Question

what is y' in Arc csc squaroot of 2x

Answer 1

\[y=\csc^{-1}(\sqrt{2x})\]?

Answer 2

if i recall correctly the derivative of \(\csc^{-1}(x)\) is \(-\frac{1}{x\sqrt{x^2-1}}\)

Answer 3

yeah i got it

Answer 4

the y'

Answer 5

so you want \(\frac{d}{dx}[\csc^{-1}(\sqrt{2x})]\)

Answer 6

what you need is the derivative of \(\csc^{-1}(x)\) and the chain rule

Answer 7

\[\frac{d}{dx}[\csc^{-1}(x)=-\frac{1}{x\sqrt{x^2-1}}\] 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