Inverse Hyperbolic Functions Formula

The hyperbolic sine function is a one-to-one function and thus has an inverse. As usual, the graph of the inverse hyperbolic sine function $sinh^{-1}(x)$ also denoted by $arcsinh(x)$ by reflecting the graph of $sinh(x)$ about the line $y=x$

For all inverse hyperbolic functions but the inverse hyperbolic cotangent and the inverse hyperbolic cosecant, the domain of the real function is connected.

Inverse hyperbolic sine (if the domain is the whole real line)

\[\large arsinh\;x=ln(x+\sqrt {x^{2}+1}\]

Inverse hyperbolic cosine (if the domain is the closed interval $(1, +\infty )$.

\[\large arcosh\;x=ln(x+\sqrt{x^{2}-1})\]

Inverse hyperbolic tangent [if the domain is the open interval (−1, 1)]

\[\large arcosh\;x=\frac{1}{2}\;ln\left(\frac{1+x}{1-x} \right )\]

Inverse hyperbolic cotangent [if the domain is the union of the open intervals (−∞, −1) and (1, +∞)]

\[\large arcosh\;x=\frac{1}{2}\;ln\left(\frac{x+1}{x-1} \right )\]

Inverse hyperbolic cosecant (if the domain is the real line with 0 removed)

$\large arcosh\;x=ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^{2}+1}}\right)=ln\left(\frac{1+\sqrt{x^{2}+1}}{x}\right)$

Inverse hyperbolic secant (if the domain is the semi-open interval 0, 1)

$\large arcosh\;x=ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^{2}+1}}\right)=ln\left(\frac{1+\sqrt{1-x^{2}}}{x}\right)$

Derivatives formula of Inverse Hyperbolic Functions

\[\large \frac{d}{dx}sinh^{-1}x=\frac{1}{\sqrt{x^{2}+1}}\]

\[\large \frac{d}{dx}cosh^{-1}x=\frac{1}{\sqrt{x^{2}-1}}\]

\[\large \frac{d}{dx}tanh^{-1}x=\frac{1}{1-x^{2}}\]

\[\large \frac{d}{dx}coth^{-1}x=\frac{1}{1-x^{2}}\]

\[\large \frac{d}{dx}sech^{-1}x=\frac{-1}{x\sqrt{1-x^{2}}}\]

\[\large \frac{d}{dx}csch^{-1}x=\frac{-1}{|x|\sqrt{1-x^{2}}}\]


Practise This Question

Which of the following statements is/are correct?
I.  Urine is hypertonic in distal convoluted tubule.
II.  When the urine pases into the collecting tubule, it becomes hypotonic.
III.  Urine is isotonic in Proximal Convoluted Tubule (PCT).
IV.  Urine becomes more and more hypotonic as it passes throught the Henle's loop.
Select the correct answer using the codes given below.