Method guideCalculus IIntermediate10 min

Derivatives of inverse trigonometric functions

Recognize the inverse-trig outer function, differentiate its input, and keep the domain restrictions visible.

LaTeX article Updated July 13, 2026

\frac{d}{dx}\arctan u=\frac{u'}{1+u^2}

Inverse trig is not reciprocal trig

The notation sin⁻¹x means arcsin x, the inverse function on a restricted domain. It does not mean csc x.

The derivative formulas follow from implicit differentiation of identities such as sin(arcsin x) = x.

\sin(\arcsin x)=x

The chain rule supplies the numerator

When the input is u(x), the base inverse-trig derivative is multiplied by u′(x). Missing that factor is the same chain-rule error that appears with powers and exponentials.

Write u and u′ separately when the inside expression is complicated. That keeps the denominator and numerator from becoming tangled.

\frac{d}{dx}\arctan(3x^2)=\frac{6x}{1+9x^4}

Domains explain the square root

For real-valued arcsin and arccos, the input must stay between −1 and 1. Their derivatives become undefined at the endpoints because the square-root denominator is zero.

Arctan accepts every real input, and 1 + u² is always positive for real u. The formula therefore has no real denominator zeros.

|u|<1\text{ for a finite }\frac{d}{dx}\arcsin u

Worked example

Common mistakes

Reading arcsin x as 1/sin x.
Forgetting the derivative of the inside function.
Ignoring where a square-root denominator becomes zero.

Keep these ideas

Inverse and reciprocal functions are different.
Every composition needs the chain rule.
The formula's denominator records domain behavior.