Question

Evaluate \(f_{x}\) and \(f_{y}\) at the given point. \(f(x, y)=\arctan \frac{y}{x}, \quad(2,-2)\)

Short answer

The partial derivative of \(f\) with respect to \(x\) evaluated at the point (2, -2) is 0.25 and with respect to \(y\) is also 0.25.

Step-by-step solution

Compute \(f_x\)

The first order partial derivative of \(f\) with respect to \(x\) is given by \(f_x = \frac{\partial}{\partial x}(\arctan \frac{y}{x}) = -\frac{y}{x^2 + y^2}\)

Compute \(f_y\)

The first order partial derivative of \(f\) with respect to \(y\) is given by \(f_y = \frac{\partial}{\partial y}(\arctan \frac{y}{x}) = \frac{x}{x^2 + y^2}\)

Evaluate \(f_x\) at (2,-2)

Using the values from the given point, \(f_x\) at (2, -2) is equal to -\frac{-2}{2^2 + (-2)^2} = -\frac{-2}{8} = 0.25 .

Evaluate \(f_y\) at (2,-2)

Using the same point, \(f_y\) at (2, -2) is equal to \frac{2}{2^2 + (-2)^2} = \frac{2}{8} = 0.25 .