Question

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:

• A. \(-a\)
• B. \(b\)
• C. \(b-a\)
• D. \(0\)

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.

Answer 1

The Correct Option is D

Step 1: Set up the composite function.
Let \(u = ax+by\), so \(w = f(u)\) is a function of the single variable \(u\), where \(u\) itself depends on both \(x\) and \(y\). To find \(\partial w/\partial x\) and \(\partial w/\partial y\), we need the chain rule for partial derivatives.

Step 2: Apply the chain rule for \(\partial w/\partial x\).
Since \(w\) depends on \(x\) only through \(u\),
\[ \frac{\partial w}{\partial x} = \frac{dw}{du}\cdot\frac{\partial u}{\partial x} = f'(u)\cdot a \] because \(\partial u/\partial x = a\) (treating \(y\), and hence \(by\), as constant).

Step 3: Apply the chain rule for \(\partial w/\partial y\).
Similarly,
\[ \frac{\partial w}{\partial y} = \frac{dw}{du}\cdot\frac{\partial u}{\partial y} = f'(u)\cdot b \] because \(\partial u/\partial y = b\).

Step 4: Substitute both derivatives into the required expression.
\[ b\frac{\partial w}{\partial x} - a\frac{\partial w}{\partial y} = b\big(a\,f'(u)\big) - a\big(b\,f'(u)\big) \] \[ = ab\,f'(u) - ab\,f'(u) = 0 \] The two terms are identical in size and opposite in sign, so they cancel completely, regardless of what the function \(f\) actually is.

Step 5: Sanity check with a concrete function.
Take \(f(u)=u^2\), so \(w=(ax+by)^2\). Then
\[ \frac{\partial w}{\partial x} = 2a(ax+by), \qquad \frac{\partial w}{\partial y} = 2b(ax+by) \] So
\[ b\frac{\partial w}{\partial x} - a\frac{\partial w}{\partial y} = 2ab(ax+by) - 2ab(ax+by) = 0 \] This confirms the general result.

Step 6: Final conclusion.
\[ \boxed{b\frac{\partial w}{\partial x} - a\frac{\partial w}{\partial y} = 0} \]
Hence, the correct option is (D).