Question

How do you implicitly differentiate `x^3 - 3xy + 2y^3 = 3`?

Answer 1

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

Explanation:

We assume that y is a function of x, ie `y=f(x)` and then differentiate each side of the equation with respect to x, then re-arrange and solve for `dy/dx`.

`d/dx(x^3-3xy+2y^3)=d/dx(3)`

`therefore 3x^2-3xdy/dx-3y+6y^2dy/dx=0`. (Used product rule and power rule)

`therefore dy/dx(-3x+6y^2)=3y-3x^2`. (Took out common factor)

`therefore dy/dx=(3y-3x^2)/(6y^2-3x)`

`therefore dy/dx=(y-x^2)/(2y^2-x)` (Divide all by 3).