17.设z=e^u sin v，而u=xy，v=x+y，(partial z)/(partial x)=（ ）A. e^xy[y sin(x+y)+cos(x+y)]B. e^xy[x sin(x+y)-cos(x+y)]C. e^xy[y sin(x+y)-cos(x+y)]D. e^xy[x sin(x+y)+cos(x+y)]

本题考查复合函数求偏导数的知识。解题思路是利用复合函数求偏导的链式法则，先分别求出$z$对$u$、$v$的偏导数，以及$u$、$v$对$x$的偏导数，再根据链式法则计算$\frac{\partial z}{\partial x}$。

步骤一：求$\frac{\partial z}{\partial u}$和$\frac{\partial z}{\partial v}$

已知$z = e^u \sin v$，对$u$求偏导数时，将$v$看作常数，根据求导公式$(e^x)^\prime=e^x$可得：
$\frac{\partial z}{\partial u}=\frac{\partial (e^u \sin v)}{\partial u}=\sin v\cdot\frac{\partial (e^u)}{\partial u}=\sin v\cdot e^u = e^u \sin v$
对$v$求偏导数时，将$u$看作常数，根据求导公式$(\sin x)^\prime=\cos x$可得：
$\frac{\partial z}{\partial v}=\frac{\partial (e^u \sin v)}{\partial v}=e^u\cdot\frac{\partial (\sin v)}{\partial v}=e^u \cos v$

步骤二：求$\frac{\partial u}{\partial x}$和$\frac{\partial v}{\partial x}$

已知$u = xy$，对$x$求偏导数时，将$y$看作常数，根据求导公式$(x^n)^\prime=nx^{n - 1}$可得：
$\frac{\partial u}{\partial x}=\frac{\partial (xy)}{\partial x}=y\cdot\frac{\partial (x)}{\partial x}=y$
已知$v = x + y$，对$x$求偏导数时，将$y$看作常数，可得：
$\frac{\partial v}{\partial x}=\frac{\partial (x + y)}{\partial x}=\frac{\partial (x)}{\partial x}+\frac{\partial (y)}{\partial x}=1 + 0 = 1$

步骤三：根据链式法则求$\frac{\partial z}{\partial x}$

根据复合函数求偏导的链式法则$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x}$，将上面所求的偏导数代入可得：
$\begin{align*}\frac{\partial z}{\partial x}&=e^u \sin v\cdot y + e^u \cos v\cdot 1\\&=e^u(y \sin v + \cos v)\end{align*}$
再将$u = xy$，$v = x + y$代回上式，得到：
$\frac{\partial z}{\partial x}=e^{xy}[y \sin(x + y) + \cos(x + y)]$