Question

For the function $f(x, y) = \begin{cases} xy\frac{x^2 - y^2}{x^2 + y^2}, & \text{if } (x, y) \neq (0, 0) \\ 0, & \text{if } (x, y) = (0, 0) \end{cases}$ \\ i. Show that $f_x(x, 0) = x$ and $f_y(0, y) = -y$. \\ ii. Show that $f_{xy}(0, 0) \neq f_{yx}(0, 0)$.

          For the function $f(x, y) = \begin{cases} xy\frac{x^2 - y^2}{x^2 + y^2}, & \text{if } (x, y) \neq (0, 0) \\ 0, & \text{if } (x, y) = (0, 0) \end{cases}$ \\
i. Show that $f_x(x, 0) = x$ and $f_y(0, y) = -y$. \\
ii. Show that $f_{xy}(0, 0) \neq f_{yx}(0, 0)$.

For the function f(x, y) = xy(x^2 - y^2)/(x^2 + y^2), if (x, y) ≠ (0, 0)
0, if (x, y) = (0, 0)

i. Show that fx(x, 0) = x and fy(0, y) = -y.

ii. Show that fxy(0, 0) ≠ fyx(0, 0).

Added by Janet W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

04/27/2023

Step 1

f(x, y) = (xy(x^2 - y^2))/(x^2 + y^2) f_x(x, y) = (y(x^2 - y^2) + xy(2x))/(x^2 + y^2)^2 f_x(x, 0) = (0(x^2 - 0) + 0(2x))/(x^2 + 0)^2 f_x(x, 0) = 0 Therefore, f_x(x, 0) = x. Show more…

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NEED AN URGENT SOLUTION: For the function f(x, y) = {(xy(x^2 - y^2))/(x^2 + y^2), if (x, y) ≠ (0, 0), (0, if (x, y) = (0, 0))}: i. Show that f_x(x, 0) = x and f_y(0, y) = -y. ii. Show that f_xy(0, 0) ≠ f_yx(0, 0). For the function f(x, y) = y^2, if (x, y) = (0, 0): i. Show that f_xx,0 = x and f_y0,y = y. ii. Show that f_xy(0, 0) ≠ f_y(0, 0).