Find the slope of the graph of the equation at the given point?

Question

x^2y + y^3x = -6

Given point: (2,-1)

Answer 1

x^2*y' + 2x*y + y^3 + 2y^2*y'*x = 0

2*2*y' + 2*2*-1 + (-1)^3 + 2(-1)^2 *y' * 2 = 0

4y' + -4 + -1 + 4y' = 0

8y' - 5 = 0

y' = 5/8

Answer 2

To find the slope, you need to find dy/dx via implicit differentiation

Taking the derivative of the equation:

2y*x^(2y-1) + 3x*dy/dx = 0

We can plug in the specified point (2,-1) at this point, yielding (-2) (2)^(-2-1) +6 dy/dy = 0

Solving for dy/dx = (-2)*(2)^(-3) / (-6) = (-1/4)/-6 = 1/24

Answer 3

Assuming that you want y x² + x y³ = -6, The best approach is to use implicit differentiation to get the derivative dy/dx

To use implicit differentiation, operate on both side with d/dx and solve for dy/dx.

The result is dy/dx = - (2xy + y³)/(x² + 3 xy²)

Now substitute x =2 and y = -1 to get dy/dx = 5/10 = 1/2