问题及解答

设 $z=xy+yf(\frac{x}{y})$, 其中 $f(u)$ 为可导函数, 证明: $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=xy+z$.

\[
\frac{\partial z}{\partial x}=y+yf'(\frac{x}{y})\cdot\frac{1}{y}=y+f'(\frac{x}{y})
\]

\[
\frac{\partial z}{\partial y}=x+1\cdot f(\frac{x}{y})+yf'(\frac{x}{y})\cdot\frac{-x}{y^2}=x+f(\frac{x}{y})-\frac{x}{y}f'(\frac{x}{y})
\]

于是

\[
\begin{split}
x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}&=x\Bigl(y+f'(\frac{x}{y})\Bigr)+y\Bigl(x+f(\frac{x}{y})-\frac{x}{y}f'(\frac{x}{y})\Bigr)\\
&=xy+xf'(\frac{x}{y})+xy+yf(\frac{x}{y})-xf'(\frac{x}{y})\\
&=xy+xy+yf(\frac{x}{y})\\
&=xy+z.
\end{split}
\]