# How do you find the derivative of the function y=cosh^-1((sqrt(x))?

Dec 30, 2016

The answer is $= \frac{1}{2 \sqrt{x} \sqrt{x - 1}}$

#### Explanation:

We need

$\left(\sqrt{x}\right) ' = \frac{1}{2 \sqrt{x}}$

$\left(\cosh x\right) ' = \sinh x$

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

Here, we have

$y = {\cosh}^{- 1} \left(\sqrt{x}\right)$

Therefore,

$\cosh y = \sqrt{x}$

Taking the derivatives on both sides

$\left(\cosh y\right) ' = \left(\sqrt{x}\right) '$

$\sinh y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x} \sinh y}$

${\cosh}^{2} y - {\sinh}^{2} y = 1$

${\sinh}^{2} y = {\cos}^{2} y - 1$

${\sinh}^{2} y = x - 1$

$\sinh y = \sqrt{x - 1}$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x} \sqrt{x - 1}}$