" (v) "e^(x)log y=sin^(-1)x+sin^(-1)y

e^xlogy=sin^(-1)x+sin^(-1)y

e^xlogy=sin^(-1)x+sin^(-1)y

Find dy/dx where e^x log y = sin^-1 x + sin^-1y

The solution of (dy)/(dx)=e^(x)(sin^(2)x+sin2x)/(y(2log y+1)) is

If y={(log)cos x sin x}{(log)_(sin x)cos x}^(-1)+sin^(-1)((2x)/(1+x^(2))) fin d (dy)/(dx)atx=(pi)/(4)

If x^(2)y^(2)=log(x+y)+sin e^(x)find(dy)/(dx)

Suppose 3 sin^(-1) ( log _(2) x) + cos^(-1) ( log _(2) y) =pi //2 and sin^(-1) ( log _(2) x ) + 2 cos^(-1) ( log_(2) y) = 11 pi //6 . then the value of 1/x^(2) + 1/y^(2) equals .