# Chapter 10: Inverse Functions and Inverse Trig Functions

> When is a function invertible, and how fast does its inverse grow? The answer to both questions lives in the derivative — no inverting required.

**Canonical:** https://calculus.kitzo.work/chapter/inverse-functions  
**Level:** Core  
**Estimated time:** 45 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Inverse function theorem
- Derivatives of inverses
- arcsin, arccos, arctan
- Inverse secant
- Inverse hyperbolic functions

**Why this matters**
Solving f(x) = y for x is often impossible in closed form, yet we routinely need the inverse's slope. Calculus sidesteps algebra entirely: one derivative check proves an inverse exists, and one reciprocal gives its slope — a superpower hiding inside the chain rule.

## 1 · The derivative proves an inverse exists

A function is invertible on an interval when it is strictly monotonic there — every output is hit exactly once. The derivative test: f′ > 0 throughout the interval forces strictly increasing, so an inverse exists; f′ < 0 does the same for decreasing. Even isolated zeros are allowed when the sign never flips (x³ is invertible; f′(0) = 0 but f′ ≥ 0). What can go wrong: f′ changing sign (x² on ℝ, sin x on ℝ, x³ − 3x), jumps, or corners — in every case the function doubles back, and the inverse fails until you restrict the domain.

## 2 · The inverse-derivative formula

Given b = f(a) with f′(a) ≠ 0, differentiate f(f⁻¹(x)) = x with the chain rule: f′(f⁻¹(x))·(f⁻¹)′(x) = 1. Therefore (f⁻¹)′(b) = 1/f′(a) — the inverse passes through (b, a) with reciprocal slope. Geometrically, reflection across y = x swaps rise and run, reciprocating every slope.

**Worked example 10.1 — Inverse slopes without inverting**
**(a) Small case:** f(x) = x³ + x, find (f⁻¹)′(2). Note f(1) = 1 + 1 = 2, so a = 1, and f′(1) = 3(1)² + 1 = 4 — hence (f⁻¹)′(2) = ¼.
**(b) Big example:** f(x) = 2x + sin x. Then f′(x) = 2 + cos x ≥ 1 > 0 everywhere, so f is invertible on ℝ despite having no elementary inverse. Since f(0) = 0, (f⁻¹)′(0) = 1/(2 + cos 0) = 1/3.
**(c) Practice pattern:** always locate a with f(a) = b *first*; then quote 1/f′(a).

## 3 · arcsin, arccos, arctan — definitions and derivatives

Trig functions are not invertible on ℝ, so each is restricted: sin on [−π/2, π/2] gives arcsin with range [−π/2, π/2]; cos on [0, π] gives arccos with range [0, π]; tan on (−π/2, π/2) gives arctan with range (−π/2, π/2). Differentiate sin y = x implicitly: cos y·y′ = 1, and cos y = +√(1 − x²) on [−π/2, π/2], so (arcsin x)′ = 1/√(1 − x²). Mirroring gives (arccos x)′ = −1/√(1 − x²). For arctan, 1 + tan²y = sec²y yields (arctan x)′ = 1/(1 + x²).

## 4 · arcsec, arccsc, arccot

The outer trio, briefly: arcsec has domain |x| ≥ 1 and (arcsec x)′ = 1/(|x|√(x² − 1)) — the absolute value keeps the sign honest on the negative branch. arccsc mirrors it with a minus: (arccsc x)′ = −1/(|x|√(x² − 1)). arccot pairs with arctan: (arccot x)′ = −1/(1 + x²). Sign conventions differ across textbooks for arcsec; the |x| version is the safest to memorise.

## 5 · Inverse hyperbolic functions

Inverting sinh gives arcsinh x = ln(x + √(x² + 1)), defined on all of ℝ, with (arcsinh x)′ = 1/√(1 + x²). Inverting cosh on x ≥ 0 gives arccosh x = ln(x + √(x² − 1)) for x ≥ 1, with (arccosh x)′ = 1/√(x² − 1). Inverting tanh on |x| < 1: arctanh x = ½ ln((1 + x)/(1 − x)), with (arctanh x)′ = 1/(1 − x²). Notice the sign pattern echoes ordinary trig derivatives, with plus signs where circles carry minus signs.

## 6 · Putting it together

The workflow never changes: (1) verify monotonicity via f′, (2) restrict the domain if f′ changes sign, (3) for slope questions, locate a with f(a) = b and take the reciprocal, (4) for arc-functions, differentiate sin y = x or tan y = x implicitly and convert the co-function via a Pythagorean identity.

**Watch out**
f⁻¹(x) is *not* 1/f(x) — the inverse function and the reciprocal are different animals. And arcsin x means inverse sine, domain [−1, 1]; writing sin⁻¹ x invites misreading as csc x. Finally, the arccos derivative carries a negative sign that arcsin does not.

**Chapter formula sheet**
(f⁻¹)′(b) = 1 / f′(a),  where b = f(a)
f′ > 0 on an interval ⇒ f invertible there
(arcsin x)′ = 1/√(1 − x²)
(arccos x)′ = −1/√(1 − x²)
(arctan x)′ = 1/(1 + x²)
(arcsec x)′ = 1/(|x|√(x² − 1))
(arcsinh x)′ = 1/√(1 + x²)
(arctanh x)′ = 1/(1 − x²)

## Common mistakes

- Reading f⁻¹(x) as 1/f(x) instead of the inverse function.

- Swapping the roles of a and b — remember (f⁻¹)′(b) uses f′ at a, the point where f(a) = b.

- Claiming every continuous function is invertible — continuity never suffices; monotonicity does.

- Differentiating arccos x with a positive sign — it is −1/√(1 − x²).

- Writing the arcsec derivative without the |x|, losing the negative-branch sign.

- Confusing √(1 − x²) with √(1 + x²) between arcsin and arcsinh derivatives.

## Practice exercises

1. Which function has an inverse defined on all of ℝ?

- A. f(x) = x²
- B. f(x) = x³ + x
- C. f(x) = sin x
- D. f(x) = |x|

2. Let f(x) = x³ + 2x. Find (f⁻¹)′(3).

3. Differentiate y = arcsin(x).

4. Evaluate the derivative of arctan(x) at x = 1.

5. Differentiate y = arctan(3x).

6. The natural domain of arcsin(x) is:

- A. All real numbers
- B. [−1, 1]
- C. (−∞, −1] ∪ [1, ∞)
- D. (0, ∞)

7. Differentiate y = arcsinh(x).

8. Evaluate the derivative of arccos(x) at x = 0.

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