If ` sin (xy) + (x)/(y) =x^(2) - y , " then " (dy)/(dx)` is equal to

Options

A. `((2x^(2) - y^(2) cos xy -1)y)/(xy^(2) cos xy - x +y^(2))`
B. `((2xy - y^(2) cos xy -1)y)/(xy^(2) cos xy - x +y^(2))`
C. `((2xy- y^(2) cos xy -1)xy)/(xy^(2) cos xy - x +y^(2))`
D. `((2x^(2)- y^(2) cos xy -1)x)/(xy^(2) cos xy - x +y^(2))`

Answer

B

Explanation

We have , ` sin (xy) + (x)/(y) = x^(2) - y`
On differentiating both sides w.r.t.x, we get
`(d)/(dx) (sin xy) + (d)/(dx) ((x)/(y)) = (d)/(dx) x^(2) - (d)/(dx) y `
`rArr xy . (d)/(dx) (xy) + (y(d)/(dx) x-x.(d)/(dx) y)/(y^(2)) = 2x- (dy)/(dx)`
`rArr cos xy.[x.(d)/(dx) y + y.(d)/(dx) .x] + (y - x(dy)/(dx))/(y^(2)) = 2x - (dy)/(dx)`
`rArr x cos xy (dy)/(dx) + y cos xy + (y)/(y^(2))-(x)/(y^(2)) (dy)/(dx) = 2x - (dy)/(dx)`
`rArr (dy)/(dx) [ x cos xy - (x)/(y^(2) ) + 1] = 2x - ycos xy - (y)/(y^(2))`
` therefore (dy)/(dx) = [ (2xy - y^(2) cos xy -1)/(y) ][(y^(2))/(xy^(2) cos xy - x + y^(2))]`
` = ((2xy - y^(2) cos xy - 1)y)/((xy^(2) cos xy - x + y^(2)))`