Question

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

Answer 1

`y'=(e^(xy^3-x(1-y^3)+y^2)/(3xy^2e^(xy^3-x)+2xy))`

Explanation:

Differentiating with respect to `x`
`0=e^(xy^3-x)(y^3+x*3y^2y'-1)-y^2-x*2y*y'`
this gives

`0=e^(xy^3-x)(y^3-1)-y^2+y'(3xy^2e^(xy^3-x)+2xy)`