Here's the question you clicked on:
Whatever1
what is y' in Arc csc squaroot of 2x
\[y=\csc^{-1}(\sqrt{2x})\]?
if i recall correctly the derivative of \(\csc^{-1}(x)\) is \(-\frac{1}{x\sqrt{x^2-1}}\)
so you want \(\frac{d}{dx}[\csc^{-1}(\sqrt{2x})]\)
what you need is the derivative of \(\csc^{-1}(x)\) and the chain rule
\[\frac{d}{dx}[\csc^{-1}(x)=-\frac{1}{x\sqrt{x^2-1}}\] i know because i just looked it up replace \(x\) by \(\sqrt{2x}\) and then multiply by the derivative of \(\sqrt{2x}\) which is \(\frac{1}{\sqrt{2x}}\) then probably some algebra to clean it up