Question

Z=f(x,y)=3x³y²-x²+y²+4x+9

Answer 1

$$𝑧 = 𝑓(𝑥, 𝑦) = 3𝑥^3𝑦^2 − 𝑥^2𝑦^3 + 4𝑥 + 9$$

$$f_x(x, y)=9x^2y^2-2xy^3+4$$

$$f_y(x, y)=6x^3y-3x^2y^2$$

$$f_x(1, 3)=9(1)^2(3)^2-2(1)(3)^3+4=31$$

$$f_y(-2, 4)=6(-2)^3(4)-3(-2)^2(4)^2=-384$$