Question

How do you implicitly differentiate `-1=1/(xy)-1-3y^2`?

Answer 1

`dy/dx=y/(x(6xy^3+1))`

Explanation:

we rely on the chain rule which states

`d/dx = dy/dx*d/dy`

`d/dx = -1/(x^2y)`
`d/dy =-1/(y^2x) -6y`

now we plug into our formula and solve
`-1/(x^2y) = (dy/dx)(-1/(y^2x) -6y)`

`dy/dx=1/(x^2y) /(1/(y^2x) +6y)`

`dy/dx=y/(x(6xy^3+1))`