differentiate equation x ^-3 y^2 + x ^2 y ^3 = 18 and find equation tangent line at the point (-1,3)

Question

Calculus solution

Answer 1

x^-3y² + x²y³ = 18

 

-3x^-4y² + 2y(dy/dx)x^-3 + 2xy³ + 3y²(dy/dx)x² = 0

 

When x = -1 and y = 3, we have -27 - 6(dy/dx)  - 54 + 27(dy/dx) = 0

 

                                                                                 21(dy/dx) = 81

 

                                                                                     dy/dx = 27/7

 

Tangent line has slope 27/7 and passes through the point (-1,3).

 

Equation of tangent line:  y - 3 = (27/7)(x - (-1))

 

                                     y - 3 = (27/7)x + 27/7

 

                                          y = (27/7)x + 48/7

                        

or  27x - 7y = -48