Develop an expression for (dy/dx), the slope of the tangent line to the curve for x^2+xy +y^3=2?

3 Answers

Relevance
  • Como
    Lv 7
    3 weeks ago

    :-

    2x + y + x dy/dx + 3y² dy/dx = 0

    ( x. + 3y² ) dy/dx = - ( y + 2x )

    dy/dx = - ( y + 2x ) / ( x + 3y² )

  • Ray S
    Lv 7
    3 weeks ago

                        x² + xy + y³ = 2

    2x + (x•y' + y•1) + 3y²•y' = 0

             2x + xy' + y + 3y²y' = 0

                          xy' + 3y²y' = -2x - y

                          (x + 3y²)y' = -(2x + y)

                                        y' = -(2x + y) / (x + 3y²)

  • x^2 + xy + y^3 = 2

    2x * dx + x * dy + y * dx + 3y^2 * dy = 0

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

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

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

Still have questions? Get your answers by asking now.