1. (5.0分) 设函数f(x,y)=x^2+2xy+y^2，则f(x,y)在点(1,1)处的全微分为?A. df=4dx+4dyB. df=2dx+2dyC. df=dx+dyD. df=3dx+3dy

本题考查二元函数全微分的计算。解题思路是先求出函数$f(x,y)$对$x$和$y$的偏导数，再将点$(1,1)$代入偏导数中，最后根据全微分公式$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$计算出函数在该点的全微分。

步骤一：求$f(x,y)$对$x$的偏导数$\frac{\partial f}{\partial x}$

根据求导公式$(X^n)^\prime=nX^{n - 1}$，对$f(x,y)=x^{2}+2xy+y^{2}$关于$x$求偏导数，此时把$y$看作常数，则：
$\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(x^{2}+2xy+y^{2})=\frac{\partial}{\partial x}(x^{2})+\frac{\partial}{\partial x}(2xy)+\frac{\partial}{\partial x}(y^{2})$
因为$\frac{\partial}{\partial x}(x^{2}) = 2x$，$\frac{\partial}{\partial x}(2xy)=2y$（$y$为常数），$\frac{\partial}{\partial x}(y^{2}) = 0$（$y$为常数），所以$\frac{\partial f}{\partial x}=2x + 2y$。

步骤二：求$f(x,y)$对$y$的偏导数$\frac{\partial f}{\partial y}$

同样根据求导公式，对$f(x,y)=x^{2}+2xy+y^{2}$关于$y$求偏导数，此时把$x$看作常数，则：
$\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}(x^{2}+2xy+y^{2})=\frac{\partial}{\partial y}(x^{2})+\frac{\partial}{\partial y}(2xy)+\frac{\partial}{\partial y}(y^{2})$
因为$\frac{\partial}{\partial y}(x^{2}) = 0$（$x$为常数），$\frac{\partial}{\partial y}(2xy)=2x$（$x$为常数），$\frac{\partial}{\partial y}(y^{2}) = 2y$，所以$\frac{\partial f}{\partial y}=2x + 2y$。

步骤三：将点$(1,1)$代入偏导数中

把$x = 1$，$y = 1$代入$\frac{\partial f}{\partial x}=2x + 2y$，可得$\frac{\partial f}{\partial x}\big|_{(1,1)} = 2\times1 + 2\times1 = 4$。
把$x = 1$，$y = 1$代入$\frac{\partial f}{\partial y}=2x + 2y$，可得$\frac{\partial f}{\partial y}\big|_{(1,1)} = 2\times1 + 2\times1 = 4$。

步骤四：计算全微分$df$

根据全微分公式$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$，将$\frac{\partial f}{\partial x}\big|_{(1,1)} = 4$，$\frac{\partial f}{\partial y}\big|_{(1,1)} = 4$代入可得：
$df\big|_{(1,1)} = 4dx + 4dy$