**Problem:** Find the expression for $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ given different functions for $u$. Specifically, 4) If $u = \cos^{-1} \left( \frac{x}{y} \right) + \tan^{-1} \left( \frac{y}{x} \right)$, find $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$. **Step-by-step solution:** 1. Recall the previous general result: $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \sin 2u$$ was derived for an expression involving $\tan u$. 2. To solve the new problem, we first differentiate $u = \cos^{-1} \left( \frac{x}{y} \right) + \tan^{-1} \left( \frac{y}{x} \right)$ with respect to $x$ and $y$. 3. Compute $\frac{\partial u}{\partial x}$: - Derivative of $\cos^{-1}\left( \frac{x}{y} \right)$ with respect to $x$ is $$-\frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \frac{1}{y} = -\frac{1}{y \sqrt{1 - \frac{x^2}{y^2}}} = -\frac{1}{y \cdot \frac{\sqrt{y^2 - x^2}}{y}} = -\frac{1}{\sqrt{y^2 - x^2}}.$$ - Derivative of $\tan^{-1} \left( \frac{y}{x} \right)$ with respect to $x$: Using chain rule: $$\frac{d}{dx} \tan^{-1} \left( \frac{y}{x} \right) = \frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2}.$$ Summing these gives: $$\frac{\partial u}{\partial x} = -\frac{1}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2}.$$ 4. Compute $\frac{\partial u}{\partial y}$: - Derivative of $\cos^{-1}\left( \frac{x}{y} \right)$ wrt $y$: $$-\frac{1}{\sqrt{1 - \left( \frac{x}{y} \right)^2}} \cdot \left(-\frac{x}{y^2}\right) = \frac{x}{y^2 \sqrt{1 - \frac{x^2}{y^2}}} = \frac{x}{y^2 \cdot \frac{\sqrt{y^2 - x^2}}{y}} = \frac{x}{y \sqrt{y^2 - x^2}}.$$ - Derivative of $\tan^{-1} \left( \frac{y}{x} \right)$ wrt $y$: $$\frac{1}{1 + \left( \frac{y}{x} \right)^2} \cdot \frac{1}{x} = \frac{1}{x + \frac{y^2}{x}} = \frac{1}{\frac{x^2 + y^2}{x}} = \frac{x}{x^2 + y^2}.$$ Summing gives: $$\frac{\partial u}{\partial y} = \frac{x}{y \sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2}.$$ 5. Now calculate $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x \left(-\frac{1}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2}\right) + y \left( \frac{x}{y \sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2} \right).$$ 6. Simplify each term: - $x \cdot -\frac{1}{\sqrt{y^2 - x^2}} = -\frac{x}{\sqrt{y^2 - x^2}}$ - $x \cdot -\frac{y}{x^2 + y^2} = -\frac{xy}{x^2 + y^2}$ - $y \cdot \frac{x}{y \sqrt{y^2 - x^2}} = \frac{x}{\sqrt{y^2 - x^2}}$ - $y \cdot \frac{x}{x^2 + y^2} = \frac{xy}{x^2 + y^2}$ 7. Sum all terms: $$-\frac{x}{\sqrt{y^2 - x^2}} - \frac{xy}{x^2 + y^2} + \frac{x}{\sqrt{y^2 - x^2}} + \frac{xy}{x^2 + y^2} = 0.$$ **Final answer:** $$\boxed{0}.$$