Evaluate \(f(x, y) = x^2y - x + 2y + 3\) at (1,2).
The partial derivatives: \[ f(x, y) = x^2y - x + 2y + 3 \]
Compute \(f_x(x, y)\): \[ f_x(x, y) = \frac{\partial}{\partial x} (x^2y - x + 2y + 3) \]
Applying the derivative: \[\begin{align*} \frac{\partial}{\partial x} (x^2y) &= 2xy, \\ \frac{\partial}{\partial x} (-x) &= -1, \\ \frac{\partial}{\partial x} (2y) &= 0, \\ \frac{\partial}{\partial x} (3) &= 0. \end{align*}\]
Which results in: \[ f_x(x, y) = 2xy - 1 \]
Find \(f_y(x, y)\): \[ f_y(x, y) = \frac{\partial}{\partial y} (x^2y - x + 2y + 3) \]
Apply the derivative: \[\begin{align*} \frac{\partial}{\partial y} (x^2y) &= x^2, \\ \frac{\partial}{\partial y} (-x) &= 0, \\ \frac{\partial}{\partial y} (2y) &= 2, \\ \frac{\partial}{\partial y} (3) &= 0. \end{align*}\]
Thus, \[ f_y(x, y) = x^2 + 2 \]
Evaluate at \((1, 2)\): \[\begin{align*} f_x(1, 2) &= 2 \times 1 \times 2 - 1 = 4 - 1 = 3, \\ f_y(1, 2) &= 1^2 + 2 = 1 + 2 = 3. \end{align*}\]
Therefore, at the point \((1, 2)\), the values of \(f_x(x, y)\) and \(f_y(x, y)\) are both 3.