设z=e的2x-y次方-搜答案-拍题作业网

df(x,y,z)/dx=[d(z^2)/dx]*y*e^x+y*z^2*(de^x/dx)=2zye^x(dz/dx)+y*z^2*e^x另,由x+y+z+xyz=0求dz/dx两边对x求偏导1+0

(太麻烦拉,给点分啊!)设v=x*x-y*y,u=exp{xy}那么dv/dx=2x(这里应该用偏导符号,代替一下),dv/dy=2y,du/dx=y*exp{xy},du/dy=x*exp{xy}那

z'x=2e^(2x+y)z'y=e^(2x+y)所以dz=2e^(2x+y)dx+e^(2x+y)dy

e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,e^y-e^0=0,则e^y=1,则y=0所以y'(

题目应是：x^2+y^2+z^2=y*e^z吧记F=x^2+y^2+z^2-y*e^z,则F'=2x,F'=2y-e^z,F'=2z-y*e^z,则z'=-F'/F'=2x/(y*e^z-2z),z'

(y^2+2xy-cos(y+z))/(e^z+cos(y+z))再问：没有过程吗？再答：求导：e^z*dz-y^2-2xy+cos(y+z)(1+dz)=0把含有dz的项移到一起：(e^z+cos(

两端对x求偏导得：-ye^(-xy)-2(z/x)+(z/x)e^z=0,所以,z/x=ye^(-xy)/(e^z-2)两端对y求偏导得：-xe^(-xy)-2(z/y)+(z/y)e^z=0,所以,

全微分,分别对x和y求微分,再综合起来

你想说这个问题?z=e^(x^2+2xy)应该是y=e^(x^2+2xy)(2x+2y）i+e^(x^2+2xy)2xj

z=arctan(x*e^x)z'={1/[1+(x*e^x)^2]}*(x*e^x)'(x*e^x)'=x'*e^x+x*(e^x)'=e^x+x*e^x=(x+1)*e^x所以dz/dx=(x+1

e^z=xyz两边对x求偏导e^z*z'(x)=y(z+x*z'(x))z'(x)=yz/(e^z-xy)∂z/∂x=yz/(e^z-xy)原式对y求偏导e^z*z'(y)=x

1e^z=xyze^zz'x=yz+xyz'xz'x=yz/(xy-e^z)=yz/(xy-xyz)=z/(x-xz)类似z'y=z/(y-yz)dz=[z/(x-xz)]dx+[z/(y-yz)]d

再问：非常感谢，还要问大侠一道题面目。曲线y=x³+3x的拐点坐标为？？？再答：y'=3x²+3y''=3x令y"=0，得x=3当x=3时，y=36所以拐点坐标（3，36）

z^x=y^z,隐函数F(x,y,z)=z^x-y^zFx=lnz*z^xFy=-zy^(z-1)Fz=x*z^(x-1)-lny*y^zбz/бx=-Fx/Fz=-(lnz*z^x)/(x*z^(x

1.3的x次方=4的y次方=6的z次方xlg3=ylg4=zlg6设：k=xlg3=ylg4=zlg6则：1/x=lg3/k1/y=lg4/k=2lg2/k1/z=lg6/k所以：1/z-1/x=(l

x+2y+z=e^(x-y-z)两边对x求偏导注意到z=z(x,y)1+z'=e^(x-y-z)*(1-z')...(1)再对x求偏导z"=e^(x-y-z)(1-z')^2-z"e^(x-y-z).

your answer here

e^(-xy)-x^2*y+e^z=z,令F(x,y,z)=e^(-xy)-x^2*y+e^z-z=0分别对F取x,y,z的偏导数,可得əF/əx=e^(-xy)*(-y)-2xy