## Thomas' Calculus 13th Edition

$\dfrac{1}{ \sqrt{(x+1)(4x+3)}}$
Given: $y=\cosh^{-1} 2\sqrt {x+1}$ Since, $\dfrac{d (\cosh^{-1} x)}{dx}=\dfrac{1}{\sqrt{x^2-1}}$ Then, $\dfrac{dy}{dx}=\dfrac{1}{\sqrt{(2\sqrt {x+1}-1)^2}} (\dfrac{2}{2 \sqrt {x+1}})=\dfrac{1}{\sqrt{4(x+1)-1}} (\dfrac{1}{ \sqrt {x+1}})$ or, $=\dfrac{1}{ \sqrt{(x+1)(4x+3)}}$