Hello, mathshelpneeded!

I'm not sure how to get it into that formShow that the derivative of: .x + y .= .x/y

. . can be written as: .y' .= .y³/x²(2 - y)

The differentiation is really simple and i got: .y' .= .(1 - y)/(x + 2y) .[1]

However i cannot make it into that form!naturally,

. . but I canhammerit into that form . . .

From the original equation, we have:

. . xy + y² .= .x . → . y² .= .x(1 - y) .[2]. → . 1 - y .= .y²/x .[3]

The original equation is: .x + y .= .x/y

Add y to both sides: .x + 2y .= .x/y + y .= .(x + y²)/y

. . From [2]: .x + 2y .= .[x + x(1 - y)]/y . → . x + 2y .= .x(2 - y)/y .[4]

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . y²/x . . . . . . . y³

Substitute [3] and [4] into [1]: .y' .= . ------------ .= .-----------

. . . . . . . . . . . . . . . . . . . . . . . . . . . x(2 - y)/y . . . .x²(2 - y)