We are tasked with finding the partial derivatives $f_x$, $f_y$, $f_x(6, -6)$, and $f_y(4, 1)$ for the given function:

$f(x, y) = \sqrt{x^2 + y^2}.$

Step 1: Compute $f_x$ and $f_y$

The partial derivatives are found using the chain rule.

Partial derivative $f_x$:

$f(x, y) = (x^2 + y^2)^{1/2}.$ $f_x = \frac{\partial}{\partial x} \sqrt{x^2 + y^2} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2}}.$

Partial derivative $f_y$:

Similarly, the partial derivative with respect to $y$ is: $f_y = \frac{\partial}{\partial y} \sqrt{x^2 + y^2} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2y = \frac{y}{\sqrt{x^2 + y^2}}.$

Step 2: Evaluate $f_x(6, -6)$ and $f_y(4, 1)$

Evaluate $f_x(6, -6)$:

Using $f_x = \frac{x}{\sqrt{x^2 + y^2}}$: $f_x(6, -6) = \frac{6}{\sqrt{6^2 + (-6)^2}} = \frac{6}{\sqrt{36 + 36}} = \frac{6}{\sqrt{72}} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}.$

Evaluate $f_y(4, 1)$:

Using $f_y = \frac{y}{\sqrt{x^2 + y^2}}$: $f_y(4, 1) = \frac{1}{\sqrt{4^2 + 1^2}} = \frac{1}{\sqrt{16 + 1}} = \frac{1}{\sqrt{17}}.$

Final Answers:

1. $f_x = \frac{x}{\sqrt{x^2 + y^2}}$,
2. $f_y = \frac{y}{\sqrt{x^2 + y^2}}$,
3. $f_x(6, -6) = \frac{1}{\sqrt{2}}$,
4. $f_y(4, 1) = \frac{1}{\sqrt{17}}$.

Would you like a detailed breakdown of any specific step?

Related Questions:

1. How can we interpret the partial derivative $f_x$ geometrically?
2. What happens to $f_x$ and $f_y$ as $x$ or $y$ approaches zero?
3. How do these derivatives relate to the gradient vector $\nabla f$?
4. What are the critical points of $f(x, y)$?
5. How does the chain rule extend to functions of more than two variables?

Tip:

For functions involving square roots, always simplify derivatives carefully to avoid algebraic mistakes.