If x 2xy-y^2=2 then at the point (1 1) dy/dx is 305896-X 2 y 2 dx 2xy dy
Separable Differential Equation Dy Dx X Sqrt 1 Y 2 Youtube The first step is to find dy/dx To do this you must first expand the brackets x 2 y 2 2xy = xy 2 Then differentiate each term with respect to x dy/dx of (x 2) = 2x dy/dx of (y 2) = 2y(dy/dx) (Using the product rule with u=y 2 and v=1) this can be explained in more detail if necessary dy/dx of (2xy) = 2y 2x(dy/dx) when x=1, y=1, so plug them i 6 2 2y' 2y' = 0 Note that there is no solution, so the tangent to the curve is vertical y' is not defined Looking at dx/dy, 6xx' 2x 2yx' 2y = 0 6x' 2 2x' 2 = 0 4x' = 0 x' = 0 just as we suspected dx/dy=0, so the slope is vertical X 2 y 2 dx 2xy dy