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.

1. Apply the chain rule: $\frac{d}{dx} [\sin^{-1}(u(x))] = \frac{1}{\sqrt{1 - (u(x))^2}} \cdot u'(x)$. 2. Simplify the expression.

1. Apply the chain rule: $\frac{d}{dx} [\tan^{-1}(u(x))] = \frac{1}{1 + (u(x))^2} \cdot u'(x)$. 2. Simplify the expression.

1. Apply the chain rule: $\frac{d}{dx} [\cos^{-1}(2x)] = -\frac{1}{\sqrt{1 - (2x)^2}} \cdot 2$. 2. Simplify: $\frac{-2}{\sqrt{1-4x^2}}$

1. Apply the chain rule: $\frac{d}{dx} [\tan^{-1}(x^2)] = \frac{1}{1 + (x^2)^2} \cdot 2x$. 2. Simplify: $\frac{2x}{1+x^4}$

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

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

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

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

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

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

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