Let w(x, y) = xy + xf(y/x) , where f / \mathbb{R} -> \mathbb{R} is a smooth function . (i) (2 pts) Write down the formulas for partial w partial x and partial w partial y . Your answers will involve f'. (ii) (2 pts) Using part (i) verify that w(x, y) satisfies the equation x partial w partial x +y partial w partial y =xy+w.

Let's break this down step by step.

### (i) Finding the partial derivatives of \( w(x, y) \):

Given the function:
\[
w(x, y) = xy + x f\left(\frac{y}{x}\right)
\]
where \( f \) is a smooth function and \( f' \) denotes its derivative.

#### 1. Partial derivative of \( w \) with respect to \( x \):
To find \( \frac{\partial w}{\partial x} \), we apply the product rule and chain rule.

\[
w(x, y) = xy + x f\left(\frac{y}{x}\right)
\]
- The derivative of \( xy \) with respect to \( x \) is simply \( y \).
- For the second term \( x f\left(\frac{y}{x}\right) \), we need to apply the product rule. First, take the derivative of \( x \) and leave the function alone, then differentiate the function \( f\left(\frac{y}{x}\right) \) using the chain rule.

Let’s differentiate:
\[
\frac{\partial}{\partial x} \left( x f\left( \frac{y}{x} \right) \right) = f\left( \frac{y}{x} \right) + x \cdot \frac{d}{dx} \left( f\left( \frac{y}{x} \right) \right)
\]

For the chain rule, we compute the derivative of \( f\left( \frac{y}{x} \right) \) with respect to \( x \):
\[
\frac{d}{dx} \left( f\left( \frac{y}{x} \right) \right) = f'\left( \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right)
\]
So the total derivative of \( x f\left( \frac{y}{x} \right) \) is:
\[
f\left( \frac{y}{x} \right) - \frac{y}{x} f'\left( \frac{y}{x} \right)
\]
Thus, the full partial derivative with respect to \( x \) is:
\[
\frac{\partial w}{\partial x} = y + f\left( \frac{y}{x} \right) - \frac{y}{x} f'\left( \frac{y}{x} \right)
\]

#### 2. Partial derivative of \( w \) with respect to \( y \):

Now, take the derivative with respect to \( y \):
\[
w(x, y) = xy + x f\left( \frac{y}{x} \right)
\]
- The derivative of \( xy \) with respect to \( y \) is \( x \).
- For the second term \( x f\left( \frac{y}{x} \right) \), we use the chain rule again. The derivative of \( f\left( \frac{y}{x} \right) \) with respect to \( y \) is:
\[
\frac{d}{dy} \left( f\left( \frac{y}{x} \right) \right) = f'\left( \frac{y}{x} \right) \cdot \frac{1}{x}
\]
So the total partial derivative with respect to \( y \) is:
\[
\frac{\partial w}{\partial y} = x + f'\left( \frac{y}{x} \right)
\]

### (ii) Verifying the equation \( x \frac{\partial w}{\partial x} + y \frac{\partial w}{\partial y

Let w(x, y) = xy + xf(y/x) , where f / ℝ -> ℝ is a smooth function. (i) Write down the formulas for partial w partial x and partial w partial y. Your answers will involve f'. (ii) Using part (i) verify that w(x, y) satisfies the equation x partial w partial x + y partial w partial y = xy + w.

This problem explores partial derivatives and the verification of the equation w(x, y) = xy + xf(y/x), involving the use of the chain rule and product rule. The detailed solution walks through the formulas for ∂w/∂x and ∂w/∂y, incorporating f' as the derivative of the smooth function f. Suitable for college-level calculus students.

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