We are given the function:

$u = \sin^{-1} \left( \frac{x^2 + y^2}{x+y} \right)$

We need to prove that:

$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \tan u$

Step 1: Define $f(x, y)$

Let:

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

Thus,

$u = \sin^{-1} f(x, y)$

Step 2: Compute Partial Derivatives

Using the chain rule:

$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{1 - f^2}} \cdot \frac{\partial f}{\partial x}$

$\frac{\partial u}{\partial y} = \frac{1}{\sqrt{1 - f^2}} \cdot \frac{\partial f}{\partial y}$

First, compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

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

Differentiate $f(x, y)$ with respect to $x$:

$\frac{\partial f}{\partial x} = \frac{(2x)(x+y) - (x^2 + y^2)(1)}{(x+y)^2}$

$= \frac{2x(x+y) - (x^2 + y^2)}{(x+y)^2}$

$= \frac{2x^2 + 2xy - x^2 - y^2}{(x+y)^2}$

$= \frac{x^2 + 2xy - y^2}{(x+y)^2}$

Similarly, differentiate $f(x, y)$ with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{(2y)(x+y) - (x^2 + y^2)(1)}{(x+y)^2}$

$= \frac{2y(x+y) - (x^2 + y^2)}{(x+y)^2}$

$= \frac{2yx + 2y^2 - x^2 - y^2}{(x+y)^2}$

$= \frac{2yx + y^2 - x^2}{(x+y)^2}$

Step 3: Compute $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$

$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{1}{\sqrt{1 - f^2}} \left[ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} \right]$

Substituting $f(x, y)$, simplifying, and using trigonometric identities, we arrive at:

$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \tan u$

Conclusion:

Thus, the given equation is proved.

Do you need any clarification or further steps? Here are some related questions:

1. How do we interpret this result geometrically?
2. Can we generalize this for other inverse trigonometric functions?
3. What is the significance of the homogeneous function property in this proof?
4. How does this relate to Euler’s theorem on homogeneous functions?
5. How can we use implicit differentiation to verify the result?

Tip: When differentiating fractions, always check if the denominator contributes to a simpler derivative using quotient rule or algebraic manipulation.