The text in the image is written in Indonesian. It seems to be asking for the proof of a partial differential equation.

The relevant part from the image reads:

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

Translation:

6) If $w = f(ax + by)$, show that: $b \frac{\partial w}{\partial x} - a \frac{\partial w}{\partial y} = 0$

Solution:

Let $u = ax + by$. Then $w = f(u)$.

Step 1: Compute $\frac{\partial w}{\partial x}$

Since $w = f(u)$ and $u = ax + by$, we apply the chain rule: $\frac{\partial w}{\partial x} = \frac{df}{du} \cdot \frac{\partial u}{\partial x}$ We have: $\frac{\partial u}{\partial x} = a$ Thus: $\frac{\partial w}{\partial x} = a \frac{df}{du}$

Step 2: Compute $\frac{\partial w}{\partial y}$

Similarly, applying the chain rule: $\frac{\partial w}{\partial y} = \frac{df}{du} \cdot \frac{\partial u}{\partial y}$ We have: $\frac{\partial u}{\partial y} = b$ Thus: $\frac{\partial w}{\partial y} = b \frac{df}{du}$

Step 3: Substitute into the original expression

We need to show: $b \frac{\partial w}{\partial x} - a \frac{\partial w}{\partial y} = 0$ Substituting the values: $b \cdot a \frac{df}{du} - a \cdot b \frac{df}{du} = 0$ This simplifies to: $ab \frac{df}{du} - ab \frac{df}{du} = 0$ Thus, the equation is satisfied, and the statement is proven.

Would you like further details or clarifications?

Here are some related questions to explore further:

1. What is the significance of the chain rule in multivariable calculus?
2. Can you generalize this proof for a more complex function?
3. How does the choice of $w = f(ax + by)$ affect the form of partial derivatives?
4. What happens if $w$ is not a function of a linear combination of $x$ and $y$?
5. How would this work if you used a different function for $w$, such as trigonometric or exponential functions?

Tip: In problems like this, look for symmetry in expressions when applying partial derivatives to simplify the proof!