题目
9.求下列函数的全微分:-|||-(1) =ysin (x+y);

题目解答
答案

解析
步骤 1:计算偏导数 $\frac{\partial z}{\partial x}$
根据函数 $z=y\sin (x+y)$,我们首先计算关于 $x$ 的偏导数。使用链式法则,我们得到:
$$
\frac{\partial z}{\partial x} = y\cos(x+y) \cdot \frac{\partial (x+y)}{\partial x} = y\cos(x+y) \cdot 1 = y\cos(x+y)
$$
步骤 2:计算偏导数 $\frac{\partial z}{\partial y}$
接下来,我们计算关于 $y$ 的偏导数。同样使用链式法则,我们得到:
$$
\frac{\partial z}{\partial y} = \sin(x+y) + y\cos(x+y) \cdot \frac{\partial (x+y)}{\partial y} = \sin(x+y) + y\cos(x+y) \cdot 1 = \sin(x+y) + y\cos(x+y)
$$
步骤 3:计算全微分 $dz$
根据全微分的定义,我们有:
$$
dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy
$$
将步骤 1 和步骤 2 中计算的偏导数代入,我们得到:
$$
dz = y\cos(x+y) dx + [\sin(x+y) + y\cos(x+y)] dy
$$
根据函数 $z=y\sin (x+y)$,我们首先计算关于 $x$ 的偏导数。使用链式法则,我们得到:
$$
\frac{\partial z}{\partial x} = y\cos(x+y) \cdot \frac{\partial (x+y)}{\partial x} = y\cos(x+y) \cdot 1 = y\cos(x+y)
$$
步骤 2:计算偏导数 $\frac{\partial z}{\partial y}$
接下来,我们计算关于 $y$ 的偏导数。同样使用链式法则,我们得到:
$$
\frac{\partial z}{\partial y} = \sin(x+y) + y\cos(x+y) \cdot \frac{\partial (x+y)}{\partial y} = \sin(x+y) + y\cos(x+y) \cdot 1 = \sin(x+y) + y\cos(x+y)
$$
步骤 3:计算全微分 $dz$
根据全微分的定义,我们有:
$$
dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy
$$
将步骤 1 和步骤 2 中计算的偏导数代入,我们得到:
$$
dz = y\cos(x+y) dx + [\sin(x+y) + y\cos(x+y)] dy
$$