问题及解答

(4)  $z=\frac{2x-y}{x+2y}$

(5)  $u=x(x+y^2+z^3)$

(4)

\[z=\frac{2x-y}{x+2y}\]

\[
\frac{\partial z}{\partial x}=\frac{2\cdot(x+2y)-(2x-y)\cdot 1}{(x+2y)^2}=\frac{5y}{(x+2y)^2}
\]

\[
\frac{\partial z}{\partial y}=\frac{-1\cdot(x+2y)-(2x-y)\cdot 2}{(x+2y)^2}=\frac{-5x}{(x+2y)^2}
\]

因此, $z$ 的全微分是

\[
\mathrm{d}x=\frac{5y}{(x+2y)^2}\mathrm{d}x+\frac{-5x}{(x+2y)^2}\mathrm{d}y=\frac{5y\mathrm{d}x-5x\mathrm{d}x}{(x+2y)^2}
\]

(5)

\[
\begin{aligned}
\frac{\partial u}{\partial x}&=1\cdot(x+y^2+z^3)+x\cdot 1=2x+y^2+z^3\\
\frac{\partial u}{\partial y}&=x\cdot 2y=2xy\\
\frac{\partial u}{\partial z}&=x\cdot 3z^2=3xz^2
\end{aligned}
\]

因此,

\[
\mathrm{d}u=\frac{\partial u}{\partial x}\mathrm{d}x+\frac{\partial u}{\partial y}\mathrm{d}y+\frac{\partial u}{\partial z}\mathrm{d}z=(2x+y^2+z^3)\mathrm{d}x+2xy\mathrm{d}y+3xz^2\mathrm{d}z.
\]