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The question you referred to (Q-5 on page 3 of the document) is:

**Q-5**:  
If \( u = \frac{y}{x^2} \) and \( v = \frac{x^2 + y^2}{x^2} \), find \( \frac{\partial(u, v)}{\partial(x, y)} \).

### Solution:
We are asked to find the Jacobian of the functions \( u \) and \( v \) with respect to \( x \) and \( y \), which is:

\[
J(u, v) = \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{vmatrix}
\]

Let's compute each of these partial derivatives step by step.

#### Step 1: Compute \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \)

Given \( u = \frac{y}{x^2} \), we compute the partial derivatives:
- \( \frac{\partial u}{\partial x} = \frac{-2y}{x^3} \)
- \( \frac{\partial u}{\partial y} = \frac{1}{x^2} \)

#### Step 2: Compute \( \frac{\partial v}{\partial x} \) and \( \frac{\partial v}{\partial y} \)

Given \( v = \frac{x^2 + y^2}{x^2} = 1 + \frac{y^2}{x^2} \), we compute:
- \( \frac{\partial v}{\partial x} = \frac{\partial}{\partial x} \left( 1 + \frac{y^2}{x^2} \right) = \frac{-2y^2}{x^3} \)
- \( \frac{\partial v}{\partial y} = \frac{\partial}{\partial y} \left( 1 + \frac{y^2}{x^2} \right) = \frac{2y}{x^2} \)

#### Step 3: Compute the Jacobian determinant

The Jacobian matrix is:

\[
J(u, v) = \begin{vmatrix}
\frac{-2y}{x^3} & \frac{1}{x^2} \\
\frac{-2y^2}{x^3} & \frac{2y}{x^2}
\end{vmatrix}
\]

The determinant is:

\[
J(u, v) = \left( \frac{-2y}{x^3} \times \frac{2y}{x^2} \right) - \left( \frac{1}{x^2} \times \frac{-2y^2}{x^3} \right)
\]

Simplifying both terms:

\[
J(u, v) = \frac{-4y^2}{x^5} + \frac{2y^2}{x^5} = \frac{-2y^2}{x^5}
\]

Thus, the Jacobian is:

\[
J(u, v) = \frac{-2y^2}{x^5}
\]

### Final Answer:
\[
\frac{\partial(u, v)}{\partial(x, y)} = \frac{-2y^2}{x^5}
\]

Would you like more details on any step?

Here are some related questions to explore:
1. How does the Jacobian relate to the concept of function independence?
2. What does it mean geometrically if the Jacobian is zero?
3. Can you explain the significance of a negative Jacobian determinant?
4. How would you solve this if the functions involved more complex terms like exponentials?
5. How can Jacobians be applied in real-world engineering problems?

**Tip:** The Jacobian matrix plays a key role in transforming variables in multivariable calculus, especially in applications like changing coordinates in integrals.

Learn how to find the Jacobian determinant for the functions u and v with respect to x and y. This solution covers key mathematical concepts such as partial derivatives and Jacobian matrices, suitable for undergraduate students studying multivariable calculus.

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