 Use Euler’s theorem on sin u, and then take the partial derivatives.3. If u = sin−1
s
x
1/3 + y
1/3
x
1/2 + y
1/2
, then show that
x
2uxx + 2xyuxy + y
2uyy =
tan u
12 
13
12
+
tan2 u
12 

To solve this problem, we need to break it down into several steps:

### 1. Expression for \( u \):
The given expression for \( u \) is:
\[
u = \sin^{-1} \left( \frac{x^{1/3} + y^{1/3}}{x^{1/2} + y^{1/2}} \right)
\]
This means \( u \) is the inverse sine (arcsine) of the given ratio.

### 2. Applying Euler's Theorem:
Euler's theorem for sine in terms of complex exponentials is:
\[
e^{i\theta} = \cos\theta + i\sin\theta
\]
Thus:
\[
\sin u = \frac{e^{iu} - e^{-iu}}{2i}
\]
However, here we are not directly concerned with expressing \( \sin u \) in terms of exponentials, since the question primarily revolves around the derivatives. But Euler’s identity can remind us how trigonometric functions are related to complex exponentials.

### 3. Implicit Differentiation:
We need to compute second-order partial derivatives of \( u \) with respect to \( x \) and \( y \).

Let’s define:
\[
z = \frac{x^{1/3} + y^{1/3}}{x^{1/2} + y^{1/2}}
\]
Thus:
\[
u = \sin^{-1}(z)
\]
To find the partial derivatives \( u_x \), \( u_y \), and second-order partials \( u_{xx} \), \( u_{xy} \), and \( u_{yy} \), we will differentiate \( u \) with respect to \( x \) and \( y \).

### 4. First-Order Partial Derivatives:
The chain rule for \( u = \sin^{-1}(z) \) gives:
\[
u_x = \frac{1}{\sqrt{1 - z^2}} \cdot z_x
\]
\[
u_y = \frac{1}{\sqrt{1 - z^2}} \cdot z_y
\]
We now need to calculate \( z_x \) and \( z_y \).

### 5. Second-Order Partial Derivatives:
For the second-order partial derivatives \( u_{xx} \), \( u_{xy} \), and \( u_{yy} \), we apply the product rule and chain rule carefully. For example, for \( u_{xx} \):
\[
u_{xx} = \frac{\partial}{\partial x} \left( \frac{1}{\sqrt{1 - z^2}} \cdot z_x \right)
\]
This involves taking the derivative of both terms (using the product and chain rules).

### 6. Substituting into the Original Equation:
The goal is to show that:
\[
x^2 u_{xx} + 2xy u_{xy} + y^2 u_{yy} = \frac{\tan u}{12} + \frac{\tan^2 u}{12}
\]
After calculating the second-order partial derivatives and substituting them into the left-hand side, you should get the required expression.

### Summary of Approach:
- **Step 1:** Write \( u \) in terms of arcsine and the function \( z \).
- **Step 2:** Use chain rule to find \( u_x \) and \( u_y \).
- **Step 3:** Use chain rule and product rule to find second-order derivatives \( u_{xx} \), \( u_{xy} \), and \( u_{yy} \).
- **Step 4:** Substitute the second-order partial derivatives into the given equation and simplify to get the desired result.

Would you like a detailed step-by-step calculation for each derivative?

### 5 Questions:
1. Would you like to see the full steps for finding \( u_x \) and \( u_y \)?
2. Are you interested in the explicit forms of \( z_x \) and \( z_y \)?
3. Would you like help simplifying the second-order partial derivatives?
4. Would you like an explanation of how Euler's Theorem is applied here?
5. Are there other identities you'd like to explore for the arcsine function?

### Tip:
When differentiating implicit functions like \( u = \sin^{-1}(z) \), always remember to apply the chain rule for both first- and second-order derivatives to ensure accurate results.

Use Euler’s theorem on sin u, and then take the partial derivatives. If u = sin−1((x^(1/3) + y^(1/3)) / (x^(1/2) + y^(1/2))), then show that x^2 * u_xx + 2xy * u_xy + y^2 * u_yy = (tan u)/12 + (tan^2 u)/12.

This solution involves using Euler's theorem on sin u and implicit differentiation to solve partial derivatives. It walks through finding first- and second-order partial derivatives and applying them to the equation x^2 * u_xx + 2xy * u_xy + y^2 * u_yy = (tan u)/12 + (tan^2 u)/12.

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