Answer to Question #208579 in Calculus for Ayesha

Question

Z=f(x,y)=3x³y²-x²+y²+4x+9

Answer 1

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

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

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

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

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