Derivatives, Integrals, and Properties of Inverse Trigonometric and Hyperbolic Functions

$\frac{d}{dx} \sin^{-1} u = \frac{1}{\sqrt{1 - u^2}} \frac{du}{dx} \quad (|u| < 1)$

$\frac{d}{dx} \cos^{-1} u = \frac{-1}{\sqrt{1 - u^2}} \frac{du}{dx} \quad (|u| < 1)$

$\frac{d}{dx} \tan^{-1} u = \frac{1}{1 + u^2} \frac{du}{dx}$

$\frac{d}{dx} \csc^{-1} u = \frac{-1}{|u|\sqrt{u^2 - 1}} \frac{du}{dx} \quad (|u| > 1)$

$\frac{d}{dx} \sec^{-1} u = \frac{1}{|u|\sqrt{u^2 - 1}} \frac{du}{dx} \quad (|u| > 1)$

$\frac{d}{dx} \cot^{-1} u = \frac{-1}{1 + u^2} \frac{du}{dx}$

$\sinh 2x = 2 \sinh x \cosh x$

$\cosh 2x = \cosh^2 x + \sinh^2 x$

$\sinh^2 x = \frac{\cosh 2x - 1}{2}$

$\cosh^2 x = \frac{\cosh 2x + 1}{2}$

$\cosh^2 x - \sinh^2 x = 1$

$\tanh^2 x = 1 - \operatorname{sech}^2 x$

$\coth^2 x = 1 + \csch^2 x$

$\int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1} \left( \frac{u}{a} \right) + C \quad \text{(Valid for } u^2 < a^2\text{)}$

$\int \frac{1}{a^2 + u^2} \, du = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \quad \text{(Valid for all } u\text{)}$

$\int \frac{1}{u\sqrt{u^2 - a^2}} \, du = \frac{1}{a} \sec^{-1} \left| \frac{u}{a} \right| + C \quad \text{(Valid for } u^2 > a^2\text{)}$

$\frac{d}{dx} \sinh u = \cosh u \frac{du}{dx}$

$\frac{d}{dx} \cosh u = \sinh u \frac{du}{dx}$

$\frac{d}{dx} \tanh u = \sech^2 u \frac{du}{dx}$

$\frac{d}{dx} \coth u = -\csch^2 u \frac{du}{dx}$

$\frac{d}{dx} \sech u = -\sech u \tanh u \frac{du}{dx}$

$\frac{d}{dx} \csch u = -\csch u \coth u \frac{du}{dx}$

$\sinh x = \frac{e^x - e^{-x}}{2}$

$\cosh x = \frac{e^x + e^{-x}}{2}$

$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

$\csch x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}$

$\sech x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$

$\coth x = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$

$\sech^{-1} x = \cosh^{-1} \left( \frac{1}{x} \right)$

$\csch^{-1} x = \sinh^{-1} \left( \frac{1}{x} \right)$

$\coth^{-1} x = \tanh^{-1} \left( \frac{1}{x} \right)$