The six inverse trig derivatives
| Function | Derivative | Function defined for | Derivative valid for |
|---|---|---|---|
| sin−1 x (arcsin) | 1 / √(1 − x²) | −1 ≤ x ≤ 1 | −1 < x < 1 |
| cos−1 x (arccos) | −1 / √(1 − x²) | −1 ≤ x ≤ 1 | −1 < x < 1 |
| tan−1 x (arctan) | 1 / (1 + x²) | All real x | All real x |
| cot−1 x (arccot) | −1 / (1 + x²) | All real x | All real x |
| sec−1 x (arcsec) | 1 / (|x| √(x² − 1)) | |x| ≥ 1 | |x| > 1 |
| csc−1 x (arccsc) | −1 / (|x| √(x² − 1)) | |x| ≥ 1 | |x| > 1 |
The derivative is valid where its denominator is real and not zero, so the endpoints x = ±1 are excluded for arcsin, arccos, arcsec and arccsc. The ranges used to define the inverses: sin−1 x is between −π/2 and π/2, cos−1 x is between 0 and π, and tan−1 x is strictly between −π/2 and π/2.
A memory aid: each "co" function (arccos, arccot, arccsc) has the same derivative as its partner, with a minus sign.
Deriving the derivative of arcsin x
The inverse function theorem says that if f is invertible and differentiable, then
- Let f(x) = sin x on [−π/2, π/2], so f−1(x) = sin−1 x and f′(x) = cos x.
- Then d/dx sin−1 x = 1 / cos(sin−1 x).
- Let y = sin−1 x, so sin y = x. From sin²y + cos²y = 1, cos y = √(1 − sin²y) = √(1 − x²). The positive root is correct because y is between −π/2 and π/2, where cosine is not negative.
- So d/dx sin−1 x = 1 / √(1 − x²).
Example value: at x = 1/2, the slope of y = sin−1 x is 1 / √(1 − 1/4) = 1 / √(3/4) = 2/√3 ≈ 1.1547. The same method, using 1 + tan²y = sec²y, gives the arctan formula.
With the chain rule
When the inside is a function g(x), multiply by g′(x):
d/dx tan−1(g(x)) = g′(x) / (1 + [g(x)]²)
Example 1: h(x) = sin−1(2x³)
- Inside: g(x) = 2x³, so g′(x) = 6x².
- Square the inside: (2x³)² = 4x6.
- h′(x) = 6x² / √(1 − 4x6).
Example 2: f(x) = tan−1(x²)
- Inside: g(x) = x², so g′(x) = 2x.
- Square the inside: (x²)² = x4.
- f′(x) = 2x / (1 + x4).
Example 3: s(t) = tan−1(1/t)
- Inside: g(t) = 1/t, so g′(t) = −1/t².
- Square the inside: 1/t².
- s′(t) = (−1/t²) / (1 + 1/t²). Multiply top and bottom by t²: −1 / (t² + 1).
At t = 1, s′(1) = −1/2. Combined with other rules: by the product rule, d/dx [x² sin−1 x] = 2x sin−1 x + x² / √(1 − x²).
The integrals that go with them
Read backward, two of these formulas give common integrals:
∫ du / (a² + u²) = (1/a) tan−1(u/a) + C
Check your work with the derivative calculator, evaluate inverse trig values with the inverse trig calculator, or review the angles on the unit circle. For integrals that lead to these forms, see trig substitution.
Sources: OpenStax Calculus Volume 1, 3.7 Derivatives of Inverse Functions; Paul's Online Notes: Derivatives of Inverse Trig Functions; OpenStax Calculus Volume 2, Table of Integrals.
Every worked example on this page was checked with Calcist's own engine and an independent reference.