Answer
$y'=\dfrac{1}{2\sqrt{x^2-x}}$
Work Step by Step
Given: $y=\cosh^{-1}{\sqrt{x}}$
Now, $y'=\dfrac{d}{dx}\cosh^{-1}{\sqrt{x}}$
Apply chain rule, we have
$y'=\dfrac{d\cosh^{-1}{\sqrt{x}}}{d\sqrt{x}}
( \dfrac{d\sqrt{x}}{dx})$
or, $y'=\dfrac{1}{\sqrt{(\sqrt x)^2-1}}(\dfrac{1}{2\sqrt x})$
or, $y'=\dfrac{1}{2\sqrt x\sqrt{x-1}}$
Thus, $y'=\dfrac{1}{2\sqrt{x^2-x}}$
