If `x = e^(x//y)` , then prove that `(dy)/(dx) = (x-y)/(xlogx)`.

Answer

N/a

Explanation

We have, `x=e^(x//y)`
` :. d/dxx=d/dxe^(x//y)`
`rArr 1=e^(x//y).d/dx(x//y)`
`rArr 1= e^(x//y).[(y.1-x.dy//dx)/(y^(2))]`
`rArr y^(2)=y.e^(x//y)-x.(dy)/(dx).e^(x//y)`
`rArr x.(dy)/(dx).e^(x//y)=ye^(x//y)-y^(2)`
`:. (dy)/(dx) = (y(e^(x//y)-y))/(x.e^(x//y))`
`= ((e^(x//y)-y))/(e^(x//y).x/y) , [:' x=e^(x//y)rArr log x=x/y]`
`= (x-y)/(x.logx)`