# More calculus help

• Consider a curve of x(y^2)-x^(3)y=6.

The derivative is: (3(x^2)y-y^2) / (2xy-x^3)

I need to find when the derivative does not exist (tangent line vertical).

So far, as we only need to see if the denominator is 0, I set 2xy-x^3=0.

Then, I factored this to get x(2y-x^2)=0.

I know from here that when x=0, there is a vertical tangent line. However, I think there is more than that. What should I do to find the rest, or if I am doing this wrong, how do I solve it?

• Curious.

Is that XYY-X^(3Y)?

(question is about the second term.  Need to know if the three and the y are powers of X.)

Let me know tonight, and I’ll help ya with it tomorrow.

• I need this for tomorrow.

It is x to the third times y, not x to the 3y.

• Solve for y:

X(Y^2)-Y(X^3)=0
Add Y(X^3) to both sides of the equality

X(Y^2)=Y(X^3)
Divide both sides by Y

XY=X^3
Divide both sides by X

Y=X^2 is your solution.

The graph of Y=X^2 exists for all Y >= 0.
The graph of Y=X^2 exists for all X  (-oo < X < oo) where oo = infinity

The derivative of Y=X^2 is:

Y’=2X

2X is a linear function and has no points of discontinuity.

Both f(x) and f’(x) are continuous functions.

BTW, you did not take the derivative correctly when you attempted to derive the tangent line off the original function.  To do this you would have to use implicit differentiation.

For instance, X^3 * Y using implicit differentiation is not 3X^2 * Y, it is 3X^2 * Y * Y’ where Y’ represents DY/DX.

• It equals 6, not zero.

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