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find dy/dx where xy^3-3x^2=7y
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(y^3 -6x) / (3y^2-7)
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?
No, this is partial differentiation, it is not like other differentiation
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.
here's the work: \[xy^3-3x^2=7y \rightarrow y^3+3xy^2\frac{dy}{dx}-6x=7\frac{dy}{dx}\] \[\rightarrow y^3-6x=7\frac{dy}{dx}-3xy^2\frac{dy}{dx}\] \[\rightarrow y^3-6x=(7-3xy^2)\frac{dy}{dx}\] \[\frac{dy}{dx}=\frac{y^3-6x}{7-3xy^2}\]
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