Functions that 'undo' the regular trigonometric functions. If $\sin(y) = x$, then $y = \sin^{-1}(x)$.

The inverse sine function, which returns the angle whose sine is x.

The inverse cosine function, which returns the angle whose cosine is x.

The inverse tangent function, which returns the angle whose tangent is x.

The inverse cosecant function, which returns the angle whose cosecant is x.

The inverse secant function, which returns the angle whose secant is x.

The inverse cotangent function, which returns the angle whose cotangent is x.

Inverse trig functions 'undo' regular trig functions. We find the derivatives of these inverse functions using implicit differentiation and trigonometric identities.

Inverse trig functions are often part of composite functions, so the chain rule is necessary to differentiate the 'inside' function.

Start with $y = \sin^{-1}(x)$, rewrite as $x = \sin(y)$, differentiate both sides with respect to x, and solve for $\frac{dy}{dx}$.

$\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2}$

$\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$

$\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{\lvert{x}\rvert\sqrt{x^{2}-1}}$

$\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2}$

$\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}$