Choose one correct option.
Given \(w = f(ax+by)\), where \(a\) and \(b\) are constants. The value of
\[
\left(b\frac{\partial w}{\partial x} - a\frac{\partial w}{\partial y}\right)
\]
is:
Show Hint
Let \(u=ax+by\) and write \(w=f(u)\); use the chain rule \(\partial w/\partial x = f'(u)\cdot a\) and \(\partial w/\partial y = f'(u)\cdot b\), then combine.
Step 1: Notice the structure without computing derivatives yet.
Whenever $w$ is a function of a single linear combination $u=ax+by$, both partial derivatives $\partial w/\partial x$ and $\partial w/\partial y$ come from the same underlying rate $f'(u)$, just scaled by different constants. This is the signature of what is called a directional or composite dependence.
Step 2: Write the gradient of $w$.
The gradient vector of $w$ with respect to $x$ and $y$ is
\[ \nabla w = \left(\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}\right) = f'(u)\,(a, b) \]
In words, the gradient of $w$ always points along the fixed direction $(a,b)$, no matter what the function $f$ is, and only its length $f'(u)$ changes.
Step 3: Interpret the expression asked for.
The quantity $b\,\partial w/\partial x - a\,\partial w/\partial y$ is exactly the cross product style combination $b(\nabla w)_x - a(\nabla w)_y$, which measures how much $\nabla w$ points away from the direction $(a,b)$. Since we just showed $\nabla w$ is always parallel to $(a,b)$, this component perpendicular to $(a,b)$ must vanish.
Step 4: Confirm algebraically.
Substituting $\partial w/\partial x = a f'(u)$ and $\partial w/\partial y = b f'(u)$:
\[ b\big(af'(u)\big) - a\big(bf'(u)\big) = abf'(u) - abf'(u) = 0 \]
Step 5: Conclude.
The expression is identically zero for any differentiable $f$.
\[ \boxed{0} \]
This is option (D).
