Calculus — Derivatives⏱ 5 min read

Inverse Trig Derivatives

d/dx(arcsin x) = 1/√(1−x²)

What is the Inverse Trig Derivatives?

These formulas give the derivatives of the inverse trigonometric functions, and arise naturally whenever an inverse trig function needs differentiating — they also appear frequently as the results of certain integrals, which is often the first place students actually encounter them, before differentiating an inverse trig function directly.

arcsin and arccos have the same derivative magnitude but opposite signs. That's not a coincidence: since arcsin(x) + arccos(x) = π/2 (a constant) for all valid x, their derivatives must sum to zero — differentiating both sides of that identity confirms it directly.

Unlike the six basic trig derivatives, none of the inverse trig derivatives contain a trig function in the result — they simplify down to purely algebraic expressions (a fraction with a square root, or a fraction with x² in the denominator). That's a genuinely useful pattern to notice on an exam: if a derivative or integral answer is supposed to involve an inverse trig function, the final expression should look algebraic, not trigonometric.

What Each Variable Means

arcsin x

Inverse sineDefined only for |x| < 1; derivative is 1/√(1−x²).

arccos x

Inverse cosineDefined only for |x| < 1; derivative is −1/√(1−x²).

arctan x

Inverse tangentDefined for all real x; derivative is 1/(1+x²).

When to Use It

Differentiating any expression involving arcsin, arccos, or arctan
Combined with the chain rule when the inverse trig function's argument is itself a function of x
Recognizing certain integral results that produce inverse trig functions

Step-by-Step Examples

Example 1: arcsin with the chain rule

Problem: Differentiate y = arcsin(3x)

Identify outer and inner functions

Outer: arcsin(u). Inner: u = 3x.

d/du[arcsin u] = 1/√(1-u²), du/dx = 3

Multiply by du/dx = 3

Apply the chain rule, substituting u = 3x back in.

dy/dx = 3 / √(1 − (3x)²)

✓

Answer: dy/dx = 3 / √(1 − 9x²)

Example 2: arctan with the chain rule

Problem: Differentiate y = arctan(x²)

Identify outer and inner functions

Outer: arctan(u). Inner: u = x².

d/du[arctan u] = 1/(1+u²), du/dx = 2x

Multiply the two derivatives

Substitute u = x² back in.

dy/dx = 2x · 1/(1 + (x²)²)

✓

Answer: dy/dx = 2x / (1 + x⁴)

Interactive Calculator

x value (between -1 and 1)

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Common Mistakes

Mistake: Forgetting the domain restriction |x| < 1 for arcsin and arccos.
Fix: arcsin x and arccos x — and their derivatives — are only defined for |x| < 1. arctan x, by contrast, is defined for all real numbers.
Mistake: Forgetting to apply the chain rule when the argument isn't just x.
Fix: d/dx[arcsin(3x)] is not simply 1/√(1-x²) — you must also multiply by the derivative of the inner function (3x)' = 3.