Derivative with two varaibles

Question

Find $y'$:

$$yx^2+y^3 = x-y$$

I tried using $\frac{dy}{dx}$ and have gotten $\frac{1}{x}+3y^2$, which isn't right. Any ideas on what I'm doing wrong?

Answer 1

You have to use implicit differentiation, as follows:

$$\frac{d}{dx}(yx^2 + y^3) = \frac{d}{dx}(x - y)$$ $$\frac{d}{dx}(yx^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(x) - \frac{d}{dx}(y)$$

Using product rule and chain rule:

$$(y'x^2 + 2xy) + 3y^2y' = 1 - y'$$ $$y'x^2 + 3y^2y' + y'= 1 - 2xy $$ $$y'(x^2 + 3y^2 + 1)= 1 - 2xy $$ $$y'= \frac{1 - 2xy}{x^2 + 3y^2 + 1} $$

Answer 2

Your first step should be to differentiate both sides using implicit differentiation, that is, noting that $y = y(x)$, and applying the chain rule as you go. We find $$ yx^2+y^3 = x-y \overset{d/dx}\implies y'x^2 + 2xy + 3y^2y' = 1 - y' $$ From there, you need to solve for $y'$ as a function of $x$ and $y$