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The problem asks for the partial derivative of the function \( f(x, y) = \arctan\left(\frac{y}{x}\right) \) with respect to \( x \), evaluated at the point \( (6, -8) \).

### Step-by-step solution:

#### 1. Partial derivative of \( f(x, y) \) with respect to \( x \):
We need to compute \( f_x(x, y) = \frac{\partial}{\partial x} \arctan\left(\frac{y}{x}\right) \).

Using the chain rule:
\[
f_x(x, y) = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{\partial}{\partial x} \left(\frac{y}{x}\right)
\]
\[
f_x(x, y) = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right)
\]
Thus, the partial derivative is:
\[
f_x(x, y) = -\frac{y}{x^2 + y^2}
\]

#### 2. Evaluate at \( (6, -8) \):
Now substitute \( x = 6 \) and \( y = -8 \) into the expression:
\[
f_x(6, -8) = -\frac{-8}{6^2 + (-8)^2}
\]
\[
f_x(6, -8) = \frac{8}{36 + 64} = \frac{8}{100} = 0.08
\]

### Final Answer:
\[
f_x(6, -8) = 0.08
\]

Would you like further details on any of the steps? Here are some related questions to explore:

1. How do you calculate partial derivatives in general?
2. What is the chain rule and how is it used in partial differentiation?
3. Why is the partial derivative of \( \arctan\left(\frac{y}{x}\right) \) dependent on both \( x \) and \( y \)?
4. Can you compute the partial derivative of \( f(x, y) \) with respect to \( y \)?
5. How does the geometry of the function \( \arctan\left(\frac{y}{x}\right) \) affect its partial derivatives?

**Tip:** When dealing with trigonometric functions and derivatives, always apply the chain rule carefully, especially if the argument involves both variables like \( \frac{y}{x} \).

Find the indicated partial derivatives. f(x, y) = arctan(y/x); f_x(6, -8)

Learn how to compute the partial derivative of the function f(x, y) = arctan(y/x) with respect to x, evaluated at the point (6, -8). This problem uses the chain rule for partial derivatives and trigonometric functions. Suitable for calculus students.

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