Inverse Hyperbolic Trigonometric Functions
10 Jan 2022
Definitions
For background on the inverse hyperbolic trig functions, you can refer to my series on trigonometry and this post . These results are very commonly used in advanced integration.
All proofs will directly evaluate the formula of the inverse functions. However, the same results could also be derived using the inverse derivative property (as we did for the standard inverse trig functions ).
Inverse Hyperbolic Sine
\[\begin{align}
\frac{d}{dx} \arcsinh (x)
&= \frac{d}{dx} \left [ \ln \left ( x + \sqrt{x^2 + 1} \right ) \right ] \\[10pt]
&= \frac{1 + \frac{x}{\sqrt{x^2+1}}}{x + \sqrt{x^2 + 1}} \\[10pt]
&= \frac{\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}}{x + \sqrt{x^2 + 1}} \\[10pt]
&= \frac{1}{\sqrt{x^2+1}}
\end{align}\]
Inverse Hyperbolic Cosine
\[\begin{align}
\frac{d}{dx} \arccosh (x)
&= \frac{d}{dx} \left [ \ln \left ( x + \sqrt{x^2 - 1} \right ) \right ] \\[10pt]
&= \frac{1 + \frac{x}{\sqrt{x^2-1}}}{x + \sqrt{x^2 - 1}} \\[10pt]
&= \frac{\frac{x + \sqrt{x^2-1}}{\sqrt{x^2-1}}}{x + \sqrt{x^2 - 1}} \\[10pt]
&= \frac{1}{\sqrt{x^2-1}}
\end{align}\]
Inverse Hyperbolic Tangent
\[\begin{align}
\frac{d}{dx} \arctanh (x)
&= \frac{d}{dx} \left [ \frac{1}{2} \ln \left ( \frac{1+x}{1-x} \right ) \right ] \\[10pt]
&= \frac{1}{2} \frac{\frac{1 \cdot (1-x) - (-1) \cdot (1+x)}{(1-x)^2}}{ \frac{1+x}{1-x} } \\[10pt]
&= \frac{\frac{1}{(1-x)^2}}{ \frac{1+x}{1-x} } \\[10pt]
&= \frac{1}{1 - x^2}
\end{align}\]
Inverse Hyperbolic Secant
\[\begin{align}
\frac{d}{dx} \arcsech (x)
&= \frac{d}{dx} \left [ \ln \left ( \frac{1}{x} + \sqrt{\frac{1}{x^2} - 1} \, \right ) \right ] \\[10pt]
&= \frac{-\frac{1}{x^2} + \frac{-\frac{1}{x^3}}{\sqrt{\frac{1}{x^2}-1}}}{\frac{1}{x} + \sqrt{\frac{1}{x^2} - 1}} \\[10pt]
&= \frac{-1 - \frac{1}{\sqrt{1 - x^2}}}{x + x\sqrt{1 - x^2}} \\[10pt]
&= \frac{- \frac{1 + \sqrt{1 - x^2}}{\sqrt{1 - x^2}}}{x(1 + \sqrt{1 - x^2})} \\[10pt]
&= \frac{- 1}{x \sqrt{1 - x^2}}
\end{align}\]
Inverse Hyperbolic Cosecant
\[\begin{align}
\frac{d}{dx} \arccsch (x)
&= \frac{d}{dx} \left [ \ln \left ( \frac{1}{x} + \sqrt{\frac{1}{x^2} + 1} \, \right ) \right ] \\[10pt]
&= \frac{-\frac{1}{x^2} + \frac{-\frac{1}{x^3}}{\sqrt{\frac{1}{x^2}+1}}}{\frac{1}{x} + \sqrt{\frac{1}{x^2} + 1}} \\[10pt]
&= \frac{-1 - \frac{1}{\sqrt{1 + x^2}}}{x + x\sqrt{1 + x^2}} \\[10pt]
&= \frac{- \frac{1 + \sqrt{1 + x^2}}{\sqrt{1 + x^2}}}{x(1 + \sqrt{1 + x^2})} \\[10pt]
&= \frac{- 1}{x \sqrt{1 + x^2}}
\end{align}\]
Inverse Hyperbolic Cotangent
\[\begin{align}
\frac{d}{dx} \arccoth (x)
&= \frac{d}{dx} \left [ \frac{1}{2} \ln \left ( \frac{x+1}{x-1} \right ) \right ] \\[10pt]
&= \frac{1}{2} \frac{\frac{1 \cdot (x-1) - 1 \cdot (x+1)}{(x-1)^2}}{ \frac{x+1}{x-1} } \\[10pt]
&= \frac{-\frac{1}{(x-1)^2}}{ \frac{x+1}{x-1} } \\[10pt]
&= \frac{1}{1 - x^2}
\end{align}\]