Question

How do you implicitly differentiate `-y=x-sqrt(x^3y-y)`?

Answer 1

`y'=(2sqrt(x^3y-y)-3x^2y)/(x^3-1-2sqrt(x^3y-y))`

Explanation:

Differentiating with respect to `x`:
`-y'=1-1/2(x^3y-y)^{-1/2}(3x^2y+x^3y'-y')`
Multiplying by `2sqrt(x^2y-y)`
`x^3y'-y'-2y'sqrt(x^3y-y)=2sqrt(x^3y-y)-3x^2y`
and this is
`y'(x^3-1-2sqrt(x^3y-y))=2sqrt(x^3y-y)-3x^2y`
so
`y'=(2sqrt(x^3y-y)-3x^2y)/(x^3-1-2sqrt(x^3y-y))`
if the denominator is not equal zero