Question

( x=sqrt{2} operatorname{cosec}^{-1} t quad y=sqrt{2^{sec ^{1} t}} )
( frac{d x}{d t}=frac{1}{2 sqrt{2^{cos e^{-1} t}}}^{2} operatorname{cosec}^{-1} t log _{e^{2}}left(frac{-1}{|x| sqrt{x^{2}-1}}right) )
( frac{d y}{d t}=frac{1}{2 sqrt{2^{sec t}}} 2^{sec ^{-1} t} log _{e^{2}}left(frac{1}{|x| sqrt{x^{2}-1}}right) )
So; clearly ( frac{d y}{d x}=frac{d y / d t}{d x / d t} )
( frac{d y}{d x}=-frac{sqrt{2^{sec ^{1} t}}}{2^{sec ^{-1} t}} times frac{2^{operatorname{cosec}^{-1} t}}{sqrt{2^{cos e c^{prime} t}}} )
( =-frac{sqrt{2^{cos e^{-1} t}}}{sqrt{2^{sec ^{-1} t}}} )
( frac{d y}{d x}=-frac{x}{y} )

# I r = J carent and va Solecito find de 2

Solution