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6) Jika \( w = f(ax + by) \), tunjukkan bahwa:  
\[
b \frac{\partial w}{\partial x} - a \frac{\partial w}{\partial y} = 0
\]

This is a problem involving multivariable calculus, specifically dealing with partial derivatives.

### Solution Outline:

1. **Step 1**: Begin with the given function \( w = f(ax + by) \).
   
   The function \( f(ax + by) \) depends on \( x \) and \( y \) through the expression \( ax + by \). This implies \( w \) is indirectly a function of both \( x \) and \( y \), and the chain rule will be needed to find the partial derivatives.

2. **Step 2**: Compute the partial derivative of \( w \) with respect to \( x \):

   Using the chain rule:
   \[
   \frac{\partial w}{\partial x} = \frac{d f}{d z} \cdot \frac{\partial (ax + by)}{\partial x}
   \]
   where \( z = ax + by \), and so:
   \[
   \frac{\partial (ax + by)}{\partial x} = a.
   \]
   Therefore:
   \[
   \frac{\partial w}{\partial x} = f'(ax + by) \cdot a.
   \]

3. **Step 3**: Compute the partial derivative of \( w \) with respect to \( y \):

   Again using the chain rule:
   \[
   \frac{\partial w}{\partial y} = \frac{d f}{d z} \cdot \frac{\partial (ax + by)}{\partial y}.
   \]
   Since \( \frac{\partial (ax + by)}{\partial y} = b \), we get:
   \[
   \frac{\partial w}{\partial y} = f'(ax + by) \cdot b.
   \]

4. **Step 4**: Plug these results into the expression to be proven:

   The expression to prove is:
   \[
   b \frac{\partial w}{\partial x} - a \frac{\partial w}{\partial y} = 0.
   \]

   Substitute the partial derivatives:
   \[
   b \cdot \left( f'(ax + by) \cdot a \right) - a \cdot \left( f'(ax + by) \cdot b \right) = 0.
   \]

5. **Step 5**: Simplify:

   Both terms contain the factor \( f'(ax + by) \), so factor it out:
   \[
   f'(ax + by) \cdot \left( b a - a b \right) = 0.
   \]

   Since \( b a - a b = 0 \), the entire expression simplifies to:
   \[
   0 = 0.
   \]

Thus, the identity is proven.

Would you like more detailed explanations or clarifications? Here are some related questions you might find helpful:

1. How do we apply the chain rule in multivariable calculus?
2. What is the geometric interpretation of the function \( w = f(ax + by) \)?
3. Can you explain the significance of partial derivatives in this context?
4. What other properties can be derived from functions of linear combinations like \( ax + by \)?
5. What are some examples of physical systems where functions of this form are important?

**Tip**: Understanding the chain rule is crucial

Jika w = f(ax + by), tunjukkan bahwa: b ∂w/∂x - a ∂w/∂y = 0

Learn how to prove the identity b ∂w/∂x - a ∂w/∂y = 0 when w = f(ax + by), using partial derivatives and the chain rule in multivariable calculus. This problem is suitable for undergraduate students studying advanced calculus topics.

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