高数强化NO9|多元函数导数的计算|数值型|抽象型|隐函数

多元函数导数的计算

数值型

$$\begin{array}{rl}& \text{解：}\frac{\partial f}{\partial x}=y{e}^y,\frac{\partial f}{\partial y}=x(1+y)e^y\\ & f(x,y)=\int \frac{\partial f}{\partial x}dx=\int y{e}^{y}dx=xy{e}^y+C(y)\\ & \\ & \frac{\partial f}{\partial y}=x\left(e^y+y{e}^y\right)+C^{\prime }(y)=x(1+y)e^y⟹C^{\prime }(y)=0,C(y)=C.\\ & \\ & \text{由}f(0,0)=0,\text{得}C=0,\text{故}f(x,y)=xy{e}^y.\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{对二元函数求不定积分的时候，思路实际上和求偏导数的时候是一致的：对一}\\ & \text{个变量求积分时，另一个变量是看成常数的。}\end{array}$$

抽象型

$$\begin{array}{rl}& \text{解：}\frac{\partial g}{\partial x}=y-\left(f_1^{\prime }+f_2^{\prime }\right),\frac{\partial g}{\partial y}=x-\left(f_1^{\prime }-f_2^{\prime }\right),\\ & \frac{{\partial }^{2}g}{\partial x^2}=-\left(f_{11}^{\prime \prime }+f_{12}^{\prime \prime }\right)-\left(f_{21}^{\prime \prime }+f_{22}^{\prime \prime }\right)\\ & \frac{{\partial }^{2}g}{\partial x\partial y}=1-\left(f_{11}^{\prime \prime }-f_{12}^{\prime \prime }+f_{21}^{\prime \prime }-f_{22}^{\prime \prime }\right),\\ & \frac{{\partial }^{2}g}{\partial y^2}=-\left(f_{11}^{\prime \prime }-f_{12}^{\prime \prime }\right)+\left(f_{21}^{\prime \prime }-f_{22}^{\prime \prime }\right)\\ & \\ & \frac{{\partial }^{2}g}{\partial x^2}+\frac{{\partial }^{2}g}{\partial x\partial y}+\frac{{\partial }^{2}g}{\partial y^2}=1-3f_{11}^{\prime \prime }-f_{22}^{\prime \prime }\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{这类题目由于变量比较多，所以先梳理变量关系；计算的过程中，注意“同路相}\\ & \text{乘、异路相加”。}\end{array}$$

$$\begin{array}{rl}& \text{解：}\frac{\partial u}{\partial x}=\frac{\partial v}{\partial x}\cdot e^{ax+by}+v\cdot a{e}^{ax+by},\frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}{e}^{ax+by}+v\cdot b{e}^{ax+by}\\ & \frac{{\partial }^{2}u}{\partial x^2}=\frac{{\partial }^{2}v}{\partial x^2}{e}^{ax+by}+2\frac{\partial v}{\partial x}\cdot a{e}^{ax+by}+v\cdot a^2{e}^{ax+by}\\ & \frac{{\partial }^{2}u}{\partial y^2}=\frac{{\partial }^{2}v}{\partial y^2}{e}^{ax+by}+2\frac{\partial v}{\partial y}\cdot b{e}^{ax+by}+v\cdot b^2{e}^{ax+by}\\ & \\ & 2\frac{{\partial }^{2}u}{\partial x^2}-2\frac{{\partial }^{2}u}{\partial y^2}+3\frac{\partial u}{\partial x}+3\frac{\partial u}{\partial y}\\ & =e^{ax+by}\left(2\frac{{\partial }^{2}v}{\partial x^2}+4a\frac{\partial v}{\partial x}+2a^{2}v-2\frac{{\partial }^{2}v}{\partial y^2}-4b\frac{\partial v}{\partial y}-2b^{2}v+3\frac{\partial v}{\partial x}+3av+3\frac{\partial v}{\partial y}+3bv\right)\\ & =e^{ax+by}\left(2\frac{{\partial }^{2}v}{\partial x^2}-2\frac{{\partial }^{2}v}{\partial y^2}+(4a+3)\frac{\partial v}{\partial x}+(3-4b)\frac{\partial v}{\partial y}+2a^{2}v-2b^{2}v+3av+3bv\right)\\ & \\ & \text{取}\{\begin{array}{l}4a+3=0\\ 3-4b=0\end{array}⟹\{\begin{array}{l}a=-\frac{3}{4}\\ b=\frac{3}{4}\end{array}\end{array}$$
$$【小结】本题首先要能够理解题目的意图,读懂题目的本意是什么,而且这一类题目的计算量也往往是比较大.$$

隐函数求导

$$\begin{array}{rl}& \text{解：将}x=0,y=1\text{代入}{e}^z+xyz+x+\cos x=2\text{，得}z=0.\\ & \text{两边对}x\text{求偏导：}\\ & e^z\cdot \frac{\partial z}{\partial x}+y\left(z+x\frac{\partial z}{\partial x}\right)+1-\sin x=0①\\ & \text{两边对}y\text{求偏导：}\\ & e^z\cdot \frac{\partial z}{\partial y}+x\left(z+y\frac{\partial z}{\partial y}\right)=0②\\ & \\ & \text{将}x=0,y=1,z=0\text{代入}①,②\text{可得：}\\ & {\frac{\partial z}{\partial x}|}_{(0,1)}=-1,{\frac{\partial z}{\partial y}|}_{(0,1)}=0.\\ & \\ & \text{故}dz{|}_{(0,1)}=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy=-dx+0\cdot dy=-dx\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1、\text{理解全微分的基本定义，即设二元函数}z=z(x,y)\text{在}(x_0,y_0)\text{点可微，则}\\ & {dz|}_{(x_0,y_0)}=z_x^{\prime }(x_0,y_0)dx+z_y^{\prime }(x_0,y_0)dy.\\ & \\ & 2、\text{多元隐函数求偏导数的方法与一元隐函数求导数类似，可结合隐函数求导学习和理解.}\end{array}$$

$$\begin{array}{rl}& \text{令}F(x,y,z)=xy-z\ln y+e^{xz}-1\\ & F_x^{\prime }=y+z{e}^{xz},F_x^{\prime }(0,1,1)=2\ne 0⟹\text{可以确定}x=x(y,z)\\ & F_y^{\prime }=x-\frac{z}{y},F_y^{\prime }(0,1,1)=1\ne 0⟹\text{可以确定}y=y(x,z)\\ & F_z^{\prime }=-\ln y+x{e}^{xz},F_z^{\prime }(0,1,1)=0=0\\ & 选D\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{定理内容实质上也不用强记，重点在于理解. 由隐函数存在定理可知}{z}_x^{\prime }=-\frac{F_x^{\prime }}{F_z^{\prime }},\\ & \text{自然能够想到分母不等于零，故而要求}{F}_z^{\prime }\ne 0.\text{同理，可讨论隐函数}x=x(y,z)\text{和}\\ & y=y(x,z).\end{array}$$