Discussion
Let \(f(u,v) = u \ln(v)\) and \(F(x,y) = f(u(x,y), v(x,y))\), where \(u = x/y\) and \(v = x - y\). Then \(\partial F/\partial y\) is
Answer: B
The derivative is \(-\dfrac{x}{y^{2}} \ln v - \dfrac{u}{v}\).
By the chain rule, with \(\partial u/\partial y = -x/y^{2}\) and \(\partial v/\partial y = -1\):
\[ \begin{aligned} \frac{\partial F}{\partial y} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} \\ &= \ln v \left(-\frac{x}{y^{2}}\right) + \frac{u}{v}(-1) \\ &= -\frac{x}{y^{2}}\ln v - \frac{u}{v} \end{aligned} \]
The first option has the right first term but a plus sign on the second, which drops the -1 from \(\partial v/\partial y\). The third and fourth lose the minus sign from differentiating x/y with respect to y.
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