Question

Iff(z,y)-lth' if (z.v) * (0,0), cal late fy(0,0) andふ(0,0), and verify that are different. (This doesn't violate Clairaut's theorem because the partials are not continuous at the origin.). Hint: In class we showed that: a)(0.0) if (z,y) # (0,0) 0 if (z,y) = (0,0)

Answer 1

F (x, y) = xy (xz^2 - y^2)/x^2 + y^2: f (x, y) not equalto (0, 0) 0: f (x, y) = (0, 0) we know f_x (x, y) = lim_h rightarrow f (x + h, y) - f (x, y)/h = {x^4 y + 4x^2 y^2 - y^5/(x^2 + y^2)^2 if (x, y) not equalto (0, 0) 0 if (x, y) = (0, 0) again we know f_xy (x, y) = lim_h rightarrow 0 f_y (x + h, y) - f_y (x, y)/h f_xy (0, 0) = lim_h rightarrow 0 f_y (h, 0) - f_y (0, 0)/h f_y (0, 0) = lim_k rightarrow 0 f (0, k) - f (0, 0)/k = lim_k rightarrow 0 0/k = 0 f_y (h, 0) = lim_k rightarrow 0f (h, k) - f (h, 0)/k = lim_k rightarrow 0 hk (h^2 - k^2)/k (h^2 + k^2) = h f_xy (0, 0) = lim_h rightarrow 0 h - 0/h = 1 \r\n again