Question

3.求由方程cosz+e^x-xy=0+所确定的二元隐函数+z=f(x,y)+的偏导数+a/a+,a/a+(

Answer 1

问：3.求由方程cosz+e^x-xy=0+所确定的二元隐函数+z=f(x,y)+的偏导数+a/a+,a/a+(

答：亲，你好根据隐函数定理，可以得到$\frac{\partial z}{\partial x}=\frac{dy/dx}{-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}+y}$，同理可得$\frac{\partial z}{\partial y}=\frac{dx/dy}{-\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z}-x}$。首先需要确定$f(x,y)$，将$x$看作自变量，$cosz+e^x-xy=0$中的$z$可以看作$y$的函数，所以$f(x,y)=cos^{-1}(-e^x+xy)$。$$\frac{\partial f}{\partial x}=\frac{y}{\sqrt{1-(-e^x+xy)^2}},\quad \frac{\partial f}{\partial y}=\frac{x}{\sqrt{1-(-e^x+xy)^2}}$$$$\frac{\partial z}{\partial x}=\frac{dy/dx}{\frac{y}{\sqrt{1-(-e^x+xy)^2}}+y},\quad \frac{\partial z}{\partial y}=\frac{dx/dy}{\frac{x}{\sqrt{1-(-e^x+xy)^2}}-x}$$其中，$$\frac{dy}{dx}=\frac{\frac{\partial}{\partial x}(cosz+e^x-xy)}{\frac{\partial}{\partial y}(cosz+e^x-xy)}=\frac{x}{\sqrt{1-cos^2z}}=\frac{x}{\sqrt{1-(-e^x+xy)^2}}$$$$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{\sqrt{1-(-e^x+xy)^2}}{x}$$因此，$$\frac{\partial z}{\partial x}=\frac{y}{1-(-e^x+xy)^2}+y,\quad \frac{\partial z}{\partial y}=\frac{\sqrt{1-(-e^x+xy)^2}}{x}-\frac{xy

问：啊 能手写发图片嘛

问：看不懂

答：将方程化简得：cosz = xy - e^x对于z=f(x,y)，可以对x、y同时求偏导数：∂z/∂x = -e^x - y*sin(z) ∂z/∂y = -x*sin(z)然后对∂z/∂x再求偏导数：a/∂(∂z/∂x)/∂x = -e^x - y*cos(z)*∂z/∂x = -e^x - y*cos(z)*(-e^x - y*sin(z)) = -e^x + y*cos(z)*e^x + y^2*cos(z)*sin(z)最后对a/∂(∂z/∂x)/∂y求偏导数：a/∂(∂z/∂x)/∂y = -cos(z)*∂z/∂y = x*cos(z)*sin(z)

答：亲，你好，老师这边没有笔和纸，解题过程已经发给你，你看最后一个