Find dy/dx of x^2 − (y^2) x = (3x − 3)y

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Discussion Overview

The discussion revolves around finding the derivative dy/dx of the equation x^2 − (y^2)x = (3x − 3)y. Participants are exploring the process of differentiation and the implications of having multiple slopes at a specific x-coordinate, particularly focusing on the points where x equals 1.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant presents the equation and seeks to derive a formula for dy/dx, proposing y'(x) = (2x - y^2 - 3y) / (2xy + 3x - 3).
  • Another participant expresses agreement with the proposed formula for dy/dx.
  • A question is raised regarding how to utilize the derived formula to find two different slopes at the same x-coordinate.
  • A participant clarifies the context by noting that finding a derivative does not involve differential equations, suggesting a shift in the discussion to a more appropriate section.
  • A later reply indicates understanding of the previous points discussed.

Areas of Agreement / Disagreement

There is agreement on the proposed formula for dy/dx, but uncertainty remains regarding the method to find two different slopes for the same x-coordinate, indicating that the discussion is not fully resolved.

Contextual Notes

The discussion does not clarify the specific values of y corresponding to the x-coordinate of 1, which may be necessary for calculating the slopes.

Moongn
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implicitdifferentiation.jpg


x^2−(y^2)x = (3x − 3)y

Using this graph and the equation I need to find two things.

(a) Formula which gives the slope dy/dx at every point (x, y) on the graph.
(b) As you can see in the picture, there are two points on the graph which have x-coordinate equal to 1. What are the exact slopes of the tangent lines at those two points?

I believe I correctly figured out (a) to be y'(x)= (2x-y^2-3y)/(2xy+3x-3)

However, I am not sure how to use this formula to find two different slopes for one value of x.
 
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Moongn said:
I believe I correctly figured out (a) to be y'(x)= (2x-y^2-3y)/(2xy+3x-3)
That looks correct

However, I am not sure how to use this formula to find two different slopes for one value of x.
You are given the value of x... what else do you need to calculate the slope at a point?
 
Since "find a derivative" is NOT differential equations, I am moving this to the Calculus section.
 
I understand now. Thank You!
 

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