\sin x) = x$ ONLY when $x$ is already in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , because $\arcsin$ can only return an angle of its range. To evaluate a trigonometric function of an inverse trigonometric function, draw the right triangle of the angle, find the missing side by Pythagoras, then fix the SIGN with the range.

The figure shows the triangle of $\theta = \arcsin\frac{x}{3}$ for $0 < x < 3$ : hypotenuse $3$ , side opposite $\theta$ equal to $x$ , so that $\sin\theta = \frac{x}{3}$ .

a) Find exactly: $\arcsin\left(\sin\frac{7\pi}{6}\right)$ , $\arccos\left(\cos\frac{5\pi}{3}\right)$ and $\arctan\left(\tan\left(-\frac{3\pi}{4}\right)\right)$ .
b) Find exactly, then to four decimals: $\cos(\arctan(-2))$ , $\sin(\arctan(-2))$ and $\tan\left(\arcsin\left(-\frac{5}{13}\right)\right)$ .
c) Using the triangle of the figure, write $\cos\left(\arcsin\frac{x}{3}\right)$ and $\tan\left(\arcsin\frac{x}{3}\right)$ as algebraic expressions in $x$ , and say for which $x$ each one holds. With a triangle of your own, write $\sec(\arctan 2x)$ in the same way. Evaluate the second expression at $x = 1$ and the third at $x = 1$ .
d) Thomas also defines $\sec^{-1}x = \cos^{-1}\frac{1}{x}$ for $|x| \ge 1$ , with values in $[0, \pi]$ except $\frac{\pi}{2}$ . Find $\sec^{-1}2$ , $\sec^{-1}(-2)$ and $\tan\left(\sec^{-1}(-2)\right)$ . Show that $\tan(\sec^{-1}x) = \sqrt{x^2 - 1}$ for $x \ge 1$ but $-\sqrt{x^2 - 1}$ for $x \le -1$ .
e) Solve for $x$ : $\arcsin(2x) = \frac{\pi}{6}$ ; $\arctan(x - 1) = -\frac{\pi}{3}$ ; $\arccos x = \frac{4\pi}{3}$ .

Type your answers, the page tells you right or wrong 0/14

a)

\arcsin\left(\sin\frac{7\pi}{6}\right)

(in multiples of

\pi

)

\arccos\left(\cos\frac{5\pi}{3}\right)

(in multiples of

\pi

)

\arctan\left(\tan\left(-\frac{3\pi}{4}\right)\right)

(in multiples of

\pi

)

b)

\cos(\arctan(-2))

\sin(\arctan(-2))

\tan\left(\arcsin\left(-\frac{5}{13}\right)\right)

c)

\tan\left(\arcsin\frac{x}{3}\right)

at

x = 1

\sec(\arctan 2x)

at

x = 1

d)

\sec^{-1}2

(in multiples of

\pi

)

\sec^{-1}(-2)

(in multiples of

\pi

)

\tan\left(\sec^{-1}(-2)\right)

e)

First equation:

x

Second equation:

x

Third equation

Show the solution

Answers

a) $-\frac{\pi}{6}$ , $\frac{\pi}{3}$ , $\frac{\pi}{4}$
b) $\frac{1}{\sqrt 5} \approx 0.4472$ , $-\frac{2}{\sqrt 5} \approx -0.8944$ , $-\frac{5}{12} \approx -0.4167$
c) $\frac{\sqrt{9 - x^2}}{3}$ on $[-3, 3]$ ; $\frac{x}{\sqrt{9 - x^2}}$ on $(-3, 3)$ ; $\sqrt{1 + 4x^2}$ for every $x$ ; values $\frac{1}{2\sqrt 2} \approx 0.3536$ and $\sqrt 5 \approx 2.2361$
d) $\frac{\pi}{3}$ , $\frac{2\pi}{3}$ , $-\sqrt 3$ ; the sign of $\tan$ follows the quadrant of the angle
e) $x = \frac{1}{4}$ ; $x = 1 - \sqrt 3$ ; no solution, $\frac{4\pi}{3}$ is not in $[0, \pi]$

a)

Compute the inner value first, then ask for the angle OF THE RANGE. $\sin\frac{7\pi}{6} = -\frac{1}{2}$ , and the angle of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with sine $-\frac{1}{2}$ is $-\frac{\pi}{6}$ . $\cos\frac{5\pi}{3} = \frac{1}{2}$ , and the angle of $[0, \pi]$ with cosine $\frac{1}{2}$ is $\frac{\pi}{3}$ . $\tan\left(-\frac{3\pi}{4}\right) = 1$ , since the tangent has period $\pi$ and $-\frac{3\pi}{4} + \pi = \frac{\pi}{4}$ ; the angle of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with tangent $1$ is $\frac{\pi}{4}$ . None of the three answers is the angle we started from: each of $\frac{7\pi}{6}$ , $\frac{5\pi}{3}$ , $-\frac{3\pi}{4}$ lies outside the range of its inverse, and writing it back as the answer is the most frequent zero of the chapter.

b)

Let $\theta = \arctan(-2)$ . Then $\tan\theta = -2$ and $\theta \in \left(-\frac{\pi}{2}, 0\right)$ , a fourth-quadrant angle: cosine POSITIVE, sine NEGATIVE. The triangle with opposite side $2$ and adjacent side $1$ has hypotenuse $\sqrt{1 + 4} = \sqrt 5$ , so $\cos\theta = \frac{1}{\sqrt 5} = \frac{\sqrt 5}{5} \approx 0.4472$ and $\sin\theta = -\frac{2}{\sqrt 5} \approx -0.8944$ . The triangle gives the SIZES, the range gives the SIGNS. Let $\varphi = \arcsin\left(-\frac{5}{13}\right)$ , also in $\left(-\frac{\pi}{2}, 0\right)$ : the triangle $5$ , $13$ has third side $\sqrt{169 - 25} = 12$ , $\cos\varphi = +\frac{12}{13}$ and $\tan\varphi = \frac{-5/13}{12/13} = -\frac{5}{12} \approx -0.4167$ . A student who takes $\cos\theta = -\frac{1}{\sqrt 5}$ because the input was negative has moved the angle to the second quadrant, where $\arctan$ never goes.

c)

In the figure, the adjacent side is $\sqrt{3^2 - x^2} = \sqrt{9 - x^2}$ by Pythagoras. It is NOT $3 - x$ : a square root does not distribute over a difference, and $\sqrt{9 - 4} = \sqrt 5 \ne 1$ settles it. So $\cos\left(\arcsin\frac{x}{3}\right) = \frac{\sqrt{9 - x^2}}{3}$ . The triangle only shows $0 < x < 3$ , but the formula holds on all of $[-3, 3]$ : $\theta$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , where the cosine is never negative, so the positive root is the right one. Then $\tan\left(\arcsin\frac{x}{3}\right) = \frac{x}{\sqrt{9 - x^2}}$ for $-3 < x < 3$ (the sign now comes from $x$ , as it should). At $x = 1$ : $\frac{1}{\sqrt 8} = \frac{\sqrt 2}{4} \approx 0.3536$ . For $\beta = \arctan 2x$ : opposite side $2x$ , adjacent $1$ , hypotenuse $\sqrt{1 + (2x)^2} = \sqrt{1 + 4x^2}$ , and since $\cos\beta > 0$ on the range, $\sec\beta = \sqrt{1 + 4x^2}$ for every $x$ . Watch the square: $(2x)^2 = 4x^2$ , not $2x^2$ . At $x = 1$ : $\sqrt 5 \approx 2.2361$ .

d)

$\sec^{-1}2 = \cos^{-1}\frac{1}{2} = \frac{\pi}{3}$ and $\sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ . The angle $\frac{2\pi}{3}$ is in the second quadrant, so $\tan\frac{2\pi}{3} = -\sqrt 3 \approx -1.7321$ . In general, let $y = \sec^{-1}x$ , so $\sec y = x$ and $\tan^2 y = \sec^2 y - 1 = x^2 - 1$ , which gives $\tan y = \pm\sqrt{x^2 - 1}$ , and the range decides. For $x \ge 1$ , $\cos y = \frac{1}{x} > 0$ , so $y \in \left[0, \frac{\pi}{2}\right)$ and $\tan y \ge 0$ : $\tan(\sec^{-1}x) = \sqrt{x^2 - 1}$ . For $x \le -1$ , $\cos y < 0$ , so $y \in \left(\frac{\pi}{2}, \pi\right]$ and $\tan y \le 0$ : $\tan(\sec^{-1}x) = -\sqrt{x^2 - 1}$ . Check at $x = -2$ : $-\sqrt 3$ , as found directly. This sign is exactly the one that will put an absolute value in the derivative of $\sec^{-1}x$ .

e)

Apply the direct function to both sides, but only after checking that the right side is in the range. $\frac{\pi}{6}$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , so $\arcsin(2x) = \frac{\pi}{6}$ gives $2x = \sin\frac{\pi}{6} = \frac{1}{2}$ and $x = \frac{1}{4}$ ; check: $\arcsin\frac{1}{2} = \frac{\pi}{6}$ . $-\frac{\pi}{3}$ is in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so $x - 1 = \tan\left(-\frac{\pi}{3}\right) = -\sqrt 3$ and $x = 1 - \sqrt 3 \approx -0.7321$ . For the third, $\frac{4\pi}{3} > \pi$ is NOT a value of $\arccos$ : the equation has no solution. Taking the cosine of both sides blindly gives $x = \cos\frac{4\pi}{3} = -\frac{1}{2}$ , and the check refutes it at once: $\arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \ne \frac{4\pi}{3}$ .

Exercise 3: Derivatives of arcsin, arccos and arctan: the formula is one line, the algebra is the rest

Thomas (section 3.9) gives, for a differentiable inner function $u$ of $x$ : $\frac{d}{dx}\sin^{-1}u = \frac{u'}{\sqrt{1 - u^2}}$ and $\frac{d}{dx}\cos^{-1}u = -\frac{u'}{\sqrt{1 - u^2}}$ for $|u| < 1$ , and $\frac{d}{dx}\tan^{-1}u = \frac{u'}{1 + u^2}$ for every $u$ .

Every one of these formulas contains $1 - u^2$ or $1 + u^2$ . When $u$ is a power, a fraction or a root, that expression is where the marks are lost: square $u$ correctly, bring $1 \pm u^2$ to ONE fraction, then take the root or invert. Exact answers, then the decimal asked.

a) Differentiate $f(x) = \arccos(x^2)$ , $g(x) = \arctan(x^2 - 1)$ and $h(x) = \arcsin\sqrt x$ , naming each inner function, simplify, and evaluate $f'\left(\frac{1}{2}\right)$ , $g'(1)$ and $h'\left(\frac{1}{2}\right)$ .
b) Differentiate $y = \arcsin\frac{x}{\sqrt{1 + x^2}}$ and show that $y' = \frac{1}{1 + x^2}$ . Which familiar function has the same derivative and the same value at $x = 0$ ? What do you conclude?
c) Differentiate $k(x) = (1 + x^2)\arctan x$ and $m(x) = \frac{\arcsin x}{x}$ , and evaluate $k'(1)$ and $m'\left(\frac{1}{2}\right)$ .
d) Find the equation of the tangent line to $y = \arccos\frac{x}{2}$ at $x = 1$ .
e) Show that $y = \arctan x$ satisfies $(1 + x^2)y'' + 2xy' = 0$ , and give $y''(1)$ .

Show the solution

Answers

a) $f' = -\frac{2x}{\sqrt{1 - x^4}}$ , $f'\left(\frac{1}{2}\right) = -\frac{4}{\sqrt{15}}$ ; $g' = \frac{2x}{x^4 - 2x^2 + 2}$ , $g'(1) = 2$ ; $h' = \frac{1}{2\sqrt{x}\sqrt{1 - x}}$ , $h'\left(\frac{1}{2}\right) = 1$
b) $y' = \frac{1}{1 + x^2}$ ; $y = \arctan x$ for every $x$
c) $k' = 2x\arctan x + 1$ , $k'(1) = \frac{\pi}{2} + 1$ ; $m' = \frac{\frac{x}{\sqrt{1 - x^2}} - \arcsin x}{x^2}$ , $m'\left(\frac{1}{2}\right) = \frac{4}{\sqrt 3} - \frac{2\pi}{3} \approx 0.2150$
d) $y = -\frac{1}{\sqrt 3}x + \frac{\pi}{3} + \frac{1}{\sqrt 3}$
e) $y'' = -\frac{2x}{(1 + x^2)^2}$ , $y''(1) = -\frac{1}{2}$

a)

$f$ : inner function $u = x^2$ , $u' = 2x$ , and $u^2 = (x^2)^2 = x^4$ , so $f'(x) = -\frac{2x}{\sqrt{1 - x^4}}$ for $|x| < 1$ . At $\frac{1}{2}$ : $1 - \frac{1}{16} = \frac{15}{16}$ , $\sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$ , and $f'\left(\frac{1}{2}\right) = -1 \cdot \frac{4}{\sqrt{15}} = -\frac{4}{\sqrt{15}} \approx -1.0328$ . The slip to avoid: $\sqrt{1 - x^2}$ , the formula of $\arccos x$ copied without replacing $x$ by $u$ . $g$ : inner function $u = x^2 - 1$ , $u' = 2x$ , $1 + u^2 = 1 + x^4 - 2x^2 + 1 = x^4 - 2x^2 + 2$ , so $g'(x) = \frac{2x}{x^4 - 2x^2 + 2}$ and $g'(1) = \frac{2}{1} = 2$ . Expanding $(x^2 - 1)^2$ as $x^4 - 1$ would give $g'(1) = 2$ by luck here but a wrong function everywhere else, for example $\frac{4}{16} = \frac{1}{4}$ instead of $\frac{4}{10}$ at $x = 2$ . $h$ : inner function $u = \sqrt x$ , $u' = \frac{1}{2\sqrt x}$ , $u^2 = x$ , so $h'(x) = \frac{1}{2\sqrt x\sqrt{1 - x}}$ for $0 < x < 1$ , and $h'\left(\frac{1}{2}\right) = \frac{1}{2 \cdot \frac{1}{\sqrt 2} \cdot \frac{1}{\sqrt 2}} = 1$ .

b)

Write $u = x(1 + x^2)^{-1/2}$ and use the product rule with the chain rule on the second factor: $u' = (1 + x^2)^{-1/2} - x^2(1 + x^2)^{-3/2}$ . Factor the SMALLEST power, $(1 + x^2)^{-3/2}$ : $u' = (1 + x^2)^{-3/2}\left[(1 + x^2) - x^2\right] = (1 + x^2)^{-3/2}$ . Then bring $1 - u^2$ to one fraction: $1 - \frac{x^2}{1 + x^2} = \frac{1 + x^2 - x^2}{1 + x^2} = \frac{1}{1 + x^2}$ , whose root is $(1 + x^2)^{-1/2}$ . Finally $y' = \frac{(1 + x^2)^{-3/2}}{(1 + x^2)^{-1/2}} = (1 + x^2)^{-1} = \frac{1}{1 + x^2}$ . This is the derivative of $\arctan x$ , and both functions are $0$ at $x = 0$ . Their difference has derivative $0$ on the whole interval $(-\infty, \infty)$ , so it is constant (a fact used as a tool, the constant being read at a point), and that constant is $0$ : $\arcsin\frac{x}{\sqrt{1 + x^2}} = \arctan x$ for every $x$ . The triangle confirms it: the angle with opposite side $x$ and adjacent side $1$ has hypotenuse $\sqrt{1 + x^2}$ .

c)

Product rule on $k$ : $k'(x) = 2x\arctan x + (1 + x^2) \cdot \frac{1}{1 + x^2} = 2x\arctan x + 1$ . The factor $1 + x^2$ was chosen to cancel the denominator; simplify it rather than leaving $\frac{1 + x^2}{1 + x^2}$ . At $x = 1$ : $k'(1) = 2 \cdot \frac{\pi}{4} + 1 = \frac{\pi}{2} + 1 \approx 2.5708$ . Quotient rule on $m$ , for $x \ne 0$ in $[-1, 1]$ : $m'(x) = \frac{x \cdot \frac{1}{\sqrt{1 - x^2}} - \arcsin x \cdot 1}{x^2}$ . At $x = \frac{1}{2}$ : $\frac{1/2}{\sqrt 3/2} = \frac{1}{\sqrt 3}$ , $\arcsin\frac{1}{2} = \frac{\pi}{6}$ , so $m'\left(\frac{1}{2}\right) = 4\left(\frac{1}{\sqrt 3} - \frac{\pi}{6}\right) = \frac{4}{\sqrt 3} - \frac{2\pi}{3} \approx 0.2150$ . The order of the numerator is 'derivative of the top times the bottom, minus the top times the derivative of the bottom': swapped, it gives $-0.2150$ .

d)

Point: $y(1) = \arccos\frac{1}{2} = \frac{\pi}{3}$ . Slope: inner function $u = \frac{x}{2}$ , $u' = \frac{1}{2}$ , $1 - u^2 = 1 - \frac{x^2}{4} = \frac{4 - x^2}{4}$ , so $y' = -\frac{1/2}{\sqrt{4 - x^2}/2} = -\frac{1}{\sqrt{4 - x^2}}$ . At $x = 1$ : $-\frac{1}{\sqrt 3} \approx -0.5774$ . Tangent: $y - \frac{\pi}{3} = -\frac{1}{\sqrt 3}(x - 1)$ , that is $y = -\frac{1}{\sqrt 3}x + \frac{\pi}{3} + \frac{1}{\sqrt 3}$ , with $y$ -intercept $\approx 1.6245$ . Two checks: the slope is negative, as it must be for an arccosine, and the point $\left(1, \frac{\pi}{3}\right)$ satisfies the tangent equation. Forgetting the inner derivative $\frac{1}{2}$ gives the slope $-\frac{2}{\sqrt 3}$ , twice too steep.

e)

$y' = \frac{1}{1 + x^2} = (1 + x^2)^{-1}$ , and by the chain rule $y'' = -(1 + x^2)^{-2} \cdot 2x = -\frac{2x}{(1 + x^2)^2}$ . Then $(1 + x^2)y'' = -\frac{2x}{1 + x^2}$ and $2xy' = \frac{2x}{1 + x^2}$ : the sum is $0$ for every $x$ . At $x = 1$ : $y''(1) = -\frac{2}{4} = -\frac{1}{2}$ . Rewriting $\frac{1}{1 + x^2}$ as a power BEFORE differentiating spares the quotient rule, and it is the rewriting that the exam rewards: the student who differentiates $\frac{1}{1 + x^2}$ as $\frac{1}{2x}$ has differentiated the denominator alone, and that is not a rule.

Exercise 4: The derivative of arcsec by implicit differentiation, and the absolute value it needs

Thomas defines $y = \sec^{-1}x$ for $|x| \ge 1$ as the angle $y$ of $[0, \pi]$ , $y \ne \frac{\pi}{2}$ , with $\sec y = x$ . Equivalently $\sec^{-1}x = \cos^{-1}\frac{1}{x}$ . The figure shows its graph: a right branch from $(1, 0)$ and a left branch ending at $(-1, \pi)$ , both approaching the dashed line $y = \frac{\pi}{2}$ , and BOTH rising from left to right.

Its derivative is the one formula of section 3.9 with an absolute value in it, and this exercise is about where that absolute value comes from.

a) Let $y = \sec^{-1}x$ with $|x| > 1$ . Differentiate $\sec y = x$ implicitly, show that $\frac{dy}{dx} = \frac{1}{\sec y\tan y}$ , then express $\sec y\tan y$ in terms of $x$ on each branch, and conclude that $\frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2 - 1}}$ . Evaluate at $x = 2$ and at $x = -2$ .
b) Check the result by a second route: differentiate $\cos^{-1}\frac{1}{x}$ with the chain rule, and simplify $\sqrt{1 - \frac{1}{x^2}}$ correctly.
c) Thomas obtains three derivatives from identities: $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$ , $\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$ , $\csc^{-1}x = \frac{\pi}{2} - \sec^{-1}x$ . Deduce the derivatives of $\cot^{-1}x$ and $\csc^{-1}x$ , and evaluate the first at $x = 1$ , the second at $x = -2$ .
d) Differentiate $f(x) = \sec^{-1}(3x)$ for $x < -\frac{1}{3}$ and evaluate $f'\left(-\frac{2}{3}\right)$ . A student writes $f'(x) = \frac{3}{3x\sqrt{9x^2 - 1}}$ : what value does he find, and how does the figure tell him he is wrong before any computation?
e) Find the equation of the tangent line to $y = \sec^{-1}x$ at $x = -2$ .

Type your answers, the page tells you right or wrong 0/9

a)

\frac{d}{dx}\sec^{-1}x

at

x = 2

\frac{d}{dx}\sec^{-1}x

at

x = -2

b)

\sqrt{x^2}

equals

c)

\frac{d}{dx}\cot^{-1}x

at

x = 1

\frac{d}{dx}\csc^{-1}x

at

x = -2

d)

f'\left(-\frac{2}{3}\right)

The value the student finds

e)

Slope of the tangent

y

-intercept of the tangent

Show the solution

Answers

a) $\sec y\tan y \cdot y' = 1$ ; $\sec y\tan y = |x|\sqrt{x^2 - 1}$ on both branches; $\frac{1}{2\sqrt 3} \approx 0.2887$ at $x = 2$ AND at $x = -2$
b) $\sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|}$ , and the chain rule gives $\frac{1}{|x|\sqrt{x^2 - 1}}$ again
c) $\frac{d}{dx}\cot^{-1}x = -\frac{1}{1 + x^2}$ , $-\frac{1}{2}$ at $1$ ; $\frac{d}{dx}\csc^{-1}x = -\frac{1}{|x|\sqrt{x^2 - 1}}$ , $-\frac{1}{2\sqrt 3}$ at $-2$
d) $f'(x) = \frac{3}{|3x|\sqrt{9x^2 - 1}}$ , $f'\left(-\frac{2}{3}\right) = \frac{\sqrt 3}{2} \approx 0.8660$ ; the student finds $-0.8660$ , a negative slope on a rising graph
e) $y = \frac{2\pi}{3} + \frac{1}{2\sqrt 3}(x + 2)$

a)

Differentiate $\sec y = x$ with respect to $x$ , $y$ being a function of $x$ (chain rule, inner function $y$ ): $\sec y\tan y \cdot y' = 1$ , so $y' = \frac{1}{\sec y\tan y}$ wherever $\tan y \ne 0$ , that is for $|x| > 1$ . Now $\sec y = x$ , and $\tan^2 y = \sec^2 y - 1 = x^2 - 1$ , so $\tan y = \pm\sqrt{x^2 - 1}$ : the RANGE chooses. On the right branch, $x > 1$ , $y \in \left(0, \frac{\pi}{2}\right)$ and $\tan y > 0$ : $\sec y\tan y = x\sqrt{x^2 - 1}$ . On the left branch, $x < -1$ , $y \in \left(\frac{\pi}{2}, \pi\right)$ and $\tan y < 0$ : $\sec y\tan y = x \cdot \left(-\sqrt{x^2 - 1}\right) = -x\sqrt{x^2 - 1}$ . In both cases the product is $|x|\sqrt{x^2 - 1}$ , positive, so $\frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2 - 1}}$ . At $x = 2$ : $\frac{1}{2\sqrt 3} = \frac{\sqrt 3}{6} \approx 0.2887$ , and at $x = -2$ exactly the same value: the two branches rise, the derivative is positive on both.

b)

With $u = \frac{1}{x}$ , $u' = -\frac{1}{x^2}$ : $\frac{d}{dx}\cos^{-1}\frac{1}{x} = -\frac{-1/x^2}{\sqrt{1 - 1/x^2}} = \frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}}$ . Bring the inside to one fraction: $1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$ . Its square root is $\frac{\sqrt{x^2 - 1}}{\sqrt{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|}$ , because $\sqrt{x^2} = |x|$ and NOT $x$ : for $x = -2$ , $\sqrt{4} = 2$ . Then $\frac{1}{x^2 \cdot \frac{\sqrt{x^2 - 1}}{|x|}} = \frac{|x|}{x^2\sqrt{x^2 - 1}} = \frac{1}{|x|\sqrt{x^2 - 1}}$ , since $x^2 = |x|^2$ . The same absolute value appears by a completely different road: it is not a convention of Thomas, it is forced by the algebra. Writing $\sqrt{x^2} = x$ in this line is the most expensive slip of section 3.9, because it gives a formula that is wrong on a whole branch.

c)

Differentiate the identities; the constant $\frac{\pi}{2}$ has derivative $0$ . $\frac{d}{dx}\cot^{-1}x = -\frac{1}{1 + x^2}$ , which is $-\frac{1}{2}$ at $x = 1$ . $\frac{d}{dx}\csc^{-1}x = -\frac{1}{|x|\sqrt{x^2 - 1}}$ for $|x| > 1$ , which is $-\frac{1}{2\sqrt 3} \approx -0.2887$ at $x = -2$ . The pattern of Thomas' table: each 'co' function has the derivative of its partner with a minus sign, because it is a constant minus the partner. Nothing new has to be memorized, and the signs are checked on the graphs: $\cot^{-1}$ decreases.

d)

Chain rule, inner function $u = 3x$ , $u' = 3$ , $u^2 = 9x^2$ : $f'(x) = \frac{3}{|3x|\sqrt{9x^2 - 1}}$ . At $x = -\frac{2}{3}$ : $|3x| = 2$ and $9x^2 - 1 = 4 - 1 = 3$ , so $f'\left(-\frac{2}{3}\right) = \frac{3}{2\sqrt 3} = \frac{\sqrt 3}{2} \approx 0.8660$ . The student's formula has $3x = -2$ in the denominator and gives $-\frac{\sqrt 3}{2} \approx -0.8660$ . The figure refutes it without any computation: $f(x) = \sec^{-1}(3x)$ is the left branch of the graph compressed horizontally, and a rising curve cannot have a negative slope. Keep the absolute value until the very last substitution, then replace $|3x|$ by $-3x$ when $x < 0$ .

e)

Point: $\sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ . Slope from a): $\frac{1}{|-2|\sqrt 3} = \frac{1}{2\sqrt 3} \approx 0.2887$ . Tangent: $y - \frac{2\pi}{3} = \frac{1}{2\sqrt 3}(x + 2)$ , that is $y = \frac{1}{2\sqrt 3}x + \frac{2\pi}{3} + \frac{1}{\sqrt 3}$ , with $y$ -intercept $\frac{2\pi}{3} + \frac{1}{\sqrt 3} \approx 2.6717$ . The point is on the left branch of the figure, at height about $2.09$ , and the tangent rises gently, as it should.

Exercise 5: The chain rule through an inverse trigonometric function, and the square root of a square

In $\sin^{-1}u$ , $\cos^{-1}u$ or $\tan^{-1}u$ , name the inner function $u$ first, write $u'$ next to it, and only then apply the formula. When $1 - u^2$ is itself a square, its root is an ABSOLUTE VALUE: $\sqrt{1 - \cos^2 x} = \sqrt{\sin^2 x} = |\sin x|$ .

Part d) uses a differentiable function $g$ known only through a table: $g(3) = \frac{1}{2}$ , $g'(3) = 6$ , $g\left(\frac{\pi}{4}\right) = 5$ , $g'\left(\frac{\pi}{4}\right) = 2$ .

a) Let $y = \arctan(e^{2x})$ . Name the inner function, differentiate, and evaluate $y'(0)$ .
b) Differentiate $y = (\arcsin x)^2$ and $z = \arctan(\ln x)$ . Evaluate $y'\left(\frac{1}{2}\right)$ and $z'(e)$ to four decimals.
c) Let $y = \arcsin(\cos x)$ . Show that $y' = -1$ for $0 < x < \pi$ , and find $y'$ for $-\pi < x < 0$ . Deduce a formula for $y$ on $[0, \pi]$ , reading the constant at $x = \frac{\pi}{2}$ , and check it at $x = \frac{\pi}{3}$ .
d) With the table of the introduction, find $h'(3)$ for $h(x) = \arcsin(g(x))$ and $k'(1)$ for $k(x) = g(\arctan x)$ .
e) Differentiate $y = \arctan(\sin 3x)$ , naming its three layers, and evaluate $y'\left(\frac{\pi}{9}\right)$ .

Type your answers, the page tells you right or wrong 0/8

a)

y'(0)

b)

y'\left(\frac{1}{2}\right)

z'(e)

c)

y'

for

-\pi < x < 0

y\left(\frac{\pi}{3}\right)

(in multiples of

\pi

)

d)

h'(3)

k'(1)

e)

y'\left(\frac{\pi}{9}\right)

Show the solution

Answers

a) $u = e^{2x}$ ; $y' = \frac{2e^{2x}}{1 + e^{4x}}$ , $y'(0) = 1$
b) $y' = \frac{2\arcsin x}{\sqrt{1 - x^2}}$ , $y'\left(\frac{1}{2}\right) = \frac{2\pi}{3\sqrt 3} \approx 1.2092$ ; $z' = \frac{1}{x(1 + \ln^2 x)}$ , $z'(e) = \frac{1}{2e} \approx 0.1839$
c) $y' = -\frac{\sin x}{|\sin x|}$ : $-1$ on $(0, \pi)$ , $+1$ on $(-\pi, 0)$ ; $y = \frac{\pi}{2} - x$ on $[0, \pi]$ , and $y\left(\frac{\pi}{3}\right) = \frac{\pi}{6}$
d) $h'(3) = \frac{6}{\sqrt{3}/2} = 4\sqrt 3 \approx 6.9282$ ; $k'(1) = g'\left(\frac{\pi}{4}\right) \cdot \frac{1}{2} = 1$
e) $y' = \frac{3\cos 3x}{1 + \sin^2 3x}$ , $y'\left(\frac{\pi}{9}\right) = \frac{6}{7} \approx 0.8571$

a)

Inner function $u = e^{2x}$ , whose derivative is $u' = 2e^{2x}$ (chain rule once more, inner $2x$ ). Then $u^2 = (e^{2x})^2 = e^{4x}$ by the law $(e^a)^b = e^{ab}$ ; it is not $e^{4x^2}$ and not $e^{2x^2}$ . So $y' = \frac{2e^{2x}}{1 + e^{4x}}$ , and $y'(0) = \frac{2}{1 + 1} = 1$ . The exponent law is the algebra of this line, and a wrong exponent gives the right value at $0$ by coincidence and a wrong function everywhere else: at $x = 1$ the correct slope is $\frac{2e^2}{1 + e^4} \approx 0.266$ .

b)

For $y = (\arcsin x)^2$ the OUTER function is the square and the inner one is $\arcsin x$ : $y' = 2\arcsin x \cdot \frac{1}{\sqrt{1 - x^2}}$ . Do not confuse it with $\arcsin(x^2)$ , whose outer function is $\arcsin$ . At $x = \frac{1}{2}$ : $\frac{2 \cdot \frac{\pi}{6}}{\frac{\sqrt 3}{2}} = \frac{2\pi}{3\sqrt 3} \approx 1.2092$ . For $z = \arctan(\ln x)$ , $x > 0$ : inner function $\ln x$ , derivative $\frac{1}{x}$ , so $z' = \frac{1/x}{1 + (\ln x)^2} = \frac{1}{x(1 + \ln^2 x)}$ . At $x = e$ , $\ln e = 1$ : $z'(e) = \frac{1}{2e} \approx 0.1839$ . Writing $\ln^2 x$ means $(\ln x)^2$ , not $\ln(x^2) = 2\ln x$ ; at $x = e$ the confusion gives $\frac{1}{3e}$ .

c)

Inner function $u = \cos x$ , $u' = -\sin x$ , $1 - u^2 = 1 - \cos^2 x = \sin^2 x$ . So $y' = \frac{-\sin x}{\sqrt{\sin^2 x}} = -\frac{\sin x}{|\sin x|}$ . On $(0, \pi)$ , $\sin x > 0$ , $|\sin x| = \sin x$ and $y' = -1$ . On $(-\pi, 0)$ , $\sin x < 0$ , $|\sin x| = -\sin x$ and $y' = +1$ . The student who writes $\sqrt{\sin^2 x} = \sin x$ finds $-1$ everywhere and misses half of the graph. On the interval $(0, \pi)$ the function $y + x$ has derivative $0$ , so it is constant there (the zero-derivative fact, used as a tool). Read it at $x = \frac{\pi}{2}$ : $\arcsin(\cos\frac{\pi}{2}) + \frac{\pi}{2} = \arcsin 0 + \frac{\pi}{2} = \frac{\pi}{2}$ . Hence $y = \frac{\pi}{2} - x$ on $(0, \pi)$ , and also at the endpoints, where $\arcsin 1 = \frac{\pi}{2}$ and $\arcsin(-1) = -\frac{\pi}{2}$ . Check at $\frac{\pi}{3}$ : $\arcsin\frac{1}{2} = \frac{\pi}{6} = \frac{\pi}{2} - \frac{\pi}{3}$ .

d)

Chain rule on $h$ , inner function $g$ : $h'(x) = \frac{g'(x)}{\sqrt{1 - g(x)^2}}$ . At $x = 3$ : $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt 3}{2}$ , so $h'(3) = \frac{6}{\sqrt 3/2} = \frac{12}{\sqrt 3} = 4\sqrt 3 \approx 6.9282$ . Chain rule on $k$ , inner function $\arctan x$ , outer function $g$ : $k'(x) = g'(\arctan x) \cdot \frac{1}{1 + x^2}$ . At $x = 1$ , $\arctan 1 = \frac{\pi}{4}$ , so $k'(1) = g'\left(\frac{\pi}{4}\right) \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1$ . The table is built with two traps: $g\left(\frac{\pi}{4}\right) = 5$ is there to be confused with $g'\left(\frac{\pi}{4}\right)$ , which gives $2.5$ ; and $g'(1)$ is not in the table because the outer derivative is evaluated at the INNER value $\arctan 1$ , not at $1$ .

e)

Three layers, from the outside: $\arctan$ , then $\sin$ , then $3x$ . Differentiate each at the layer below it: $y' = \frac{1}{1 + \sin^2 3x} \cdot \cos 3x \cdot 3 = \frac{3\cos 3x}{1 + \sin^2 3x}$ . At $x = \frac{\pi}{9}$ , $3x = \frac{\pi}{3}$ : $\cos\frac{\pi}{3} = \frac{1}{2}$ and $\sin^2\frac{\pi}{3} = \frac{3}{4}$ , so $y' = \frac{3/2}{7/4} = \frac{6}{7} \approx 0.8571$ . The factor $3$ of the innermost layer is the one that disappears on exams; the check is to count the layers, three, and the factors in the answer, three.

Exercise 6: Two functions that agree on one side only: arcsec from the arctan key

Most scientific calculators, including the models approved for the course, have no $\sec^{-1}$ key. This exercise builds a formula that computes $\sec^{-1}x$ with the $\tan^{-1}$ key, and proves it with the tool of the chapter: if $F'(x) = 0$ at every point of an INTERVAL, then $F$ is constant on that interval (admitted), and the constant is read at one convenient point OF THAT INTERVAL.

Let $g(x) = \tan^{-1}\sqrt{x^2 - 1}$ and $h(x) = \sec^{-1}x$ , both defined for $|x| \ge 1$ . Recall from section 3.9 that $h'(x) = \frac{1}{|x|\sqrt{x^2 - 1}}$ for $|x| > 1$ . The figure shows $h$ in blue and $g$ in dashed orange: they coincide on the right and separate on the left.

a) Show that $g$ is even and that $h$ is not. Compute $g(2)$ , $g(-2)$ , $h(2)$ and $h(-2)$ exactly.
b) Differentiate $g$ for $|x| > 1$ and show that $g'(x) = \frac{1}{x\sqrt{x^2 - 1}}$ . Compare $g'$ with $h'$ on each side, and evaluate both at $x = -2$ .
c) Show that $g - h$ is constant on $(1, \infty)$ and find the constant.
d) Show that $g + h$ is constant on $(-\infty, -1)$ and find the constant. Why is it wrong to read this constant at $x = 2$ ?
e) Conclude: write $\sec^{-1}x$ in terms of $\tan^{-1}$ for $x \ge 1$ and for $x \le -1$ (check the endpoints $\pm 1$ directly), then compute $\sec^{-1}(-3)$ to four decimals with the $\tan^{-1}$ key, and check it with $\cos^{-1}\left(-\frac{1}{3}\right)$ .

Type your answers, the page tells you right or wrong 0/7

a)

g(-2)

(in multiples of

\pi

)

h(-2)

(in multiples of

\pi

)

b)

g'(-2)

h'(-2)

c)

The constant

g - h

on

(1, \infty)

d)

The constant

g + h

on

(-\infty, -1)

(in multiples of

\pi

)

e)

\sec^{-1}(-3)

Show the solution

Answers

a) $g(-x) = g(x)$ ; $h(2) = \frac{\pi}{3} \ne h(-2) = \frac{2\pi}{3}$ ; $g(2) = g(-2) = \frac{\pi}{3}$
b) $g'(x) = \frac{1}{x\sqrt{x^2 - 1}}$ : $g' = h'$ for $x > 1$ , $g' = -h'$ for $x < -1$ ; $g'(-2) = -\frac{1}{2\sqrt 3}$ , $h'(-2) = \frac{1}{2\sqrt 3}$
c) $(g - h)' = 0$ on $(1, \infty)$ , constant $g(2) - h(2) = 0$
d) $(g + h)' = 0$ on $(-\infty, -1)$ , constant $g(-2) + h(-2) = \pi$ ; $x = 2$ is not in that interval
e) $\sec^{-1}x = \tan^{-1}\sqrt{x^2 - 1}$ for $x \ge 1$ , $\pi - \tan^{-1}\sqrt{x^2 - 1}$ for $x \le -1$ ; $\sec^{-1}(-3) \approx 1.9106$

a)

$g(-x) = \tan^{-1}\sqrt{(-x)^2 - 1} = \tan^{-1}\sqrt{x^2 - 1} = g(x)$ : $g$ only sees $x^2$ , so it is even and its graph is symmetric in the $y$ -axis, as the orange curve of the figure is. $h$ is not even: $h(2) = \cos^{-1}\frac{1}{2} = \frac{\pi}{3}$ but $h(-2) = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ . And $g(2) = g(-2) = \tan^{-1}\sqrt 3 = \frac{\pi}{3}$ . The values already say it: $g = h$ at $x = 2$ , but not at $x = -2$ , where $g(-2) + h(-2) = \pi$ .

b)

Inner function $u = \sqrt{x^2 - 1} = (x^2 - 1)^{1/2}$ , $u' = \frac{1}{2}(x^2 - 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}}$ . Now the gesture of the chapter: $1 + u^2 = 1 + (x^2 - 1) = x^2$ , since squaring a square root of a nonnegative number gives the number back. So $g'(x) = \frac{x/\sqrt{x^2 - 1}}{x^2} = \frac{1}{x\sqrt{x^2 - 1}}$ , with $x$ and not $|x|$ : here no root of a square was taken, and the sign of $x$ stays. For $x > 1$ , $x = |x|$ and $g'(x) = h'(x)$ . For $x < -1$ , $x = -|x|$ and $g'(x) = -h'(x)$ . At $x = -2$ : $g'(-2) = \frac{1}{-2\sqrt 3} \approx -0.2887$ and $h'(-2) = \frac{1}{2\sqrt 3} \approx 0.2887$ . The figure agrees: on the left, the orange curve goes down while the blue one goes up.

c)

On $(1, \infty)$ , $(g - h)'(x) = g'(x) - h'(x) = 0$ at every point, and $(1, \infty)$ is an interval, so $g - h$ is constant there. Read the constant at a point of THIS interval, $x = 2$ : $g(2) - h(2) = \frac{\pi}{3} - \frac{\pi}{3} = 0$ . So $g(x) = h(x)$ for every $x > 1$ , which is the coincidence of the two curves on the right of the figure.

d)

On $(-\infty, -1)$ , $(g + h)'(x) = g'(x) + h'(x) = -h'(x) + h'(x) = 0$ , so $g + h$ is constant on this interval. Read it at $x = -2$ : $g(-2) + h(-2) = \frac{\pi}{3} + \frac{2\pi}{3} = \pi$ . So $h(x) = \pi - g(x)$ for $x < -1$ : the left branches of the figure are mirror images in the dashed line $y = \frac{\pi}{2}$ . Reading the constant at $x = 2$ would give $g(2) + h(2) = \frac{2\pi}{3}$ , a wrong answer: the fact 'zero derivative gives a constant' holds on ONE interval, and $x = 2$ belongs to the other one. The domain $|x| > 1$ is made of two separate intervals, and each gets its own constant, read at its own point.

e)

Collecting c) and d): $\sec^{-1}x = \tan^{-1}\sqrt{x^2 - 1}$ for $x > 1$ and $\sec^{-1}x = \pi - \tan^{-1}\sqrt{x^2 - 1}$ for $x < -1$ . The endpoints are checked directly: $\sec^{-1}1 = 0 = \tan^{-1}0$ and $\sec^{-1}(-1) = \pi = \pi - \tan^{-1}0$ . With the calculator in radians: $\sec^{-1}(-3) = \pi - \tan^{-1}\sqrt 8 \approx 3.141593 - 1.230959 = 1.910633$ , that is $1.9106$ . Check: $\cos^{-1}\left(-\frac{1}{3}\right) \approx 1.9106$ , the same. The tempting shortcut $\tan^{-1}\sqrt 8 \approx 1.2310$ is the formula of the RIGHT branch used on the left; it gives an angle below $\frac{\pi}{2}$ , impossible for $\sec^{-1}$ of a negative number.

Exercise 7: Horizontal tangents and prescribed slopes: one denominator, then factor

Let $f(x) = \sin^{-1}x - 2\sqrt{1 - x^2}$ on $[-1, 1]$ , whose graph is in the figure, and $g(x) = 4\tan^{-1}x - x^2$ for every real $x$ .

A horizontal tangent is a point where the derivative is $0$ ; a tangent parallel to a line $y = mx + b$ is a point where the derivative is $m$ . In both cases the equation is solved on ONE fraction: put the derivative over a common denominator, then factor the numerator. Exact answers first, then decimals.

a) Show that $f'(x) = \frac{1 + 2x}{\sqrt{1 - x^2}}$ for $-1 < x < 1$ , and find the point of the graph of $f$ with a horizontal tangent, exactly and to four decimals.
b) What does $f'(x)$ do as $x \to 1^-$ and as $x \to -1^+$ ? Describe the tangent to the graph at each endpoint.
c) Find every point of the graph of $f$ where the tangent has slope $2$ . Squaring is allowed, but every candidate must be checked.
d) Show that $g'(x) = \frac{4 - 2x - 2x^3}{1 + x^2}$ and that the graph of $g$ has exactly one horizontal tangent. Give the point.
e) Find every point of the graph of $g$ where the tangent is parallel to the line $y = 4x$ .

Show the solution

Answers

a) Horizontal tangent at $\left(-\frac{1}{2}, -\frac{\pi}{6} - \sqrt 3\right) \approx (-0.5, -2.2556)$
b) $f'(x) \to +\infty$ as $x \to 1^-$ and $-\infty$ as $x \to -1^+$ : vertical tangents at both endpoints
c) One point, $x = \frac{-1 + \sqrt 7}{4} \approx 0.4114$ ; $x = \frac{-1 - \sqrt 7}{4}$ rejected
d) $x^3 + x - 2 = (x - 1)(x^2 + x + 2)$ : only $x = 1$ , the point $(1, \pi - 1)$
e) $-2x(x + 1)^2 = 0$ : at $(0, 0)$ and $(-1, -\pi - 1)$

a)

Differentiate term by term: $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$ , and by the chain rule $\frac{d}{dx}\left(-2\sqrt{1 - x^2}\right) = -2 \cdot \frac{-2x}{2\sqrt{1 - x^2}} = \frac{2x}{\sqrt{1 - x^2}}$ . The two terms have the SAME denominator, so $f'(x) = \frac{1 + 2x}{\sqrt{1 - x^2}}$ . A fraction is $0$ when its numerator is $0$ and its denominator is not: $1 + 2x = 0$ gives $x = -\frac{1}{2}$ , where $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt 3}{2} \ne 0$ . Then $f\left(-\frac{1}{2}\right) = -\frac{\pi}{6} - 2 \cdot \frac{\sqrt 3}{2} = -\frac{\pi}{6} - \sqrt 3 \approx -2.2556$ . The figure agrees: the lowest point of the curve is near $(-0.5, -2.26)$ . Leaving $f'$ as two fractions and setting each to $0$ separately is the mistake that finds nothing.

b)

As $x \to 1^-$ , the numerator $1 + 2x \to 3$ and the denominator $\sqrt{1 - x^2} \to 0^+$ , so $f'(x) \to +\infty$ . As $x \to -1^+$ , the numerator tends to $-1$ and the denominator to $0^+$ , so $f'(x) \to -\infty$ . The slopes grow without bound: the graph arrives VERTICALLY at both endpoints $\left(1, \frac{\pi}{2}\right)$ and $\left(-1, -\frac{\pi}{2}\right)$ , which is what the figure shows. Numerically, $f'(0.99) \approx 21.1$ and $f'(-0.99) \approx -6.9$ .

c)

Solve $\frac{1 + 2x}{\sqrt{1 - x^2}} = 2$ , that is $1 + 2x = 2\sqrt{1 - x^2}$ . The right side is positive, so any solution must have $1 + 2x > 0$ : note it BEFORE squaring. Squaring: $1 + 4x + 4x^2 = 4 - 4x^2$ , so $8x^2 + 4x - 3 = 0$ and $x = \frac{-4 \pm \sqrt{16 + 96}}{16} = \frac{-1 \pm \sqrt 7}{4}$ . The candidate $\frac{-1 + \sqrt 7}{4} \approx 0.4114$ has $1 + 2x \approx 1.82 > 0$ : kept, and $f'(0.4114) \approx 2.000$ . The candidate $\frac{-1 - \sqrt 7}{4} \approx -0.9114$ has $1 + 2x \approx -0.82 < 0$ : at that point $f' = -2$ , not $2$ . Squaring merged the equations $f' = 2$ and $f' = -2$ ; the check separates them again. One point only.

d)

$g'(x) = \frac{4}{1 + x^2} - 2x = \frac{4 - 2x(1 + x^2)}{1 + x^2} = \frac{4 - 2x - 2x^3}{1 + x^2}$ , over the common denominator $1 + x^2$ , never $0$ . Horizontal tangent: $4 - 2x - 2x^3 = 0$ , that is $x^3 + x - 2 = 0$ after dividing by $-2$ . $x = 1$ is a root ( $1 + 1 - 2 = 0$ ), so factor: $x^3 + x - 2 = (x - 1)(x^2 + x + 2)$ , and $x^2 + x + 2$ has discriminant $1 - 8 < 0$ , no real root. Exactly one horizontal tangent, at $x = 1$ , where $g(1) = 4 \cdot \frac{\pi}{4} - 1 = \pi - 1 \approx 2.1416$ .

e)

Parallel to $y = 4x$ means $g'(x) = 4$ : $\frac{4 - 2x - 2x^3}{1 + x^2} = 4$ , so $4 - 2x - 2x^3 = 4 + 4x^2$ , that is $-2x^3 - 4x^2 - 2x = 0$ . Factor $-2x$ first, the common factor that a division by $x$ would lose: $-2x(x^2 + 2x + 1) = -2x(x + 1)^2 = 0$ . Two points: $x = 0$ , with $g(0) = 0$ , and $x = -1$ , with $g(-1) = 4\left(-\frac{\pi}{4}\right) - 1 = -\pi - 1 \approx -4.1416$ . Check: $g'(-1) = \frac{4}{2} + 2 = 4$ . Dividing both sides by $x$ at the start keeps only $x = -1$ and loses the tangent $y = 4x$ at the origin itself.

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 203 assignment, and each is false. Say what is wrong, give a counterexample or the correct computation, and write the correct statement.

a) $\sin^{-1}\left(\sin\frac{5\pi}{6}\right) = \frac{5\pi}{6}$ , since $\sin^{-1}$ undoes $\sin$ .
b) $\frac{d}{dx}\sec^{-1}x = \frac{1}{x\sqrt{x^2 - 1}}$ for every $x$ with $|x| > 1$ .
c) $\tan^{-1}x = \frac{\sin^{-1}x}{\cos^{-1}x}$ for $-1 < x < 1$ , because $\tan = \frac{\sin}{\cos}$ .
d) $\frac{d}{dx}\tan^{-1}\frac{1}{x} = \frac{1}{1 + \frac{1}{x^2}} = \frac{x^2}{x^2 + 1}$ .
e) The domain of $\sin^{-1}(3x)$ is $[-1, 1]$ , and $\frac{d}{dx}\sin^{-1}(3x) = \frac{3}{\sqrt{1 - 3x^2}}$ .

Type your answers, the page tells you right or wrong 0/10

a)

The statement is

The true value (in multiples of

\pi

)

b)

The statement is

The true derivative at

x = -2

c)

The statement is

\tan^{-1}\frac{1}{2}

d)

The statement is

The true derivative at

x = 1

e)

The statement is

The true derivative at

x = 0.3

Show the solution

Answers

a) False: $\sin\frac{5\pi}{6} = \frac{1}{2}$ and $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$
b) False for $x < -1$ : the derivative is $\frac{1}{|x|\sqrt{x^2 - 1}}$ , equal to $+\frac{1}{2\sqrt 3}$ at $x = -2$
c) False: at $x = \frac{1}{2}$ the right side is $\frac{1}{2}$ , while $\tan^{-1}\frac{1}{2} \approx 0.4636$
d) False: the chain factor $-\frac{1}{x^2}$ is missing; $\frac{d}{dx}\tan^{-1}\frac{1}{x} = -\frac{1}{1 + x^2}$
e) False twice: domain $\left[-\frac{1}{3}, \frac{1}{3}\right]$ , derivative $\frac{3}{\sqrt{1 - 9x^2}}$

a)

$\sin^{-1}$ undoes $\sin$ only on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , and $\frac{5\pi}{6}$ is not in that interval. Compute from the inside: $\sin\frac{5\pi}{6} = \frac{1}{2}$ , and the angle of the range with sine $\frac{1}{2}$ is $\frac{\pi}{6}$ . So $\sin^{-1}\left(\sin\frac{5\pi}{6}\right) = \frac{\pi}{6}$ . Correct statement: $\sin^{-1}(\sin x) = x$ for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ only, while $\sin(\sin^{-1}x) = x$ for every $x$ of $[-1, 1]$ . The student's answer fails the one-second test: $\frac{5\pi}{6} > \frac{\pi}{2}$ is not a possible value of $\sin^{-1}$ .

b)

The formula is right for $x > 1$ and wrong for $x < -1$ . At $x = -2$ it gives $\frac{1}{-2\sqrt 3} < 0$ , but $\sec^{-1}$ rises on its left branch, so its slope there is positive. The implicit derivation gives $\frac{1}{\sec y\tan y}$ with $\tan y < 0$ on the left branch, and $\sec y\tan y = x \cdot \left(-\sqrt{x^2 - 1}\right) = |x|\sqrt{x^2 - 1}$ . Correct statement: $\frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^2 - 1}}$ for $|x| > 1$ , which is $\frac{1}{2\sqrt 3} \approx 0.2887$ at $x = -2$ .

c)

The identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$ is about ONE angle $\theta$ . The three inverse functions of the statement return three DIFFERENT angles for the same $x$ , so nothing allows dividing them. Counterexample at $x = \frac{1}{2}$ : $\frac{\sin^{-1}(1/2)}{\cos^{-1}(1/2)} = \frac{\pi/6}{\pi/3} = \frac{1}{2}$ , while $\tan^{-1}\frac{1}{2} \approx 0.4636$ . At $x = 1$ it is worse: $\cos^{-1}1 = 0$ , a division by $0$ . What IS true, by the triangle of the angle $\theta = \sin^{-1}x$ : $\tan(\sin^{-1}x) = \frac{x}{\sqrt{1 - x^2}}$ for $-1 < x < 1$ , a trigonometric function of ONE angle.

d)

The formula $\frac{1}{1 + u^2}$ was applied with $u = \frac{1}{x}$ , but the factor $u' = -\frac{1}{x^2}$ of the chain rule was dropped. Correctly: $\frac{d}{dx}\tan^{-1}\frac{1}{x} = \frac{-1/x^2}{1 + 1/x^2}$ , and multiplying top and bottom by $x^2$ (the complex fraction gesture) gives $-\frac{1}{x^2 + 1}$ for $x \ne 0$ . At $x = 1$ the student finds $+\frac{1}{2}$ and the true value is $-\frac{1}{2}$ : a sign error that a single look at $\tan^{-1}\frac{1}{x}$ reveals, since $\frac{1}{x}$ decreases on $(0, \infty)$ and so does its arctangent.

e)

Two errors. The domain: $\sin^{-1}u$ needs $-1 \le u \le 1$ , here $-1 \le 3x \le 1$ , so $-\frac{1}{3} \le x \le \frac{1}{3}$ . The derivative: $u^2 = (3x)^2 = 9x^2$ , the square applies to the $3$ as well, so $\frac{d}{dx}\sin^{-1}(3x) = \frac{3}{\sqrt{1 - 9x^2}}$ for $|x| < \frac{1}{3}$ . At $x = 0.3$ the student gets $\frac{3}{\sqrt{0.73}} \approx 3.511$ and the true value is $\frac{3}{\sqrt{0.19}} \approx 6.8825$ . The student's formula is even defined at $x = 0.5$ , a point where the function itself does not exist: a derivative can never have a larger domain than its function.

Exercise 9: The angle under which a painting is seen, and its rate with distance

In a gallery, a painting $2$ m tall hangs on a wall with its lower edge $1$ m above the eye level of a visitor and its upper edge $3$ m above. The visitor stands at a horizontal distance $x$ metres from the wall, $x > 0$ . The angle $\theta$ under which the painting is seen is the angle at the eye $E$ between the line of sight to the upper edge and the line of sight to the lower edge, as in the figure.

Scientific calculator allowed; angles in radians unless degrees are asked. For b), recall $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$ . This exercise studies $\theta$ as a function of $x$ and its rate $\frac{d\theta}{dx}$ ; where the angle is LARGEST is an optimization question for section 4.6, not asked here.

a) Explain why $\theta(x) = \tan^{-1}\frac{3}{x} - \tan^{-1}\frac{1}{x}$ . Compute $\theta(2)$ in radians, to four decimals, and in degrees, to two decimals.
b) Show that $\tan\theta = \frac{2x}{x^2 + 3}$ , and deduce that $\theta(x) = \tan^{-1}\frac{2x}{x^2 + 3}$ . Why is it legitimate to apply $\tan^{-1}$ here? Give $\tan\theta$ at $x = 2$ .
c) Show that $\frac{d}{dx}\tan^{-1}\frac{a}{x} = -\frac{a}{x^2 + a^2}$ for a constant $a > 0$ , then write $\theta'(x)$ as a single fraction and evaluate $\theta'(1)$ and $\theta'(2)$ .
d) Give the units of $\theta'(1)$ and its meaning in one sentence, then convert it to degrees per metre. At $x = 2$ , does the angle grow or shrink when the visitor steps back a little?
e) Differentiate the one-arctangent form of b) directly, and show that the two routes agree by factoring $x^4 + 10x^2 + 9$ . Then evaluate $\theta'(3)$ .

Show the solution

Answers

a) $\theta = \beta - \alpha$ with $\tan\beta = \frac{3}{x}$ , $\tan\alpha = \frac{1}{x}$ ; $\theta(2) \approx 0.5191$ rad $\approx 29.74°$
b) $\tan\theta = \frac{2/x}{1 + 3/x^2} = \frac{2x}{x^2 + 3}$ , and $0 < \theta < \frac{\pi}{2}$ ; $\tan\theta(2) = \frac{4}{7}$
c) $\theta'(x) = \frac{1}{x^2 + 1} - \frac{3}{x^2 + 9} = \frac{2(3 - x^2)}{(x^2 + 1)(x^2 + 9)}$ ; $\theta'(1) = 0.2$ , $\theta'(2) = -\frac{2}{65} \approx -0.0308$
d) Radians per metre: at $1$ m, the angle grows by about $0.2$ rad per metre stepped back, $\approx 11.46°$ per metre; at $2$ m it shrinks
e) $x^4 + 10x^2 + 9 = (x^2 + 1)(x^2 + 9)$ ; $\theta'(3) = -\frac{1}{15} \approx -0.0667$

a)

Let $\beta$ be the angle between the horizontal line of sight and the line to the upper edge, and $\alpha$ the angle to the lower edge. In the right triangles of the figure, the side opposite $\beta$ is $3$ and the adjacent side is $x$ , so $\tan\beta = \frac{3}{x}$ ; likewise $\tan\alpha = \frac{1}{x}$ . Both angles are in $\left(0, \frac{\pi}{2}\right)$ , the range of $\tan^{-1}$ , so $\beta = \tan^{-1}\frac{3}{x}$ and $\alpha = \tan^{-1}\frac{1}{x}$ , and $\theta = \beta - \alpha$ . At $x = 2$ : $\tan^{-1}1.5 - \tan^{-1}0.5 \approx 0.982794 - 0.463648 = 0.519146$ , so $\theta(2) \approx 0.5191$ rad. In degrees: $0.519146 \cdot \frac{180}{\pi} \approx 29.74°$ . Keep the six-digit values until the end: subtracting two rounded angles loses a digit.

b)

With $\tan\beta = \frac{3}{x}$ and $\tan\alpha = \frac{1}{x}$ : $\tan\theta = \frac{\frac{3}{x} - \frac{1}{x}}{1 + \frac{3}{x^2}}$ . This is a complex fraction, the gesture that costs marks in MATH 203: multiply its top and bottom by $x^2$ , the common denominator of every small fraction inside it. The top becomes $3x - x = 2x$ and the bottom $x^2 + 3$ , so $\tan\theta = \frac{2x}{x^2 + 3}$ . Since $0 < \alpha < \beta < \frac{\pi}{2}$ , the angle $\theta = \beta - \alpha$ lies in $\left(0, \frac{\pi}{2}\right)$ , inside the range of $\tan^{-1}$ ; so applying $\tan^{-1}$ gives back $\theta$ itself: $\theta = \tan^{-1}\frac{2x}{x^2 + 3}$ . Without that check the step could be false, exactly as $\tan^{-1}\left(\tan\frac{3\pi}{4}\right) \ne \frac{3\pi}{4}$ . At $x = 2$ : $\tan\theta = \frac{4}{7} \approx 0.5714$ , and $\tan^{-1}\frac{4}{7} \approx 0.5191$ , as in a).

c)

Inner function $u = \frac{a}{x} = ax^{-1}$ , $u' = -\frac{a}{x^2}$ . So $\frac{d}{dx}\tan^{-1}\frac{a}{x} = \frac{-a/x^2}{1 + a^2/x^2}$ ; multiply top and bottom by $x^2$ : $-\frac{a}{x^2 + a^2}$ . With $a = 3$ and $a = 1$ : $\theta'(x) = -\frac{3}{x^2 + 9} + \frac{1}{x^2 + 1}$ . One fraction, over $(x^2 + 1)(x^2 + 9)$ : the numerator is $-3(x^2 + 1) + (x^2 + 9) = 6 - 2x^2$ , so $\theta'(x) = \frac{2(3 - x^2)}{(x^2 + 1)(x^2 + 9)}$ . Watch the distribution of $-3$ over $x^2 + 1$ : writing $-3x^2 + 1$ gives $10 - 2x^2$ and a wrong function. $\theta'(1) = \frac{4}{2 \cdot 10} = 0.2$ and $\theta'(2) = \frac{-2}{5 \cdot 13} = -\frac{2}{65} \approx -0.0308$ .

d)

$\theta$ is in radians and $x$ in metres, so $\theta'(1)$ is in radians per metre. Meaning: when the visitor stands $1$ m from the wall, the viewing angle increases at the rate of $0.2$ rad for each metre of stepping back, that is, about $0.02$ rad for a step of $10$ cm. In degrees: $0.2 \cdot \frac{180}{\pi} \approx 11.46°$ per metre. At $x = 2$ , $\theta'(2) \approx -0.0308 < 0$ : stepping back a little makes the painting look slightly SMALLER. So the angle grows when one steps back from $1$ m and shrinks when one steps back from $2$ m; where exactly the change of behaviour happens, and why the angle is largest there, belongs to the optimization chapter.

e)

With $u = \frac{2x}{x^2 + 3}$ , the quotient rule gives $u' = \frac{2(x^2 + 3) - 2x \cdot 2x}{(x^2 + 3)^2} = \frac{6 - 2x^2}{(x^2 + 3)^2}$ , and $1 + u^2 = \frac{(x^2 + 3)^2 + 4x^2}{(x^2 + 3)^2} = \frac{x^4 + 10x^2 + 9}{(x^2 + 3)^2}$ . In the quotient $\frac{u'}{1 + u^2}$ the factors $(x^2 + 3)^2$ cancel: $\theta'(x) = \frac{6 - 2x^2}{x^4 + 10x^2 + 9}$ . Factor the denominator as a quadratic in $x^2$ : two numbers with product $9$ and sum $10$ are $1$ and $9$ , so $x^4 + 10x^2 + 9 = (x^2 + 1)(x^2 + 9)$ , and the two routes agree. Then $\theta'(3) = \frac{6 - 18}{10 \cdot 18} = -\frac{12}{180} = -\frac{1}{15} \approx -0.0667$ .

Exercise 10: A final exam question: one function, two corners, three formulas

Let $f(x) = \sin^{-1}\frac{2x}{1 + x^2}$ . The figure shows its graph: it climbs to $\frac{\pi}{2}$ at $x = 1$ , where it turns sharply, and it is symmetric with respect to the origin. A final exam often takes one such function and asks every question of the chapter about it: domain, values, derivative with a square root to simplify, differentiability, and an identity proved by a zero derivative.

Exact answers, then decimals when asked. The fact 'a zero derivative on an interval gives a constant' is used as a tool.

a) Prove that $-1 \le \frac{2x}{1 + x^2} \le 1$ for every real $x$ , so that $f$ is defined on $\mathbb{R}$ . Compute $f(1)$ , $f(-1)$ and $f(\sqrt 3)$ exactly.
b) Show that $1 - \left(\frac{2x}{1 + x^2}\right)^2 = \frac{(1 - x^2)^2}{(1 + x^2)^2}$ , deduce that $f'(x) = \frac{2(1 - x^2)}{(1 + x^2)\,|1 - x^2|}$ for $x \ne \pm 1$ , and write $f'$ without absolute value on each interval. Evaluate $f'(0)$ and $f'(2)$ .
c) Which slopes do the two pieces of the graph approach at $x = 1$ , from the left and from the right? Is $f$ differentiable at $x = 1$ ? Describe the graph there.
d) Compare $f'$ with the derivative of $2\tan^{-1}x$ . Deduce, interval by interval, that $f(x) = 2\tan^{-1}x$ on $[-1, 1]$ and $f(x) = \pi - 2\tan^{-1}x$ for $x \ge 1$ , reading each constant at a point of its own interval. Find the formula for $x \le -1$ .
e) Solve $f(x) = \frac{\pi}{3}$ and explain why there are two solutions although $\sin^{-1}$ returns only one angle. Then compute $f(3)$ to four decimals in two ways.

Type your answers, the page tells you right or wrong 0/13

a)

f(1)

(in multiples of

\pi

)

f(-1)

(in multiples of

\pi

)

f(\sqrt 3)

(in multiples of

\pi

)

b)

f'(0)

f'(2)

c)

Slope from the left

Slope from the right

Is

f

differentiable at

x = 1

?

d)

f(x) + 2\tan^{-1}x

for

x > 1

(in multiples of

\pi

)

f(x) + 2\tan^{-1}x

for

x < -1

(in multiples of

\pi

)

e)

The smaller solution

The larger solution

f(3)

Show the solution

Answers

a) $1 + x^2 \mp 2x = (1 \mp x)^2 \ge 0$ ; $f(1) = \frac{\pi}{2}$ , $f(-1) = -\frac{\pi}{2}$ , $f(\sqrt 3) = \frac{\pi}{3}$
b) $(1 + x^2)^2 - 4x^2 = (1 - x^2)^2$ , root $|1 - x^2|$ ; $f' = \frac{2}{1 + x^2}$ for $|x| < 1$ , $-\frac{2}{1 + x^2}$ for $|x| > 1$ ; $f'(0) = 2$ , $f'(2) = -\frac{2}{5}$
c) Left slope $1$ , right slope $-1$ : a corner, $f$ is not differentiable at $1$
d) $f = 2\tan^{-1}x$ on $[-1, 1]$ ; $f + 2\tan^{-1}x = \pi$ on $[1, \infty)$ ; $f = -\pi - 2\tan^{-1}x$ on $(-\infty, -1]$
e) $x = \frac{1}{\sqrt 3}$ and $x = \sqrt 3$ ; $f(3) = \sin^{-1}\frac{3}{5} = \pi - 2\tan^{-1}3 \approx 0.6435$

a)

Since $1 + x^2 > 0$ , the double inequality is equivalent to $-(1 + x^2) \le 2x \le 1 + x^2$ . On the right: $1 + x^2 - 2x = (1 - x)^2 \ge 0$ . On the left: $1 + x^2 + 2x = (1 + x)^2 \ge 0$ . Both hold for every real $x$ , so $\frac{2x}{1 + x^2}$ is always in $[-1, 1]$ and $f$ is defined on $\mathbb{R}$ ; the values $\pm 1$ are reached exactly at $x = \pm 1$ . Then $f(1) = \sin^{-1}1 = \frac{\pi}{2}$ , $f(-1) = \sin^{-1}(-1) = -\frac{\pi}{2}$ , and $f(\sqrt 3) = \sin^{-1}\frac{2\sqrt 3}{4} = \sin^{-1}\frac{\sqrt 3}{2} = \frac{\pi}{3}$ . Recognizing a perfect square is the whole proof: expanding and 'testing some values' proves nothing.

b)

$1 - \frac{4x^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2}$ , and the numerator is $1 + 2x^2 + x^4 - 4x^2 = 1 - 2x^2 + x^4 = (1 - x^2)^2$ . Its square root is $\frac{|1 - x^2|}{1 + x^2}$ : the root of a square is an ABSOLUTE VALUE, and here $1 - x^2$ is negative as soon as $|x| > 1$ . The inner derivative, by the quotient rule: $u' = \frac{2(1 + x^2) - 2x \cdot 2x}{(1 + x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}$ . So $f'(x) = \frac{2(1 - x^2)}{(1 + x^2)^2} \cdot \frac{1 + x^2}{|1 - x^2|} = \frac{2(1 - x^2)}{(1 + x^2)\,|1 - x^2|}$ for $x \ne \pm 1$ . For $|x| < 1$ , $|1 - x^2| = 1 - x^2$ and $f'(x) = \frac{2}{1 + x^2}$ ; for $|x| > 1$ , $|1 - x^2| = x^2 - 1$ and $f'(x) = -\frac{2}{1 + x^2}$ . $f'(0) = 2$ and $f'(2) = -\frac{2}{5}$ . The student who writes $\sqrt{(1 - x^2)^2} = 1 - x^2$ finds $\frac{2}{1 + x^2}$ everywhere, a positive slope at $x = 2$ where the figure goes down.

c)

Just left of $1$ , $f'(x) = \frac{2}{1 + x^2}$ , which approaches $\frac{2}{2} = 1$ . Just right of $1$ , $f'(x) = -\frac{2}{1 + x^2}$ , which approaches $-1$ . The two pieces arrive at the point $\left(1, \frac{\pi}{2}\right)$ with slopes $1$ and $-1$ : the graph has a CORNER there, and $f$ is continuous but not differentiable at $x = 1$ . In the formula of b), the division by $|1 - x^2| = 0$ at $x = 1$ was the warning. By the symmetry of the graph, the same happens at $x = -1$ .

d)

$\frac{d}{dx}\left(2\tan^{-1}x\right) = \frac{2}{1 + x^2}$ . On the interval $(-1, 1)$ , $f' - 2(\tan^{-1})' = 0$ , so $f(x) - 2\tan^{-1}x$ is constant there; at $x = 0$ it is $0 - 0 = 0$ . So $f = 2\tan^{-1}x$ on $(-1, 1)$ , and also at $\pm 1$ : $f(1) = \frac{\pi}{2} = 2 \cdot \frac{\pi}{4}$ . On $(1, \infty)$ , $f' + 2(\tan^{-1})' = 0$ , so $f(x) + 2\tan^{-1}x$ is constant; read it at $x = \sqrt 3$ , a point of THAT interval: $\frac{\pi}{3} + 2 \cdot \frac{\pi}{3} = \pi$ . So $f(x) = \pi - 2\tan^{-1}x$ for $x \ge 1$ . On $(-\infty, -1)$ , same derivative relation, constant read at $x = -\sqrt 3$ : $f(-\sqrt 3) + 2\tan^{-1}(-\sqrt 3) = -\frac{\pi}{3} - \frac{2\pi}{3} = -\pi$ , so $f(x) = -\pi - 2\tan^{-1}x$ for $x \le -1$ . Three intervals, three constants $0$ , $\pi$ , $-\pi$ : reading the first one at $x = 0$ and extending it to the whole line would claim $f(\sqrt 3) = \frac{2\pi}{3}$ , a value that $\sin^{-1}$ can never take, since it exceeds $\frac{\pi}{2}$ .

e)

$f(x) = \frac{\pi}{3}$ means $\frac{2x}{1 + x^2} = \sin\frac{\pi}{3} = \frac{\sqrt 3}{2}$ , legitimate since $\frac{\pi}{3}$ is in the range of $\sin^{-1}$ . Then $4x = \sqrt 3(1 + x^2)$ , that is $\sqrt 3x^2 - 4x + \sqrt 3 = 0$ , with discriminant $16 - 12 = 4$ : $x = \frac{4 \pm 2}{2\sqrt 3}$ , so $x = \sqrt 3 \approx 1.7321$ or $x = \frac{1}{\sqrt 3} \approx 0.5774$ . The formulas of d) agree: $2\tan^{-1}x = \frac{\pi}{3}$ gives $\frac{1}{\sqrt 3}$ on $[-1, 1]$ , and $\pi - 2\tan^{-1}x = \frac{\pi}{3}$ gives $\sqrt 3$ on $[1, \infty)$ . $\sin^{-1}$ returns one angle, but the INNER function $\frac{2x}{1 + x^2}$ takes the value $\frac{\sqrt 3}{2}$ twice, once on each side of $x = 1$ . Finally $f(3) = \sin^{-1}\frac{6}{10} = \sin^{-1}0.6 \approx 0.6435$ , and $\pi - 2\tan^{-1}3 \approx 3.141593 - 2.498092 = 0.643501$ : the same, to four decimals $0.6435$ .

Corrected exercises: inverse trigonometric functions (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: Restricted ranges: the one angle an inverse function returns

Exercise 2: Compositions and the reference triangle: the range fixes the sign

Exercise 3: Derivatives of arcsin, arccos and arctan: the formula is one line, the algebra is the rest

Exercise 4: The derivative of arcsec by implicit differentiation, and the absolute value it needs

Exercise 5: The chain rule through an inverse trigonometric function, and the square root of a square

Part B: problems and reasoning (/50)

Exercise 6: Two functions that agree on one side only: arcsec from the arctan key

Exercise 7: Horizontal tangents and prescribed slopes: one denominator, then factor

Exercise 8: Five statements to correct

Exercise 9: The angle under which a painting is seen, and its rate with distance

Exercise 10: A final exam question: one function, two corners, three formulas

See also

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