3.4 Differentiating Inverse Trigonometric Functions

Core inverse trig derivatives

Memorize these: d/dx(arcsin x) = 1/√(1−x²), d/dx(arccos x) = −1/√(1−x²), d/dx(arctan x) = 1/(1+x²), d/dx(arccot x) = −1/(1+x²), d/dx(arcsec x) = 1/(|x|√(x²−1)), d/dx(arccsc x) = −1/(|x|√(x²−1)). Notice arcsin/arccos and arctan/arccot pairs are negatives of each other.

Always apply the chain rule

When the inverse trig function has an inner function u(x), multiply by u′(x): d/dx(arcsin u) = u′/√(1−u²), d/dx(arctan u) = u′/(1+u²). Forgetting the chain rule factor is the most common error on these problems.

Where these formulas come from

These derivatives are derived using implicit differentiation on the original trig equation. For example, if y = arcsin x, then sin y = x; differentiating implicitly gives cos y · y′ = 1, and using a right triangle (or the Pythagorean identity) to express cos y in terms of x yields y′ = 1/√(1−x²).

Key terms

d/dx(arcsin x)

1/√(1−x²)

d/dx(arccos x)

−1/√(1−x²)

d/dx(arctan x)

1/(1+x²)

d/dx(arcsec x)

1/(|x|√(x²−1))

How to differentiate arcsin(u(x))

Apply the chain rule: d/dx[arcsin u] = u′/√(1−u²).

Try an AP-style question

d/dx arcsin x = ?

1. −1/√(1 − x²)
2. 1/(x√(x² − 1))
3. 1/√(1 − x²)
4. 1/(1 + x²)