复合函数微分法
求法
设\(u=\varphi \left( x,y \right) ,\,\,v=\psi \left( x,y \right)\)在点\(\left( x,y \right)\)处有对x及对v的偏导数,函数\(z=f\left( u,v \right)\)在对应点\(\left( u,v \right)\)处有连续偏导数,则\(z=f\left[ \varphi \left( x,y \right) ,\,\,\psi \left( x,y \right) \right]\)在点\(\left( x,y \right)\)处的两个偏导数存在,且有
\[\frac{\text{d}z}{\text{d}x}=\frac{\partial z}{\partial u}\frac{\text{d}u}{\text{d}x}+\frac{\partial z}{\partial v}\frac{\text{d}v}{\text{d}x} \] \[\frac{\text{d}z}{\text{d}y}=\frac{\partial z}{\partial u}\frac{\text{d}u}{\text{d}y}+\frac{\partial z}{\partial v}\frac{\text{d}v}{\text{d}y} \]这个公式不能死记硬背,需要用树形图来推导
全微分形式的不变性
设函数\(z=f\left( u,v \right),u=\varphi \left( x,y \right) ,\,\,v=\psi \left( x,y \right)\)都有连续的一阶偏导数,则复合函数\(z=f\left[ \varphi \left( x,y \right) ,\,\,\psi \left( x,y \right) \right]\)在点\(\left( x,y \right)\)的全微分
\[\text{d}z=\frac{\partial z}{\partial u}\text{d}u+\frac{\partial z}{\partial v}\text{d}v=\frac{\partial z}{\partial x}\text{d}x+\frac{\partial z}{\partial y}\text{d}y \]隐函数微分法
由方程F=(x,y)确定的隐函数y=y(x)
\[\frac{\text{d}y}{\text{d}x}=-\frac{F_{x}^{}}{F_{y}^{}} \]由方程F=(x,y,z)确定的隐函数z=z(x,y)
隐函数存在定理:若\(F\left( x,y,z \right)\)在点\(\left( x_0,y_0,z_0 \right)\)的某一领域内有连续偏导数,且\(F\left( x,y,z \right)=0\),\(F_{z}^{'}\left( x,y,z \right) \ne 0\),则方程\(F\left( x,y,z \right)=0\)在点\(\left( x_0,y_0,z_0 \right)\)的某领域可唯一确定一个有连续偏导数的z=z(x,y),并有
\[\frac{\partial z}{\partial x}=-\frac{F_x}{F_z} \] \[\frac{\partial z}{\partial y}=-\frac{F_y}{F_z} \]隐函数存在定理最重要的是打高亮部分的,题设是对z的偏导数,若对x的偏导数为不等于0的话,则是x关于y和z的函数,对y类似