what is the curve given by the equation x^3=3xy^2+y^3=1
what is dy/dx?
To find the curve given by the equation x^3 = 3xy^2 + y^3 = 1, we need to differentiate the equation implicitly with respect to x.
First, let's differentiate the equation with respect to x:
d/dx (x^3) = d/dx (3xy^2 + y^3)
Applying the power rule, we get:
3x^2 = 3y^2 + 6xy(dy/dx) + 3y^2(dy/dx)
Next, let's isolate the terms involving dy/dx:
6xy(dy/dx) + 3y^2(dy/dx) = 3x^2 - 3y^2
Factoring out dy/dx:
(dy/dx)(6xy + 3y^2) = 3x^2 - 3y^2
Now, we can solve for dy/dx by dividing both sides by (6xy + 3y^2):
dy/dx = (3x^2 - 3y^2) / (6xy + 3y^2)
Thus, dy/dx is given by the equation (3x^2 - 3y^2) / (6xy + 3y^2).