Answer
$ f_x(2, -1) = 11$
$f_y(2, -1) = -8$
Work Step by Step
Step 1 \[ f(x, y) = 2x^2 - 3xy + y^2 \] From theorem (13.3.1): \[ f_x(2, -1) = \lim_{{\Delta x \to 0}} \frac{f(2 + \Delta x, -1) - f(2, -1)}{\Delta x} = \lim_{{\Delta x \to 0}} \frac{2(2 + \Delta x)^2 - 3(2 + \Delta x)(-1) + (-1)^2 - 15\Delta x}{\Delta x} \] \[ = \lim_{{\Delta x \to 0}} \frac{8 + 8\Delta x + 2\Delta x^2 + 6 + 3\Delta x + 1 - 15\Delta x}{\Delta x} \] \[ = \lim_{{\Delta x \to 0}} \frac{11\Delta x + 2\Delta x^2}{\Delta x} \] \[ = \lim_{{\Delta x \to 0}} (11 + 2\Delta x) = 11 \] Step 2 From theorem (13.3.1): \[ f_y(2, -1) = \lim_{{\Delta y \to 0}} \frac{f(2, -1 + \Delta y) - f(2, -1)}{\Delta y} = \lim_{{\Delta y \to 0}} \frac{2(2)^2 - 3(2)(-1 + \Delta y) + (-1 + \Delta y)^2 - 15\Delta y}{\Delta y} \] \[ = \lim_{{\Delta y \to 0}} \frac{8 + 6 - 6\Delta y + 1 - 2\Delta y + \Delta y^2 - 15\Delta y}{\Delta y} \] \[ = \lim_{{\Delta y \to 0}} \frac{-8\Delta y + \Delta y^2}{\Delta y} \] \[ = \lim_{{\Delta y \to 0}} (-8 + \Delta y) = -8 \] \[ f_x(2, -1) = 11 \quad f_y(2, -1) = -8 \] Step 3 We can find the derivative of \(f(x, y) = 2x^2 - 3xy + y^2\) by the usual way: \[ f_x(x, y) = 4x - 3y \quad f_x(2, -1) = 4(2) - 3(-1) = 8 + 3 = 11 \] \[ f_y(x, y) = -3x + 2y \quad f_y(2, -1) = -3(2) + 2(-1) = -8 \] Result \[ f_x(2, -1) = 11 \quad f_y(2, -1) = -8 \]
