题目
设 =(e)^xy+ln ((x)^2+(y)^2) 则（ ）A =(e)^xy+ln ((x)^2+(y)^2)B =(e)^xy+ln ((x)^2+(y)^2)C =(e)^xy+ln ((x)^2+(y)^2)D =(e)^xy+ln ((x)^2+(y)^2)
设 
则（ ）
A 
B 
C 
D 
题目解答
答案
根据复合函数求导法则，设
在
处可导，
在对应点处可导，则复合函数
在处可导，且
，在求 关于的导数时，将
视为常量：
，在求 关于的导数时，将视为常量：
，故答案为C
解析
步骤 1：求 $\dfrac {\partial z}{\partial x}$
根据复合函数求导法则，设$u=\varphi (x)$在处可导，$=f(u)$在对应点处可导，则复合函数$[ (x)d] f=f$在处可导，且$\dfrac {dy}{dx}=\dfrac {dy}{du}\times \dfrac {du}{dx}=f'(u)\varphi '(x)$，在求 $y={e}^{xy}+\ln ({x}^{2}+{y}^{2})$关于的导数时，将视为常量：
$\dfrac {\partial z}{\partial x}=\dfrac {\partial [ {e}^{xy}+\ln ({x}^{2}+{y}^{2})] }{\partial x}$ $=\dfrac {\partial {e}^{xy}}{\partial x}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial x}$ $=\dfrac {\partial {e}^{xy}}{\partial (xy)}\times \dfrac {\partial xy}{\partial x}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial ({x}^{2}+{y}^{2})}\times \dfrac {\partial ({x}^{2}+{y}^{2})}{\partial x}$ $={e}^{xy}\times y+\dfrac {1}{({x}^{2}+{y}^{2})}\times 2x$ $=y{e}^{xy}+\dfrac {2x}{({x}^{2}+{y}^{2})}$
步骤 2：求 $\dfrac {\partial z}{\partial y}$
在求 关于的导数时，将视为常量：
$\dfrac {\partial z}{\partial y}=\dfrac {\partial [ {e}^{xy}+\ln ({x}^{2}+{y}^{2})] }{\partial y}$ $=\dfrac {\partial {e}^{xy}}{\partial y}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial y}$ $=\dfrac {\partial {e}^{xy}}{\partial (xy)}\times \dfrac {\partial xy}{\partial y}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial ({x}^{2}+{y}^{2})}\times \dfrac {\partial ({x}^{2}+{y}^{2})}{\partial y}$ $={e}^{xy}\times x+\dfrac {1}{({x}^{2}+{y}^{2})}\times 2y$ $=x{e}^{xy}+\dfrac {2y}{({x}^{2}+{y}^{2})}$
根据复合函数求导法则，设$u=\varphi (x)$在处可导，$=f(u)$在对应点处可导，则复合函数$[ (x)d] f=f$在处可导，且$\dfrac {dy}{dx}=\dfrac {dy}{du}\times \dfrac {du}{dx}=f'(u)\varphi '(x)$，在求 $y={e}^{xy}+\ln ({x}^{2}+{y}^{2})$关于的导数时，将视为常量：
$\dfrac {\partial z}{\partial x}=\dfrac {\partial [ {e}^{xy}+\ln ({x}^{2}+{y}^{2})] }{\partial x}$ $=\dfrac {\partial {e}^{xy}}{\partial x}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial x}$ $=\dfrac {\partial {e}^{xy}}{\partial (xy)}\times \dfrac {\partial xy}{\partial x}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial ({x}^{2}+{y}^{2})}\times \dfrac {\partial ({x}^{2}+{y}^{2})}{\partial x}$ $={e}^{xy}\times y+\dfrac {1}{({x}^{2}+{y}^{2})}\times 2x$ $=y{e}^{xy}+\dfrac {2x}{({x}^{2}+{y}^{2})}$
步骤 2：求 $\dfrac {\partial z}{\partial y}$
在求 关于的导数时，将视为常量：
$\dfrac {\partial z}{\partial y}=\dfrac {\partial [ {e}^{xy}+\ln ({x}^{2}+{y}^{2})] }{\partial y}$ $=\dfrac {\partial {e}^{xy}}{\partial y}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial y}$ $=\dfrac {\partial {e}^{xy}}{\partial (xy)}\times \dfrac {\partial xy}{\partial y}+\dfrac {\partial \ln ({x}^{2}+{y}^{2})}{\partial ({x}^{2}+{y}^{2})}\times \dfrac {\partial ({x}^{2}+{y}^{2})}{\partial y}$ $={e}^{xy}\times x+\dfrac {1}{({x}^{2}+{y}^{2})}\times 2y$ $=x{e}^{xy}+\dfrac {2y}{({x}^{2}+{y}^{2})}$