作业帮多元微分 多元微分 多元微分
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/23 14:48:15
多元微分 多元微分 多元微分
设z=z(x,y)是由方程x^2+y^2-z=f(x+y+z)所确定的函数,其中f具有2阶导数,求dz
（【请用“两边同时取微分”的方法做这道题】）
设z=z(x,y)是由方程x^2+y^2-z=f(x+y+z)所确定的函数,其中f具有2阶导数,求dz
（【请用“两边同时取微分”的方法做这道题】）
x^2+y^2-z = f(x+y+z),
两边对x求偏导数,得
2x-z'=(1+z')f'(x+y+z), 解得 z' = [2x-f'(x+y+z)]/[1+f'(x+y+z)];
两边对y求偏导数,得
2y-z'=(1+z')f'(x+y+z), 解得 z' = [2y-f'(x+y+z)]/[1+f'(x+y+z)].
则 dz = {[2x-f'(x+y+z)]dx+[2y-f'(x+y+z)]dy} / [1+f'(x+y+z)].
再问： （【请用“两边同时取微分”的方法做这道题】）
再答： x^2+y^2-z = f(x+y+z),
两边对x求微分，得
2xdx-z'dx=(dx+z'dx)f'(x+y+z),
解得 z'dx = [2x-f'(x+y+z)]dx/[1+f'(x+y+z)];
两边对y求微分，得
2ydy-z'dy=(dy+z'dy)f'(x+y+z),
解得 z'dy = [2y-f'(x+y+z)]dy/[1+f'(x+y+z)].
则 dz={[2x-f'(x+y+z)]dx+[2y-f'(x+y+z)]dy}/[1+f'(x+y+z)].
两边对x求偏导数,得
2x-z'=(1+z')f'(x+y+z), 解得 z' = [2x-f'(x+y+z)]/[1+f'(x+y+z)];
两边对y求偏导数,得
2y-z'=(1+z')f'(x+y+z), 解得 z' = [2y-f'(x+y+z)]/[1+f'(x+y+z)].
则 dz = {[2x-f'(x+y+z)]dx+[2y-f'(x+y+z)]dy} / [1+f'(x+y+z)].
再问： （【请用“两边同时取微分”的方法做这道题】）
再答： x^2+y^2-z = f(x+y+z),
两边对x求微分，得
2xdx-z'dx=(dx+z'dx)f'(x+y+z),
解得 z'dx = [2x-f'(x+y+z)]dx/[1+f'(x+y+z)];
两边对y求微分，得
2ydy-z'dy=(dy+z'dy)f'(x+y+z),
解得 z'dy = [2y-f'(x+y+z)]dy/[1+f'(x+y+z)].
则 dz={[2x-f'(x+y+z)]dx+[2y-f'(x+y+z)]dy}/[1+f'(x+y+z)].