Derivatives of inverse functions and inverse trig functions

Worked solutions

Problem 1

1. $f'(x) = 3x^2 + 1 \ge 1 > 0$Power rule; $3x^2 \ge 0$.
2. So $f$ is increasing on $\mathbb{R}$, hence one-to-one, with an inverseAn increasing function never takes the same value twice.
3. $f(1) = 1 + 1 + 1 = 3$, so $f^{-1}(3) = 1$Find the $x$ that $f$ sends to $3$ by inspection.
4. $(f^{-1})'(3) = \dfrac{1}{f'(1)} = \dfrac14$The inverse function rule, with $f'$ taken at $x = 1$, not at $3$.
5. $f(0) = 1$, so $f^{-1}(1) = 0$ and $(f^{-1})'(1) = \dfrac{1}{f'(0)} = 1$$f'(0) = 1$.
6. $(f^{-1})'(3) = \tfrac14$ and $(f^{-1})'(1) = 1$Solving $x^3 + x + 1 = y$ for $x$ needs the cubic formula, but the slope of the inverse needs only one point of $f$ and its slope there. A finite-difference slope of a numerical inverse agrees with both.

Problem 2

1. $f(0) = e^0 + 0 = 1$, so $f^{-1}(1) = 0$The point $(0, 1)$ is on the graph of $f$, so $(1, 0)$ is on the graph of $f^{-1}$.
2. $f'(x) = 2e^{2x} + 1$, so $f'(0) = 3$Chain rule on $e^{2x}$.
3. $(f^{-1})'(1) = \tfrac13$One over the slope of $f$ at the matching point.
4. Tangent: $x = 0 + \tfrac13(y - 1)$Point-slope form for the inverse, written with $y$ as its input.
5. $(f^{-1})'(1) = \tfrac13$; the tangent line to $f^{-1}$ at $(1, 0)$ is $x = \tfrac13(y - 1)$The tangent to $f$ at $(0, 1)$ is $y = 1 + 3x$; solving it for $x$ gives the same line, the reflection of $f$'s tangent in $y = x$.

Problem 3

1. $y = \arccos x$ means $\cos y = x$ with $0 \le y \le \pi$The principal branch.
2. $-\sin y \cdot y' = 1$, so $y' = -\dfrac{1}{\sin y}$Differentiate both sides in $x$; the chain rule puts $y'$ on the left.
3. $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$On $[0, \pi]$, $\sin y \ge 0$, so the positive root is the right one.
4. $(\arccos x)' = -\dfrac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$At $x = \pm1$, $\sin y = 0$ and the slope is infinite.
5. $y = \arctan x$: $\sec^2 y \cdot y' = 1$$(\tan y)' = \sec^2 y$.
6. $\sec^2 y = 1 + \tan^2 y = 1 + x^2$The Pythagorean identity divided by $\cos^2 y$; no square root, so no sign to choose.
7. $(\arccos x)' = -\dfrac{1}{\sqrt{1 - x^2}}$, which is $-\dfrac{2}{\sqrt3} \approx -1.1547$ at $\tfrac12$; $(\arctan x)' = \dfrac{1}{1 + x^2}$, which is $\tfrac45$ at $\tfrac12$$\arccos$ falls from $\pi$ to $0$ as $x$ goes from $-1$ to $1$, so its slope must be negative. Its derivative is minus that of $\arcsin$, because $\arcsin x + \arccos x = \tfrac\pi2$.

Problem 4

1. $\dfrac{d}{dx}\arctan(3x) = \dfrac{1}{1 + 9x^2}\cdot 3$Chain rule: the outer derivative at $u = 3x$, times $u' = 3$.
2. At $x = \tfrac13$: $\dfrac{3}{1 + 1} = \tfrac32$$9 \cdot \tfrac19 = 1$.
3. $\dfrac{d}{dx}\arcsin(x^2) = \dfrac{2x}{\sqrt{1 - x^4}}$Outer derivative at $u = x^2$, so $1 - u^2 = 1 - x^4$; then $u' = 2x$.
4. At $x = \tfrac12$: $\dfrac{1}{\sqrt{15/16}} = \dfrac{4}{\sqrt{15}} \approx 1.0328$$2x = 1$ and $1 - \tfrac1{16} = \tfrac{15}{16}$.
5. $\dfrac{d}{dx}\arctan(\sqrt x) = \dfrac{1}{1 + x}\cdot\dfrac{1}{2\sqrt x}$$(\sqrt x)^2 = x$ inside, then $(\sqrt x)' = 1/(2\sqrt x)$.
6. $\dfrac{3}{1 + 9x^2} = \tfrac32$ at $\tfrac13$; $\dfrac{2x}{\sqrt{1 - x^4}} \approx 1.0328$ at $\tfrac12$; $\dfrac{1}{2\sqrt x(1 + x)} = \tfrac14$ at $1$Each is the outer derivative evaluated at the inside, times the derivative of the inside. Finite differences agree with all three values.

Problem 5

1. $(x\arcsin x)' = \arcsin x + \dfrac{x}{\sqrt{1 - x^2}}$Product rule.
2. $\big(\sqrt{1 - x^2}\big)' = \dfrac{-2x}{2\sqrt{1 - x^2}} = -\dfrac{x}{\sqrt{1 - x^2}}$Chain rule.
3. $F'(x) = \arcsin x$The two fractions cancel.
4. $F(\tfrac12) = \tfrac12\cdot\tfrac\pi6 + \tfrac{\sqrt3}{2}$ and $F(0) = 1$$\arcsin\tfrac12 = \tfrac\pi6$ and $\sqrt{1 - \tfrac14} = \tfrac{\sqrt3}2$.
5. $F' = \arcsin x$, and $\displaystyle\int_0^{1/2}\arcsin x\,dx = \tfrac{\pi}{12} + \tfrac{\sqrt3}{2} - 1 \approx 0.1278$The fundamental theorem: $F(\tfrac12) - F(0)$. The antiderivative comes from integration by parts with $u = \arcsin x$, $dv = dx$, and differentiating it is the check.

Problem 6

1. $\sigma(\operatorname{logit}(p)) = p$$\sigma(t) = p$ solves to $e^{-t} = (1 - p)/p$, that is $t = \ln\frac{p}{1 - p}$.
2. $\operatorname{logit}'(p) = \dfrac{1}{\sigma'(t)}$ with $t = \operatorname{logit}(p)$The inverse function rule.
3. $\sigma'(t) = \sigma(t)(1 - \sigma(t)) = p(1 - p)$At the matching point, $\sigma(t) = p$.
4. Directly: $\dfrac{d}{dp}\big(\ln p - \ln(1 - p)\big) = \dfrac1p + \dfrac{1}{1 - p}$Split the log of a quotient.
5. $\dfrac1p + \dfrac1{1 - p} = \dfrac{1 - p + p}{p(1 - p)}$Common denominator: the two routes agree.
6. $\operatorname{logit}'(p) = \dfrac{1}{p(1 - p)}$: $6.25$ at $p = 0.2$ and $4$ at $p = 0.5$The sigmoid is flattest near $0$ and $1$, so its inverse is steepest there: near a probability of $0$ or $1$, a small change in $p$ is a large change in log-odds.

Problem 7

1. $g'(y) = \dfrac{1}{f'(g(y))}$The inverse function rule, written as a function of $y$.
2. $g''(y) = -\dfrac{f''(g(y))\,g'(y)}{f'(g(y))^2}$Differentiate $1/u$ with $u = f'(g(y))$: $(1/u)' = -u'/u^2$, and $u' = f''(g(y))\,g'(y)$ by the chain rule.
3. $g''(y) = -\dfrac{f''(x)}{f'(x)^3}$Substitute $g'(y) = 1/f'(x)$ from step 1.
4. $f(0) = 1$, so $g(1) = 0$; $f'(x) = 1 + e^x$ and $f''(x) = e^x$$f'(0) = 2$, $f''(0) = 1$.
5. $g'(1) = \tfrac12$ and $g''(1) = -\tfrac18$$f$ is convex and increasing, so its reflection is concave, and $g'' < 0$ agrees. The inverse here has no elementary formula (it needs the Lambert W function), and a second difference of a numerical inverse gives $-0.125$.

Problem 8

1. $f$ is decreasing on $(-\infty, 0]$, with range $[0, \infty)$$f'(x) = 2x < 0$ for $x < 0$.
2. $y = x^2$ with $x \le 0$ gives $x = -\sqrt y$, so $f^{-1}(y) = -\sqrt y$Of the two square roots, the branch with $x \le 0$ is the negative one.
3. From the formula: $(f^{-1})'(y) = -\dfrac{1}{2\sqrt y}$, so $-\tfrac14$ at $y = 4$Power rule.
4. With the rule: $f^{-1}(4) = -2$ and $f'(-2) = -4$The matching point is $x = -2$, not $x = 2$.
5. $f^{-1}(y) = -\sqrt y$ and $(f^{-1})'(4) = -\tfrac14$, by both routesThe inverse of a decreasing function is decreasing. The point $x = 2$ is not in the domain of this $f$, and using it gives $+\tfrac14$ (Mistake 5).

Problem 9

1. $\theta = \arctan\dfrac{h}{500}$The right triangle with legs $h$ (up) and $500$ (along the ground).
2. $\dfrac{d\theta}{dt} = \dfrac{1}{1 + (h/500)^2}\cdot\dfrac{1}{500}\dfrac{dh}{dt}$Chain rule through $\arctan$, then through $h/500$, then through $h(t)$.
3. At $h = 1000$: $\dfrac{1}{1 + 4} = \tfrac15$$h/500 = 2$.
4. $\tfrac15\cdot\tfrac{100}{500} = \tfrac{1}{25}$$dh/dt = 100$.
5. $d\theta/dt = 0.04$ rad/s, about $2.3$ degrees per secondThe angle changes more slowly the higher the rocket goes: the factor $1/(1 + (h/500)^2)$ shrinks as $h$ grows, though the rocket's speed is constant.

Problem 10

1. $\dfrac{d}{dx}\arctan\dfrac ax = \dfrac{1}{1 + a^2/x^2}\cdot\Big(-\dfrac{a}{x^2}\Big) = -\dfrac{a}{x^2 + a^2}$Chain rule, then multiply top and bottom by $x^2$.
2. $\theta'(x) = -\dfrac{3}{x^2 + 9} + \dfrac{1}{x^2 + 1}$Step 1 with $a = 3$ and $a = 1$.
3. $\theta' = 0$ when $x^2 + 9 = 3(x^2 + 1)$Cross-multiply.
4. $2x^2 = 6$, so $x = \sqrt3 \approx 1.732$Only the positive root is a distance.
5. $\theta'(1) = 0.2 > 0$ and $\theta'(3) \approx -0.067 < 0$The angle rises, then falls: a maximum.
6. $\theta(\sqrt3) = \arctan\sqrt3 - \arctan\tfrac{1}{\sqrt3} = \tfrac\pi3 - \tfrac\pi6$$\tan\tfrac\pi3 = \sqrt3$ and $\tan\tfrac\pi6 = 1/\sqrt3$.
7. Stand $\sqrt3 \approx 1.73$ m away, where the angle is $\pi/6$, that is $30°$Close up the painting is seen almost edge on, and far away it shrinks; in between, the best distance is the geometric mean $\sqrt{1 \cdot 3}$ of the two heights.