Question

How do you find the derivative of `y^2=2+xy`?

Answer 1

`(dy)/(dx)=y/(2y-x)`

Explanation:

`y^2=2+xy`

`rarr color(white)("XX")y^2-xy=2`

`rarr color(white)("XX")(d(y^2-xy))/(dx)=(d (2))/(dx)`

`rarr color(white)("XX")(d(y^2))/(dx)-(d(xy))/(dx)=0`

`rarr color(white)("XX")2y(dy)/(dx)-(y+x(dy)/(dx))=0`
`color(white)("XXXXXXXXXXXX")`(using the Chain and Product rules)

`rarr color(white)("XX")2y(dy)/(dx)-x(dy)/(dx)=y`

`rarr color(white)("XX")(dy)/(dx)(2y-x)=y`

`rarr color(white)("XX")(dy)/(dx)=y/(2y-x)`