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Concept Review

Differential Calculus

Inverse Functions and Inverse Trigonometric Derivatives

Inverse trigonometric functions exist so that we can solve equations like sin⁡θ=0.4\sin\theta = 0.4 or tan⁡θ=3\tan\theta = 3 for the angle θ\theta. Their derivatives are remarkable: although the functions are transcendental, every one of their derivatives is a purely algebraic expression. This section sits on top of the chain rule and implicit differentiation, and the six formulas below are exactly the ones you will later run backwards when you integrate 11+x2\dfrac{1}{1+x^2} and 11−x2\dfrac{1}{\sqrt{1-x^2}}.

One-to-one functions and the horizontal line test

A function ff is one-to-one (injective) on a set DD if f(x1)=f(x2) ⇒ x1=x2for all x1,x2∈D.f(x_1)=f(x_2)\ \Rightarrow\ x_1=x_2 \qquad \text{for all } x_1,x_2\in D.

Horizontal line test. ff is one-to-one exactly when no horizontal line y=cy=c meets the graph of ff more than once. (It is the graphical restatement of the definition: two intersection points are two inputs with the same output.)

A sufficient condition. If ff is differentiable on an interval II and f′(x)>0f'(x)>0 for every x∈Ix\in I (or f′(x)<0f'(x)<0 throughout II), then ff is strictly monotone on II and therefore one-to-one on II. This is sufficient, not necessary: f(x)=x3f(x)=x^3 is one-to-one on all of R\mathbf{R} even though f′(0)=0f'(0)=0. Also note the condition is about a single interval: f(x)=1/xf(x)=1/x has f′<0f'<0 on its whole domain and happens to be one-to-one, but f(x)=x2f(x)=x^2 on (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty) shows why "monotone on each piece" is not enough by itself.

The inverse function

If f:D→Rf:D\to R is one-to-one with range R=f(D)R=f(D), its inverse f−1:R→Df^{-1}:R\to D is defined by f−1(y)=xexactly whenf(x)=y.f^{-1}(y)=x \quad\text{exactly when}\quad f(x)=y .

  • domain(f−1)=range(f)\text{domain}(f^{-1})=\text{range}(f) and range(f−1)=domain(f)\text{range}(f^{-1})=\text{domain}(f) — domains and ranges swap.
  • f−1(f(x))=xf^{-1}(f(x))=x for x∈Dx\in D, and f(f−1(y))=yf(f^{-1}(y))=y for y∈Ry\in R.
  • The graph of f−1f^{-1} is the reflection of the graph of ff across the line y=xy=x.
  • Notation warning: f−1f^{-1} means the inverse function, not 1/f1/f. In particular sin⁡−1x≠1sin⁡x\sin^{-1}x \ne \dfrac{1}{\sin x}.

Derivative of an inverse function

Theorem (derivative of an inverse). Let ff be continuous and one-to-one on an open interval II, and let g=f−1g=f^{-1}, defined on the interval J=f(I)J=f(I). Suppose that

  1. a∈Ia\in I and ff is differentiable at aa, and
  2. f′(a)≠0f'(a)\ne 0.

Put b=f(a)b=f(a). Then gg is differentiable at bb and g′(b)=1f′(a)=1f′(g(b)).g'(b)=\frac{1}{f'(a)}=\frac{1}{f'(g(b))}.

Every hypothesis earns its keep. Continuity plus injectivity on an interval forces ff to be strictly monotone, which makes gg continuous — that continuity is what the proof of the limit needs. If f′(a)=0f'(a)=0 the conclusion is false: gg is not differentiable at bb, and the graph of gg has a vertical tangent there. Example: f(x)=x3f(x)=x^3, a=0a=0, b=0b=0, g(y)=y1/3g(y)=y^{1/3}, and g′(0)g'(0) does not exist.

How to use it. You never need a formula for f−1f^{-1}. To get (f−1)′(b)(f^{-1})'(b): find the number aa with f(a)=bf(a)=b (usually by inspection), then compute 1/f′(a)1/f'(a).

Second derivative. Add the hypothesis that ff is twice differentiable on a neighbourhood of aa with f′≠0f'\ne 0 there. Differentiating g′(x)=1f′(g(x))g'(x)=\dfrac{1}{f'(g(x))} with the chain rule then gives g′′(x)=−f′′(g(x))[f′(g(x))]3,sog′′(b)=−f′′(a)[f′(a)]3.g''(x)=-\frac{f''(g(x))}{[f'(g(x))]^{3}}, \qquad\text{so}\qquad g''(b)=-\frac{f''(a)}{[f'(a)]^{3}}.

Branch conventions for the inverse trig functions

Every trig function is periodic, hence badly non-injective, so each must be restricted to an interval on which it is one-to-one before it can be inverted. The choices below are the principal branches used in this course.

Inverse functionTrig function restricted toDomain of the inverseRange (principal values)
arcsin⁡\arcsinsin⁡\sin on [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right][−1,1][-1,1][−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
arccos⁡\arccoscos⁡\cos on [0,π][0,\pi][−1,1][-1,1][0,π][0,\pi]
arctan⁡\arctantan⁡\tan on (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)all real numbers(−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
arccot\text{arccot}cot⁡\cot on (0,π)(0,\pi)all real numbers(0,π)(0,\pi)
arcsec\text{arcsec}sec⁡\sec on [0,π2)∪(π2,π]\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]∣x∣≥1\lvert x\rvert\ge 1[0,π2)∪(π2,π]\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]
arccsc\text{arccsc}csc⁡\csc on [−π2,0)∪(0,π2]\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]∣x∣≥1\lvert x\rvert\ge 1[−π2,0)∪(0,π2]\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]

Some books choose a different branch for arcsec\text{arcsec} (range [0,π2)∪[π,3π2)\left[0,\frac{\pi}{2}\right)\cup\left[\pi,\frac{3\pi}{2}\right) — note π\pi is included, so that x=−1x=-1 still has a value), which removes the absolute value from its derivative. Everything below uses the table above.

Useful identities (all valid on the stated domains): arcsin⁡x+arccos⁡x=π2  (−1≤x≤1),arctan⁡x+arccot x=π2  (all x),\arcsin x+\arccos x=\frac{\pi}{2}\ \ (-1\le x\le 1),\qquad \arctan x+\text{arccot}\,x=\frac{\pi}{2}\ \ (\text{all }x), arcsec x=arccos⁡(1x),arccsc x=arcsin⁡(1x)(∣x∣≥1).\text{arcsec}\,x=\arccos\left(\frac{1}{x}\right),\qquad \text{arccsc}\,x=\arcsin\left(\frac{1}{x}\right)\qquad (\lvert x\rvert\ge 1).

Cancellation trap. sin⁡(arcsin⁡x)=x\sin(\arcsin x)=x for all x∈[−1,1]x\in[-1,1], but arcsin⁡(sin⁡θ)=θ\arcsin(\sin\theta)=\theta only when θ∈[−π2,π2]\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. For instance arcsin⁡(sin⁡5π6)=arcsin⁡12=π6\arcsin\left(\sin\frac{5\pi}{6}\right)=\arcsin\frac12=\frac{\pi}{6}, not 5π6\frac{5\pi}{6}.

Deriving ddxarcsin⁡x\dfrac{d}{dx}\arcsin x by implicit differentiation

Let y=arcsin⁡xy=\arcsin x with −1<x<1-1<x<1, so sin⁡y=x\sin y=x and y∈(−π2,π2)y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). Differentiate both sides with respect to xx: cos⁡y⋅dydx=1dydx=1cos⁡y.\begin{aligned}\cos y\cdot \frac{dy}{dx} &= 1\\ \frac{dy}{dx} &= \frac{1}{\cos y}.\end{aligned} Now convert cos⁡y\cos y into xx. From sin⁡2y+cos⁡2y=1\sin^2y+\cos^2y=1 we get cos⁡y=±1−sin⁡2y=±1−x2\cos y=\pm\sqrt{1-\sin^2 y}=\pm\sqrt{1-x^2}, and the branch decides the sign: on (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) we have cos⁡y>0\cos y>0, so the ++ sign is correct. Hence ddxarcsin⁡x=11−x2,−1<x<1.\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},\qquad -1<x<1 . At x=±1x=\pm1 we would need cos⁡y=0\cos y=0: the derivative does not exist there (vertical tangent), matching the theorem's hypothesis f′(a)≠0f'(a)\ne 0 with f=sin⁡f=\sin, a=±π2a=\pm\frac{\pi}{2}.

The same argument for arcsec\text{arcsec} (where the sign analysis really bites). Let y=arcsec xy=\text{arcsec}\,x, so sec⁡y=x\sec y=x and y∈[0,π2)∪(π2,π]y\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]. Then sec⁡ytan⁡y⋅y′=1\sec y\tan y\cdot y'=1, so y′=1sec⁡ytan⁡yy'=\dfrac{1}{\sec y\tan y}, and tan⁡y=±sec⁡2y−1=±x2−1\tan y=\pm\sqrt{\sec^2y-1}=\pm\sqrt{x^2-1}.

  • If y∈(0,π2)y\in\left(0,\frac{\pi}{2}\right) then x=sec⁡y>1x=\sec y>1 and tan⁡y>0\tan y>0, so sec⁡ytan⁡y=xx2−1=∣x∣x2−1\sec y\tan y=x\sqrt{x^2-1}=\lvert x\rvert\sqrt{x^2-1}.
  • If y∈(π2,π)y\in\left(\frac{\pi}{2},\pi\right) then x=sec⁡y<−1x=\sec y<-1 and tan⁡y<0\tan y<0, so sec⁡ytan⁡y=x(−x2−1)=−xx2−1=∣x∣x2−1\sec y\tan y=x\left(-\sqrt{x^2-1}\right)=-x\sqrt{x^2-1}=\lvert x\rvert\sqrt{x^2-1}.

Both cases collapse to the same thing, which is why the absolute value appears: ddxarcsec x=1∣x∣x2−1,∣x∣>1.\frac{d}{dx}\text{arcsec}\,x=\frac{1}{\lvert x\rvert\sqrt{x^2-1}},\qquad \lvert x\rvert>1 .

The six derivatives and their chain-rule versions

Throughout, u=u(x)u=u(x) is differentiable and u′=dudxu'=\dfrac{du}{dx}.

F(x)F(x)F′(x)F'(x)Valid forChain-rule version ddxF(u)\dfrac{d}{dx}F(u)
arcsin⁡x\arcsin x11−x2\dfrac{1}{\sqrt{1-x^2}}−1<x<1-1<x<1u′1−u2\dfrac{u'}{\sqrt{1-u^2}}
arccos⁡x\arccos x−11−x2-\dfrac{1}{\sqrt{1-x^2}}−1<x<1-1<x<1−u′1−u2-\dfrac{u'}{\sqrt{1-u^2}}
arctan⁡x\arctan x11+x2\dfrac{1}{1+x^2}all real xxu′1+u2\dfrac{u'}{1+u^2}
arccot x\text{arccot}\,x−11+x2-\dfrac{1}{1+x^2}all real xx−u′1+u2-\dfrac{u'}{1+u^2}
arcsec x\text{arcsec}\,x1∣x∣x2−1\dfrac{1}{\lvert x\rvert\sqrt{x^2-1}}∣x∣>1\lvert x\rvert>1u′∣u∣u2−1\dfrac{u'}{\lvert u\rvert\sqrt{u^2-1}}
arccsc x\text{arccsc}\,x−1∣x∣x2−1-\dfrac{1}{\lvert x\rvert\sqrt{x^2-1}}∣x∣>1\lvert x\rvert>1−u′∣u∣u2−1-\dfrac{u'}{\lvert u\rvert\sqrt{u^2-1}}

Pattern to memorise: the three "co-" functions (arccos⁡\arccos, arccot\text{arccot}, arccsc\text{arccsc}) have exactly the negatives of the derivatives of their partners. That follows instantly from the identities arccos⁡x=π2−arcsin⁡x\arccos x=\frac{\pi}{2}-\arcsin x, etc.

Worked example 1 — derivative of an inverse

Let f(x)=x5+2x+1f(x)=x^5+2x+1. Show ff is invertible on R\mathbf{R} and find (f−1)′(4)\left(f^{-1}\right)'(4).

f′(x)=5x4+2≥2>0f'(x)=5x^4+2\ge 2>0 for all xx, so ff is strictly increasing, hence one-to-one on the interval R\mathbf{R}; being a polynomial it is continuous and differentiable everywhere. So g=f−1g=f^{-1} exists.

We need aa with f(a)=4f(a)=4. Try a=1a=1: 15+2(1)+1=41^5+2(1)+1=4. Since ff is one-to-one this aa is the only one, so g(4)=1g(4)=1. Then f′(1)=5(1)4+2=7≠0f'(1)=5(1)^4+2=7\ne0, and the theorem applies: g′(4)=1f′(g(4))=1f′(1)=17.g'(4)=\frac{1}{f'(g(4))}=\frac{1}{f'(1)}=\frac{1}{7}.

Worked example 2 — chain rule, with a surprise

Differentiate h(x)=arctan⁡(1−x1+x)h(x)=\arctan\left(\dfrac{1-x}{1+x}\right) for x≠−1x\ne -1.

Set u=1−x1+xu=\dfrac{1-x}{1+x}. By the quotient rule, u′=(−1)(1+x)−(1−x)(1)(1+x)2=−1−x−1+x(1+x)2=−2(1+x)2.u'=\frac{(-1)(1+x)-(1-x)(1)}{(1+x)^2}=\frac{-1-x-1+x}{(1+x)^2}=\frac{-2}{(1+x)^2}. Also 1+u2=1+(1−x)2(1+x)2=(1+x)2+(1−x)2(1+x)2=1+2x+x2+1−2x+x2(1+x)2=2+2x2(1+x)2.1+u^2=1+\frac{(1-x)^2}{(1+x)^2}=\frac{(1+x)^2+(1-x)^2}{(1+x)^2}=\frac{1+2x+x^2+1-2x+x^2}{(1+x)^2}=\frac{2+2x^2}{(1+x)^2}. Therefore h′(x)=u′1+u2=−2(1+x)2⋅(1+x)22(1+x2)=−11+x2.h'(x)=\frac{u'}{1+u^2}=\frac{-2}{(1+x)^2}\cdot\frac{(1+x)^2}{2(1+x^2)}=\frac{-1}{1+x^2}. So h′(x)=−11+x2h'(x)=-\dfrac{1}{1+x^2}, i.e. hh and −arctan⁡x-\arctan x have the same derivative. On the interval (−1,∞)(-1,\infty) this forces h(x)=C−arctan⁡xh(x)=C-\arctan x, and putting x=0x=0 gives h(0)=arctan⁡1=π4h(0)=\arctan 1=\frac{\pi}{4}, so C=π4C=\frac{\pi}{4} there. (On (−∞,−1)(-\infty,-1) the constant is different, namely −3π4-\frac{3\pi}{4} — the zero-derivative argument only works one interval at a time.)

Common mistakes

WrongRightWhy
(f−1)′(b)=1f′(b)\left(f^{-1}\right)'(b)=\dfrac{1}{f'(b)}(f−1)′(b)=1f′(a)\left(f^{-1}\right)'(b)=\dfrac{1}{f'(a)} where f(a)=bf(a)=bYou must feed f′f' the input aa, not the output bb.
ddxarcsec x=1xx2−1\dfrac{d}{dx}\text{arcsec}\,x=\dfrac{1}{x\sqrt{x^2-1}}1∣x∣x2−1\dfrac{1}{\lvert x\rvert\sqrt{x^2-1}}Without the absolute value the formula gives a negative slope at x=−2x=-2, but arcsec\text{arcsec} is increasing there.
ddxarctan⁡(3x)=11+9x2\dfrac{d}{dx}\arctan(3x)=\dfrac{1}{1+9x^2}31+9x2\dfrac{3}{1+9x^2}The chain-rule factor u′=3u'=3 is not optional.
ddxarcsin⁡(u)=u′1−x2\dfrac{d}{dx}\arcsin(u)=\dfrac{u'}{\sqrt{1-x^2}}u′1−u2\dfrac{u'}{\sqrt{1-u^2}}The uu must go inside the radical too.
sin⁡−1x=1sin⁡x\sin^{-1}x=\dfrac{1}{\sin x}sin⁡−1x=arcsin⁡x\sin^{-1}x=\arcsin xf−1f^{-1} is the inverse function, not a reciprocal.
arcsin⁡(sin⁡5π6)=5π6\arcsin\left(\sin\frac{5\pi}{6}\right)=\frac{5\pi}{6}π6\frac{\pi}{6}The output must land in [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right].
sec⁡2y−1=tan⁡y\sqrt{\sec^2y-1}=\tan ysec⁡2y−1=∣tan⁡y∣\sqrt{\sec^2 y-1}=\lvert\tan y\rvertt2=∣t∣\sqrt{t^2}=\lvert t\rvert; you must check the branch before dropping the bars.
"f′(a)=0f'(a)=0, so just write 1/0=∞1/0=\infty"(f−1)′(b)\left(f^{-1}\right)'(b) does not existThe inverse has a vertical tangent; the theorem's hypothesis fails.

Two more domain slips worth naming: quoting ddxarcsin⁡x=11−x2\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}} "for −1≤x≤1-1\le x\le1" (it fails at the endpoints, where the expression is undefined), and forgetting that arccos⁡′\arccos' carries a minus sign while arcsin⁡′\arcsin' does not.

Key terms

  • one-to-one (injective) function
  • horizontal line test
  • strictly monotone
  • inverse function
  • reflection across the line y = x
  • derivative of an inverse function
  • vertical tangent
  • principal branch
  • restricted domain
  • arcsin
  • arccos
  • arctan
  • arccot
  • arcsec
  • arccsc
  • implicit differentiation
  • chain rule
  • Pythagorean identity
  • domain restriction
  • constancy theorem (zero derivative on an interval)

Practice Problems

Use the horizontal line test to decide whether each function is one-to-one on all of R\mathbf{R}:

(a) f(x)=x3+2xf(x)=x^3+2x (b) g(x)=x2−4x+1g(x)=x^2-4x+1

For any function that fails, give the largest interval containing x=3x=3 on which it is one-to-one.

Show hint

A horizontal line meeting the graph twice means two different inputs share an output. The sign of the derivative tells you whether the graph can ever turn around.

Show answer

(a) Differentiate: f′(x)=3x2+2f'(x)=3x^2+2. Since 3x2≥03x^2\ge 0, we have f′(x)≥2>0f'(x)\ge 2>0 for every real xx, so ff is strictly increasing on the interval R\mathbf{R}.

A strictly increasing function satisfies: if x1<x2x_1<x_2 then f(x1)<f(x2)f(x_1)<f(x_2), so f(x1)≠f(x2)f(x_1)\ne f(x_2). Hence no horizontal line can meet the graph twice, and ff is one-to-one on R\mathbf{R}.

(b) gg is a parabola opening upward with vertex where g′(x)=2x−4=0g'(x)=2x-4=0, i.e. x=2x=2. To exhibit a failure concretely, take the horizontal line y=1y=1: g(0)=0−0+1=1,g(4)=16−16+1=1.g(0)=0-0+1=1,\qquad g(4)=16-16+1=1 . The line y=1y=1 meets the graph at (0,1)(0,1) and (4,1)(4,1), so gg is not one-to-one on R\mathbf{R}.

Largest interval containing x=3x=3: g′(x)=2x−4>0g'(x)=2x-4>0 exactly when x>2x>2, and g′(x)<0g'(x)<0 when x<2x<2, so the graph turns around at x=2x=2. On [2,∞)[2,\infty) the function is strictly increasing, hence one-to-one, and 3∈[2,∞)3\in[2,\infty). Any larger interval containing 33 would have to contain points on both sides of 22, and by symmetry about x=2x=2 it would then contain a pair 2−h2-h, 2+h2+h with g(2−h)=g(2+h)g(2-h)=g(2+h).

Answer: [2,∞)[2,\infty).

Evaluate exactly, using the principal branches:

(a) arcsin⁡(−12)\arcsin\left(-\tfrac12\right) (b) arccos⁡(−32)\arccos\left(-\tfrac{\sqrt3}{2}\right) (c) arctan⁡(−1)\arctan(-1)

(d) arcsec(−2)\text{arcsec}(-2) (e) arcsin⁡(sin⁡5π6)\arcsin\left(\sin\tfrac{5\pi}{6}\right) (f) cos⁡(arcsin⁡35)\cos\left(\arcsin\tfrac35\right)

Show hint

For each one, write down the required output interval first, then find the angle in that interval with the right trig value.

Show answer

(a) We need y∈[−π2,π2]y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] with sin⁡y=−12\sin y=-\frac12. The angle is y=−π6y=-\frac{\pi}{6} (and −π6-\frac{\pi}{6} is indeed in the range). arcsin⁡(−12)=−π6\arcsin\left(-\frac12\right)=-\dfrac{\pi}{6}.

(b) We need y∈[0,π]y\in[0,\pi] with cos⁡y=−32\cos y=-\frac{\sqrt3}{2}. The reference angle is π6\frac{\pi}{6}, and cosine is negative in the second quadrant, so y=π−π6=5π6y=\pi-\frac{\pi}{6}=\frac{5\pi}{6}. arccos⁡(−32)=5π6\arccos\left(-\frac{\sqrt3}{2}\right)=\dfrac{5\pi}{6}.

(c) We need y∈(−π2,π2)y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) with tan⁡y=−1\tan y=-1: y=−π4y=-\frac{\pi}{4}. arctan⁡(−1)=−π4\arctan(-1)=-\dfrac{\pi}{4}. (Note 3π4\frac{3\pi}{4} also has tangent −1-1 but is outside the principal range.)

(d) We need y∈[0,π2)∪(π2,π]y\in\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right] with sec⁡y=−2\sec y=-2, i.e. cos⁡y=−12\cos y=-\frac12. In [0,π][0,\pi] that gives y=2π3y=\frac{2\pi}{3}, which lies in (π2,π]\left(\frac{\pi}{2},\pi\right]. arcsec(−2)=2π3\text{arcsec}(-2)=\dfrac{2\pi}{3}.

(e) First the inside: sin⁡5π6=sin⁡(π−5π6)=sin⁡π6=12\sin\frac{5\pi}{6}=\sin\left(\pi-\frac{5\pi}{6}\right)=\sin\frac{\pi}{6}=\frac12. Then arcsin⁡12=π6\arcsin\frac12=\frac{\pi}{6}. The answer is π6\dfrac{\pi}{6}, not 5π6\frac{5\pi}{6}, because 5π6∉[−π2,π2]\frac{5\pi}{6}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right] so the cancellation law does not apply.

(f) Let θ=arcsin⁡35\theta=\arcsin\frac35, so sin⁡θ=35\sin\theta=\frac35 and θ∈[−π2,π2]\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. Then cos⁡2θ=1−sin⁡2θ=1−925=1625 ⇒ cos⁡θ=±45.\cos^2\theta=1-\sin^2\theta=1-\frac{9}{25}=\frac{16}{25}\ \Rightarrow\ \cos\theta=\pm\frac45 . On [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] cosine is ≥0\ge 0, so we take the ++ sign. cos⁡(arcsin⁡35)=45\cos\left(\arcsin\frac35\right)=\dfrac45.

Differentiate, and state the interval of xx on which your formula is valid:

(a) y=arcsin⁡(3x)y=\arcsin(3x) (b) y=arctan⁡(x2)y=\arctan\left(x^2\right) (c) y=arccos⁡(x)y=\arccos\left(\sqrt{x}\right)

Show hint

Each is a chain rule: identify the inside function uu, compute u′u', and remember that uu (not xx) goes into the derivative formula.

Show answer

(a) Let u=3xu=3x, so u′=3u'=3. Using ddxarcsin⁡u=u′1−u2\dfrac{d}{dx}\arcsin u=\dfrac{u'}{\sqrt{1-u^2}}: y′=31−(3x)2=31−9x2.y'=\frac{3}{\sqrt{1-(3x)^2}}=\frac{3}{\sqrt{1-9x^2}} . Valid where 1−9x2>01-9x^2>0, i.e. 9x2<19x^2<1, i.e. −13<x<13-\dfrac13<x<\dfrac13.

(b) Let u=x2u=x^2, so u′=2xu'=2x. Using ddxarctan⁡u=u′1+u2\dfrac{d}{dx}\arctan u=\dfrac{u'}{1+u^2}: y′=2x1+(x2)2=2x1+x4.y'=\frac{2x}{1+\left(x^2\right)^2}=\frac{2x}{1+x^4}. Valid for all real xx (the denominator 1+x4≥11+x^4\ge 1 is never zero).

(c) Let u=x=x1/2u=\sqrt{x}=x^{1/2}, so u′=12xu'=\dfrac{1}{2\sqrt{x}} (for x>0x>0). Using ddxarccos⁡u=−u′1−u2\dfrac{d}{dx}\arccos u=-\dfrac{u'}{\sqrt{1-u^2}}, and u2=xu^2=x: y′=−12x1−x=−12x1−x=−12x−x2.y'=-\frac{\dfrac{1}{2\sqrt x}}{\sqrt{1-x}}=-\frac{1}{2\sqrt{x}\sqrt{1-x}}=-\frac{1}{2\sqrt{x-x^2}} . Valid where x>0x>0 and 1−x>01-x>0, i.e. 0<x<10<x<1. (At x=0x=0 the factor u′u' blows up; at x=1x=1 the arccos⁡\arccos derivative blows up.)

Let f(x)=x3+3x+1f(x)=x^3+3x+1. Show that ff has an inverse gg on R\mathbf{R}, and compute g′(5)g'(5) without finding a formula for gg.

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First check the sign of f′f' to justify invertibility, then hunt for the input whose output is 55.

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Step 1 — invertibility. f′(x)=3x2+3≥3>0f'(x)=3x^2+3\ge 3>0 for all xx, so ff is strictly increasing on the interval R\mathbf{R} and therefore one-to-one; it is a polynomial, so it is continuous and differentiable everywhere. Hence g=f−1g=f^{-1} exists on the range of ff, which is all of R\mathbf{R} (a cubic with positive leading coefficient has lim⁡x→−∞f(x)=−∞\lim_{x\to-\infty}f(x)=-\infty and lim⁡x→∞f(x)=∞\lim_{x\to\infty}f(x)=\infty, so by the Intermediate Value Theorem it attains every real value).

Step 2 — find aa with f(a)=5f(a)=5. Try small integers: f(1)=1+3+1=5f(1)=1+3+1=5. Since ff is one-to-one this is the only such input, so g(5)=1g(5)=1.

Step 3 — apply the theorem. ff is differentiable at a=1a=1 and f′(1)=3(1)2+3=6≠0f'(1)=3(1)^2+3=6\ne 0, so the derivative-of-an-inverse theorem applies: g′(5)=1f′(g(5))=1f′(1)=16.g'(5)=\frac{1}{f'(g(5))}=\frac{1}{f'(1)}=\frac{1}{6}.

Check. Near x=1x=1, f(1.01)=1.030301+3.03+1=5.060301f(1.01)=1.030301+3.03+1=5.060301, so the average rate of change of ff is about 0.0603010.01=6.03≈f′(1)\frac{0.060301}{0.01}=6.03\approx f'(1). The inverse's slope is the reciprocal, about 0.16580.1658, and 16=0.16‾\frac16=0.1\overline{6}. Consistent.

Answer: g′(5)=16g'(5)=\dfrac16.

Let f(x)=2x+cos⁡xf(x)=2x+\cos x.

(a) Explain why ff is invertible on R\mathbf{R}.

(b) Find the equation of the tangent line to y=f−1(x)y=f^{-1}(x) at the point where x=1x=1.

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Bound f′f' using −1≤sin⁡x≤1-1\le\sin x\le 1. For (b) you need both the point on the inverse's graph and the slope there.

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(a) f′(x)=2−sin⁡xf'(x)=2-\sin x. Since −1≤sin⁡x≤1-1\le\sin x\le 1 for all xx, we get f′(x)=2−sin⁡x ≥ 2−1=1>0for all x.f'(x)=2-\sin x\ \ge\ 2-1=1>0 \qquad\text{for all }x . So ff is strictly increasing on the interval R\mathbf{R}, hence one-to-one; it is continuous and differentiable everywhere. Therefore g=f−1g=f^{-1} exists.

(b) Point. We need g(1)g(1), i.e. the aa with f(a)=1f(a)=1. Try a=0a=0: f(0)=2(0)+cos⁡0=0+1=1f(0)=2(0)+\cos 0=0+1=1. Good, so g(1)=0g(1)=0 and the point of tangency is (1,0)(1,0).

Slope. f′(0)=2−sin⁡0=2−0=2≠0f'(0)=2-\sin 0=2-0=2\ne0, so the theorem gives g′(1)=1f′(g(1))=1f′(0)=12.g'(1)=\frac{1}{f'(g(1))}=\frac{1}{f'(0)}=\frac12 .

Line. Point-slope form through (1,0)(1,0) with slope 12\frac12: y−0=12(x−1)⟹y=x−12.y-0=\frac12(x-1)\qquad\Longrightarrow\qquad y=\frac{x-1}{2}.

Sanity check. The tangent to y=f(x)y=f(x) at (0,1)(0,1) is y=1+2xy=1+2x, and reflecting a line of slope 22 through (0,1)(0,1) across y=xy=x gives a line of slope 12\frac12 through (1,0)(1,0) — exactly what we found.

Answer: y=x−12y=\dfrac{x-1}{2}.

Derive ddxarctan⁡x=11+x2\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2} by implicit differentiation. State the branch you use, and explain why — unlike the arcsin⁡\arcsin derivation — no sign discussion is needed here. Then explain why the formula is valid for every real xx.

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Write y=arctan⁡xy=\arctan x, take tangent of both sides, and differentiate. Which Pythagorean identity converts sec⁡2y\sec^2 y into something in xx?

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Set-up. Let y=arctan⁡xy=\arctan x. By the principal-branch definition this means tan⁡y=xwithy∈(−π2,π2),\tan y=x \qquad\text{with}\qquad y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), and xx may be any real number.

Differentiate implicitly with respect to xx, remembering ddytan⁡y=sec⁡2y\dfrac{d}{dy}\tan y=\sec^2 y and the chain rule: ddx(tan⁡y)=ddx(x)sec⁡2y⋅dydx=1dydx=1sec⁡2y.\begin{aligned}\frac{d}{dx}\left(\tan y\right)&=\frac{d}{dx}(x)\\ \sec^2 y\cdot\frac{dy}{dx}&=1\\ \frac{dy}{dx}&=\frac{1}{\sec^2 y}.\end{aligned}

Convert to xx. The Pythagorean identity sec⁡2y=1+tan⁡2y\sec^2y=1+\tan^2 y gives sec⁡2y=1+tan⁡2y=1+x2.\sec^2 y=1+\tan^2y=1+x^2 . Therefore ddxarctan⁡x=11+x2.\frac{d}{dx}\arctan x=\frac{1}{1+x^2}.

Why no sign discussion? In the arcsin⁡\arcsin derivation we met cos⁡y\cos y to the first power and had to recover it from cos⁡y=±1−sin⁡2y\cos y=\pm\sqrt{1-\sin^2y}, choosing the sign from the branch. Here the identity delivers sec⁡2y\sec^2 y — an even power — directly in terms of tan⁡y=x\tan y=x. No square root is ever taken, so no ambiguous sign arises.

Why valid for all xx? Two things must hold. First, tan⁡\tan is differentiable at every y∈(−π2,π2)y\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). Second, the derivative-of-an-inverse theorem needs ddytan⁡y≠0\dfrac{d}{dy}\tan y\ne 0: indeed sec⁡2y=1+tan⁡2y≥1>0\sec^2y=1+\tan^2y\ge 1>0, so it is never zero on the branch. Since tan⁡\tan maps (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right) onto all of R\mathbf{R}, every real xx is covered, and 1+x2≠01+x^2\ne0 always. Contrast this with arcsin⁡\arcsin, whose derivative fails at x=±1x=\pm1 precisely because cos⁡(±π2)=0\cos\left(\pm\frac{\pi}{2}\right)=0.

Differentiate f(x)=arcsin⁡(x)1−x2f(x)=\arcsin(x)\sqrt{1-x^2} and simplify. On what interval is your formula valid? Evaluate f′(0)f'(0) and interpret it.

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This is a product; you will need the chain rule (or the power rule with (1−x2)1/2(1-x^2)^{1/2}) on the second factor.

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Product rule. Write f=uvf=uv with u=arcsin⁡xu=\arcsin x and v=(1−x2)1/2v=\left(1-x^2\right)^{1/2}.

u′=11−x2u'=\dfrac{1}{\sqrt{1-x^2}} for −1<x<1-1<x<1.

v′=12(1−x2)−1/2⋅(−2x)=−x1−x2v'=\dfrac12\left(1-x^2\right)^{-1/2}\cdot(-2x)=\dfrac{-x}{\sqrt{1-x^2}} for −1<x<1-1<x<1.

Then f′(x)=u′v+uv′=11−x2⋅1−x2  +  arcsin⁡x⋅−x1−x2=1−xarcsin⁡x1−x2.\begin{aligned} f'(x)&=u'v+uv'\\ &=\frac{1}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2}\;+\;\arcsin x\cdot\frac{-x}{\sqrt{1-x^2}}\\ &=1-\frac{x\arcsin x}{\sqrt{1-x^2}} .\end{aligned}

Validity. Both u′u' and v′v' require 1−x2>01-x^2>0, so the formula holds on −1<x<1-1<x<1. At x=±1x=\pm1 the function ff is still defined (f(±1)=arcsin⁡(±1)⋅0=0f(\pm1)=\arcsin(\pm1)\cdot 0=0), but the derivative expression is undefined. Only one one-sided derivative is even available at each endpoint, and it fails to exist: with x=1−εx=1-\varepsilon we have 1−x2=ε(2−ε)≈2ε\sqrt{1-x^2}=\sqrt{\varepsilon(2-\varepsilon)}\approx\sqrt{2\varepsilon}, so f(1)−f(1−ε)ε≈−π2⋅2εε→−∞,\frac{f(1)-f(1-\varepsilon)}{\varepsilon}\approx-\frac{\pi}{2}\cdot\frac{\sqrt{2\varepsilon}}{\varepsilon}\to-\infty , and the same computation at x=−1x=-1 also gives −∞-\infty. The graph meets the xx-axis vertically at both ends.

Value at 00. f′(0)=1−0⋅arcsin⁡01=1−0=1f'(0)=1-\dfrac{0\cdot\arcsin 0}{\sqrt{1}}=1-0=1. Since f(0)=arcsin⁡(0)⋅1=0f(0)=\arcsin(0)\cdot 1=0, the curve passes through the origin with slope exactly 11, i.e. it is tangent to the line y=xy=x there.

Numerical check at x=0.5x=0.5. The formula gives f′(0.5)=1−0.5⋅0.52359880.8660254=1−0.3022999=0.6977001.f'(0.5)=1-\frac{0.5\cdot 0.5235988}{0.8660254}=1-0.3022999=0.6977001 . A symmetric difference quotient: f(0.51)=0.5351848⋅0.8601744=0.4603523f(0.51)=0.5351848\cdot0.8601744=0.4603523 and f(0.49)=0.5120898⋅0.8717224=0.4464001f(0.49)=0.5120898\cdot0.8717224=0.4464001, so f(0.51)−f(0.49)0.02=0.01395220.02=0.69761,\frac{f(0.51)-f(0.49)}{0.02}=\frac{0.0139522}{0.02}=0.69761, which agrees with 0.697700.69770 to the accuracy a step of 0.010.01 can deliver.

Answer: f′(x)=1−xarcsin⁡x1−x2f'(x)=1-\dfrac{x\arcsin x}{\sqrt{1-x^2}} on (−1,1)(-1,1), and f′(0)=1f'(0)=1.

Differentiate y=arcsec(3x)y=\text{arcsec}(3x) using the branch convention range(arcsec)=[0,π2)∪(π2,π]\text{range}(\text{arcsec})=\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]. Simplify the absolute value fully and state the domain of the derivative. Then evaluate y′(−1)y'(-1) and check the sign is plausible.

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Apply the chain rule with u=3xu=3x, keeping the ∣u∣\lvert u\rvert exactly as the formula demands before you simplify it.

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Chain rule. With u=3xu=3x we have u′=3u'=3, and ddxarcsec(u)=u′∣u∣u2−1.\frac{d}{dx}\text{arcsec}(u)=\frac{u'}{\lvert u\rvert\sqrt{u^2-1}} . So y′=3∣3x∣(3x)2−1=33∣x∣9x2−1=1∣x∣9x2−1,y'=\frac{3}{\lvert 3x\rvert\sqrt{(3x)^2-1}}=\frac{3}{3\lvert x\rvert\sqrt{9x^2-1}}=\frac{1}{\lvert x\rvert\sqrt{9x^2-1}}, using ∣3x∣=3∣x∣\lvert 3x\rvert=3\lvert x\rvert (legitimate because 3>03>0).

Domain. The formula for arcsec(u)\text{arcsec}(u) requires ∣u∣>1\lvert u\rvert>1, i.e. ∣3x∣>1\lvert 3x\rvert>1, i.e. ∣x∣>13,that isx∈(−∞,−13)∪(13,∞).\lvert x\rvert>\frac13,\qquad\text{that is}\qquad x\in\left(-\infty,-\tfrac13\right)\cup\left(\tfrac13,\infty\right). (The function y=arcsec(3x)y=\text{arcsec}(3x) itself is also defined at x=±13x=\pm\frac13, but has a vertical tangent there.)

Value at x=−1x=-1. y′(−1)=1∣−1∣9(1)−1=11⋅8=122=24≈0.35355.y'(-1)=\frac{1}{\lvert -1\rvert\sqrt{9(1)-1}}=\frac{1}{1\cdot\sqrt{8}}=\frac{1}{2\sqrt2}=\frac{\sqrt2}{4}\approx 0.35355 .

Sign check. Under this branch, arcsec t=arccos⁡(1/t)\text{arcsec}\,t=\arccos(1/t). For t<−1t<-1, as tt increases toward −1-1, 1/t1/t decreases toward −1-1, and arccos⁡\arccos is decreasing, so arcsec t\text{arcsec}\,t increases (from π2\frac{\pi}{2} up to π\pi). So the derivative must be positive at x=−1x=-1 — and it is. Note that had we dropped the absolute value and written y′=33x9x2−1=1x9x2−1y'=\dfrac{3}{3x\sqrt{9x^2-1}}=\dfrac{1}{x\sqrt{9x^2-1}}, we would have obtained y′(−1)=−18=−24y'(-1)=-\dfrac{1}{\sqrt8}=-\dfrac{\sqrt2}{4}, the wrong sign.

Numerical check. arcsec(3x)=arccos⁡(13x)\text{arcsec}(3x)=\arccos\left(\frac{1}{3x}\right). At x=−1.001x=-1.001: arccos⁡(−0.3330003)=1.9102801\arccos(-0.3330003)=1.9102801; at x=−0.999x=-0.999: arccos⁡(−0.3336670)=1.9109872\arccos(-0.3336670)=1.9109872. The difference quotient is 1.9109872−1.91028010.002=0.35355\frac{1.9109872-1.9102801}{0.002}=0.35355, matching 24=0.3535534\frac{\sqrt2}{4}=0.3535534.

Answer: y′=1∣x∣9x2−1y'=\dfrac{1}{\lvert x\rvert\sqrt{9x^2-1}} for ∣x∣>13\lvert x\rvert>\frac13, and y′(−1)=24y'(-1)=\dfrac{\sqrt2}{4}.

Prove that arcsin⁡x+arccos⁡x=π2\arcsin x+\arccos x=\dfrac{\pi}{2} for every x∈[−1,1]x\in[-1,1]. Be careful to say which theorem you use, on which interval, and how you handle the endpoints x=±1x=\pm1 where the derivatives do not exist.

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Give the sum a name and differentiate it on the open interval. Then you need a theorem that upgrades "zero derivative" to "constant".

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Set-up. Define h(x)=arcsin⁡x+arccos⁡xh(x)=\arcsin x+\arccos x on [−1,1][-1,1]. Both arcsin⁡\arcsin and arccos⁡\arccos are continuous on the closed interval [−1,1][-1,1] (they are inverses of functions continuous and strictly monotone on [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] and [0,π][0,\pi] respectively), so hh is continuous on [−1,1][-1,1].

Differentiate on the open interval. For −1<x<1-1<x<1, h′(x)=11−x2+(−11−x2)=0.h'(x)=\frac{1}{\sqrt{1-x^2}}+\left(-\frac{1}{\sqrt{1-x^2}}\right)=0 .

Apply the constancy theorem. The relevant statement is the corollary of the Mean Value Theorem: if hh is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and h′(x)=0h'(x)=0 for all x∈(a,b)x\in(a,b), then hh is constant on [a,b][a,b]. Note that this version requires differentiability only on the open interval, plus continuity on the closed one — exactly the situation here with [a,b]=[−1,1][a,b]=[-1,1]. So hh is constant on all of [−1,1][-1,1], endpoints included, even though h′(±1)h'(\pm1) does not exist.

Identify the constant. Evaluate at the convenient point x=0x=0: h(0)=arcsin⁡0+arccos⁡0=0+π2=π2.h(0)=\arcsin 0+\arccos 0=0+\frac{\pi}{2}=\frac{\pi}{2}.

Therefore arcsin⁡x+arccos⁡x=π2\arcsin x+\arccos x=\dfrac{\pi}{2} for all x∈[−1,1]x\in[-1,1]. ■\blacksquare

Direct endpoint verification (optional). h(1)=π2+0=π2,h(−1)=−π2+π=π2.h(1)=\frac{\pi}{2}+0=\frac{\pi}{2},\qquad h(-1)=-\frac{\pi}{2}+\pi=\frac{\pi}{2}. Both agree, as they must.

Corollary. Differentiating arccos⁡x=π2−arcsin⁡x\arccos x=\frac{\pi}{2}-\arcsin x gives ddxarccos⁡x=−11−x2\dfrac{d}{dx}\arccos x=-\dfrac{1}{\sqrt{1-x^2}} — this is the cleanest way to remember why the "co-" functions pick up the minus sign.

Let F(x)=arctan⁡x+arctan⁡(1x)F(x)=\arctan x+\arctan\left(\dfrac{1}{x}\right), defined for x≠0x\ne 0.

(a) Compute F′(x)F'(x) and simplify.

(b) Is FF constant on its whole domain? Determine FF explicitly, and explain carefully what the zero-derivative theorem does and does not give you here.

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Differentiate with the chain rule on the second term and clear the compound fraction. Then look hard at the shape of the domain before concluding anything about constancy.

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(a) The second term has inside function u=1x=x−1u=\dfrac1x=x^{-1}, so u′=−x−2=−1x2u'=-x^{-2}=-\dfrac{1}{x^2}. Then F′(x)=11+x2+u′1+u2=11+x2+−1x21+1x2.\begin{aligned} F'(x)&=\frac{1}{1+x^2}+\frac{u'}{1+u^2}\\ &=\frac{1}{1+x^2}+\frac{-\dfrac{1}{x^2}}{1+\dfrac{1}{x^2}} .\end{aligned} Multiply numerator and denominator of the second fraction by x2x^2 (valid since x≠0x\ne0): −1x21+1x2=−1x2+1.\frac{-\dfrac{1}{x^2}}{1+\dfrac{1}{x^2}}=\frac{-1}{x^2+1}. Hence F′(x)=11+x2−11+x2=0for every x≠0.F'(x)=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0\qquad\text{for every }x\ne 0 .

(b) The domain is (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty) — not an interval. The MVT corollary says: if h′=0h'=0 on an interval, then hh is constant on that interval. It says nothing across a gap in the domain. So we may conclude only that FF is constant on (0,∞)(0,\infty) and constant on (−∞,0)(-\infty,0), possibly with different constants.

On (0,∞)(0,\infty): evaluate at x=1x=1: F(1)=arctan⁡1+arctan⁡1=π4+π4=π2.F(1)=\arctan 1+\arctan 1=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}.

On (−∞,0)(-\infty,0): evaluate at x=−1x=-1: F(−1)=arctan⁡(−1)+arctan⁡(−1)=−π4−π4=−π2.F(-1)=\arctan(-1)+\arctan(-1)=-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2}.

The two constants genuinely differ, so FF is not constant on its whole domain: F(x)={π2,x>0−π2,x<0F(x)=\begin{cases}\dfrac{\pi}{2}, & x>0\\ -\dfrac{\pi}{2}, & x<0\end{cases}

Spot-check with a second point. At x=2x=2: arctan⁡2≈1.1071487\arctan 2\approx 1.1071487 and arctan⁡12≈0.4636476\arctan\frac12\approx 0.4636476; their sum is 1.5707963≈π21.5707963\approx\frac{\pi}{2}. At x=−3x=-3: arctan⁡(−3)≈−1.2490458\arctan(-3)\approx-1.2490458 and arctan⁡(−13)≈−0.3217506\arctan\left(-\frac13\right)\approx-0.3217506; sum ≈−1.5707964≈−π2\approx-1.5707964\approx-\frac{\pi}{2}. Both confirm the case split.

Moral. "Derivative zero everywhere on the domain" implies "constant" only when the domain is a single interval. Writing F(x)=π2F(x)=\frac{\pi}{2} for all x≠0x\ne0 is the classic error here.

Using the branch convention range(arccsc)=[−π2,0)∪(0,π2]\text{range}(\text{arccsc})=\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right], derive ddx arccsc x=−1∣x∣x2−1,∣x∣>1,\frac{d}{dx}\,\text{arccsc}\,x=-\frac{1}{\lvert x\rvert\sqrt{x^2-1}},\qquad \lvert x\rvert>1, by implicit differentiation. Treat the two pieces of the range separately and show explicitly how the absolute value appears. Then confirm the result a second way using an identity.

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Start from csc⁡y=x\csc y=x and differentiate. When you replace cot⁡y\cot y by ±csc⁡2y−1\pm\sqrt{\csc^2y-1}, decide the sign separately on each piece of the range.

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Set-up. Let y=arccsc xy=\text{arccsc}\,x, so csc⁡y=x,y∈[−π2,0)∪(0,π2],∣x∣≥1.\csc y=x,\qquad y\in\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right],\qquad \lvert x\rvert\ge1 .

Implicit differentiation. Since ddycsc⁡y=−csc⁡ycot⁡y\dfrac{d}{dy}\csc y=-\csc y\cot y, −csc⁡ycot⁡y⋅dydx=1dydx=−1csc⁡ycot⁡y.\begin{aligned}-\csc y\cot y\cdot\frac{dy}{dx}&=1\\ \frac{dy}{dx}&=-\frac{1}{\csc y\cot y}.\end{aligned} This requires csc⁡ycot⁡y≠0\csc y\cot y\ne0, i.e. cot⁡y≠0\cot y\ne 0, i.e. y≠±π2y\ne\pm\frac{\pi}{2}, i.e. x≠±1x\ne\pm1. So from the outset the formula is restricted to ∣x∣>1\lvert x\rvert>1.

Convert to xx. From 1+cot⁡2y=csc⁡2y1+\cot^2y=\csc^2y we get cot⁡y=±csc⁡2y−1=±x2−1,\cot y=\pm\sqrt{\csc^2y-1}=\pm\sqrt{x^2-1}, and the branch decides the sign.

Case 1: y∈(0,π2)y\in\left(0,\frac{\pi}{2}\right). Here sin⁡y>0\sin y>0 and cos⁡y>0\cos y>0, so cot⁡y>0\cot y>0 and cot⁡y=+x2−1\cot y=+\sqrt{x^2-1}. Also x=csc⁡y=1sin⁡y>1>0x=\csc y=\frac{1}{\sin y}>1>0, so x=∣x∣x=\lvert x\rvert. Then csc⁡ycot⁡y=xx2−1=∣x∣x2−1.\csc y\cot y=x\sqrt{x^2-1}=\lvert x\rvert\sqrt{x^2-1}.

Case 2: y∈(−π2,0)y\in\left(-\frac{\pi}{2},0\right). Here sin⁡y<0\sin y<0 and cos⁡y>0\cos y>0, so cot⁡y<0\cot y<0 and cot⁡y=−x2−1\cot y=-\sqrt{x^2-1}. Also x=csc⁡y<−1<0x=\csc y<-1<0, so −x=∣x∣-x=\lvert x\rvert. Then csc⁡ycot⁡y=x(−x2−1)=(−x)x2−1=∣x∣x2−1.\csc y\cot y=x\left(-\sqrt{x^2-1}\right)=(-x)\sqrt{x^2-1}=\lvert x\rvert\sqrt{x^2-1}.

Both cases give the same expression, which is exactly why an absolute value — and not a ±\pm — appears: ddx arccsc x=−1∣x∣x2−1,∣x∣>1.■\frac{d}{dx}\,\text{arccsc}\,x=-\frac{1}{\lvert x\rvert\sqrt{x^2-1}},\qquad \lvert x\rvert>1 . \qquad\blacksquare

Second derivation via an identity. For ∣x∣≥1\lvert x\rvert\ge1 the two branch conventions in use satisfy arcsec x+arccsc x=π2\text{arcsec}\,x+\text{arccsc}\,x=\frac{\pi}{2} (check: at x=2x=2, arcsec 2=π3\text{arcsec}\,2=\frac{\pi}{3} and arccsc 2=π6\text{arccsc}\,2=\frac{\pi}{6}; at x=−2x=-2, arcsec(−2)=2π3\text{arcsec}(-2)=\frac{2\pi}{3} and arccsc(−2)=−π6\text{arccsc}(-2)=-\frac{\pi}{6}; both sum to π2\frac{\pi}{2}). Differentiating and using ddxarcsec x=1∣x∣x2−1\dfrac{d}{dx}\text{arcsec}\,x=\dfrac{1}{\lvert x\rvert\sqrt{x^2-1}} gives ddxarccsc x=0−1∣x∣x2−1,\frac{d}{dx}\text{arccsc}\,x=0-\frac{1}{\lvert x\rvert\sqrt{x^2-1}}, the same answer.

Sign sanity check. At x=2x=2 the formula gives −123≈−0.2887-\frac{1}{2\sqrt3}\approx-0.2887. Numerically, arccsc x=arcsin⁡1x\text{arccsc}\,x=\arcsin\frac1x: at x=2.001x=2.001, arcsin⁡(0.4997501)=0.5233103\arcsin(0.4997501)=0.5233103; at x=1.999x=1.999, arcsin⁡(0.5002501)=0.5238876\arcsin(0.5002501)=0.5238876. The difference quotient is 0.5233103−0.52388760.002≈−0.28865,\frac{0.5233103-0.5238876}{0.002}\approx-0.28865, which agrees with −123=−0.2886751-\frac{1}{2\sqrt3}=-0.2886751 to the precision the seven-decimal inputs allow. Matches.

Let f(x)=x5+2x3+3xf(x)=x^5+2x^3+3x and let g=f−1g=f^{-1}.

(a) Justify that gg exists and is twice differentiable near x=6x=6.

(b) Find g′(6)g'(6).

(c) Derive the formula g′′(b)=−f′′(a)[f′(a)]3g''(b)=-\dfrac{f''(a)}{[f'(a)]^{3}} (where b=f(a)b=f(a)) and use it to compute g′′(6)g''(6) exactly.

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For (c), differentiate the identity g′(x)=1f′(g(x))g'(x)=\dfrac{1}{f'(g(x))} with the chain rule, then substitute what you already know from (b).

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(a) f′(x)=5x4+6x2+3f'(x)=5x^4+6x^2+3. Since 5x4≥05x^4\ge0 and 6x2≥06x^2\ge0, we have f′(x)≥3>0f'(x)\ge 3>0 for all xx, so ff is strictly increasing on R\mathbf{R}, hence one-to-one. As a polynomial it is infinitely differentiable, and lim⁡x→±∞f(x)=±∞\lim_{x\to\pm\infty}f(x)=\pm\infty, so by the Intermediate Value Theorem its range is all of R\mathbf{R}. Since f′f' is never zero, the derivative-of-an-inverse theorem applies at every point, and because g′(x)=1/f′(g(x))g'(x)=1/f'(g(x)) is itself a composition/quotient of differentiable functions with non-vanishing denominator, gg is twice differentiable everywhere — in particular near x=6x=6.

(b) Find aa with f(a)=6f(a)=6: try a=1a=1, giving 1+2+3=61+2+3=6. Since ff is one-to-one, g(6)=1g(6)=1. Now f′(1)=5(1)4+6(1)2+3=5+6+3=14,f'(1)=5(1)^4+6(1)^2+3=5+6+3=14 , so g′(6)=1f′(g(6))=1f′(1)=114.g'(6)=\frac{1}{f'(g(6))}=\frac{1}{f'(1)}=\frac{1}{14}.

(c) Deriving the second-derivative formula. Start from the identity valid on an interval around bb: g′(x)=1f′(g(x))=[f′(g(x))]−1.g'(x)=\frac{1}{f'(g(x))}=\left[f'(g(x))\right]^{-1}. Differentiate with the chain rule (outer power −1-1, then the derivative of f′∘gf'\circ g): g′′(x)=−[f′(g(x))]−2⋅f′′(g(x))⋅g′(x).g''(x)=-\left[f'(g(x))\right]^{-2}\cdot f''(g(x))\cdot g'(x). Now substitute g′(x)=1f′(g(x))g'(x)=\dfrac{1}{f'(g(x))}: g′′(x)=−f′′(g(x))[f′(g(x))]2⋅1f′(g(x))=−f′′(g(x))[f′(g(x))]3.g''(x)=-\frac{f''(g(x))}{\left[f'(g(x))\right]^{2}}\cdot\frac{1}{f'(g(x))}=-\frac{f''(g(x))}{\left[f'(g(x))\right]^{3}} . Evaluating at x=bx=b where g(b)=ag(b)=a: g′′(b)=−f′′(a)[f′(a)]3.g''(b)=-\frac{f''(a)}{\left[f'(a)\right]^{3}} .

Now compute. f′′(x)=20x3+12xf''(x)=20x^3+12x, so f′′(1)=20+12=32f''(1)=20+12=32. With a=1a=1, b=6b=6, f′(1)=14f'(1)=14: g′′(6)=−32143=−322744=−4343≈−0.011662.g''(6)=-\frac{32}{14^{3}}=-\frac{32}{2744}=-\frac{4}{343}\approx -0.011662 . (The reduction: gcd⁡(32,2744)=8\gcd(32,2744)=8, and 32÷8=432\div8=4, 2744÷8=3432744\div8=343.)

Check of the formula on a case we can solve. Take f(x)=x2f(x)=x^2 on x>0x>0, so g(y)=yg(y)=\sqrt y and g′′(y)=−14y−3/2g''(y)=-\frac{1}{4}y^{-3/2}. The formula with a=ya=\sqrt y gives −f′′(a)[f′(a)]3=−2(2y)3=−28y3/2=−14y3/2-\dfrac{f''(a)}{[f'(a)]^3}=-\dfrac{2}{(2\sqrt y)^3}=-\dfrac{2}{8y^{3/2}}=-\dfrac{1}{4y^{3/2}}. Agreed.

Answers: g′(6)=114g'(6)=\dfrac{1}{14} and g′′(6)=−4343g''(6)=-\dfrac{4}{343}.