## anonymous 5 years ago find dy/dx where xy^3-3x^2=7y

1. anonymous

(y^3 -6x) / (3y^2-7)

2. anonymous

I am not sure how to work it. I just started working on these kind of problems. My examples are confusing. I have (x+3xy) -3x^2=7 to start off with, is that correct?

3. anonymous

No, this is partial differentiation, it is not like other differentiation

4. anonymous

Its Implicit differentiation, partials are when you hold one variable as a constant. In Implicit differentiation you take the derivative of both variables with respect to x.

5. anonymous

here's the work: $xy^3-3x^2=7y \rightarrow y^3+3xy^2\frac{dy}{dx}-6x=7\frac{dy}{dx}$ $\rightarrow y^3-6x=7\frac{dy}{dx}-3xy^2\frac{dy}{dx}$ $\rightarrow y^3-6x=(7-3xy^2)\frac{dy}{dx}$ $\frac{dy}{dx}=\frac{y^3-6x}{7-3xy^2}$