设y=f(x)是由方程arctany x=ln根号下x² y²确定的隐函数
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方程两边同时求x对y的导:y+xdy/dx+1/x+2ydy/dx=0,dy/dx=-(y+1/x)/(x+2y),dy=-(y+1/x)dx/(x+2y)
方程两边微分就行了dx*y+x*dy+e^y*dy=2xdx得dy/dx=(2x-y)/(x+e^y)
dz=d[xyP(z)]=yP(z)dx+xP(z)dy+xyP'(z)dz所以dz=[yP(z)dx+xP(z)dy]/[1-xyP'(z)]du=df(x,z)=f'x(x,z)dx+f'z(x,
∂z/∂x=(∂f/∂x)+(∂f/∂y)(dy/dx)//:g(y)+y=xg'(y)y'+y'=1y'=1/[1+g'(y)
y=1+xe^y两边对x求导得y'=e^y+xe^y*y'(是对x求导那么e^y就是一个复合函数了所以最后要在对y求导)(1-xe^y)y'=e^y∴y'=e^y/(1-xe^y)再问:还不是很明白这
两边对x求导:2cos(x^2+y)*(-sin(x^2+y))*(2x+y')=1所以y'=-1/sin(2x^2+2y)-2x再问:求f'(x)```再答:y'就是f'(x)啊。。。。。
两边对x求导有y'e^y=y+xy'整理解得y‘=dy/dx=x/(e^y-x)
e^z-z+xy^3=0偏z/偏x:z'e^z-z'+y^3=0y^3=z'(1-e^z)z'=y^3/(1-e^z)偏z/偏y:z'e^z-z'+3xy^2=0z'=3xy^2/(1-e^z)偏z/
两边对x求导:2yy'f(x)+y^2f'(x)+f(y)+xy'f(y)=2x则y'=[2x-f(y)-y^2f'(x)]/[2yf(x)+xf(y)]再问:给的那个f(x)是x可微函数什么意思再答
这个题目要利用隐函数的求导法则.则sin(x^2+y)=xy(两边同时求导,还要结合复合函数的求导法则)cos(x^2+y)*(2x+y′)=y+xy′2xcos(x^2+y)-y=xy′-y′cos
两边对x求导:y'e^y+(1+y')cos(x+y)=0,1)这里可得到y'=-cos(x+y)/[e^y+cos(x+y)]再对1)求导:y"e^y+(y')^2e^y+y"cos(x+y)-(1
xy+y^2-2x=0y+xy'+2yy'-2=0(x+2y)y'=2-yy'=(2-y)/(x+2y)dy/dx=(2-y)/(x+2y)
两端对x求导数(把y看作x的函数),则1-y'=e^(xy)*(1*y+x*y')y'[xe^(xy)+1]=1-ye^(xy)dy/dx=y'=[1-ye^(xy)]/[xe^(xy)+1]
dz=-dx-dy
两边对x求导得:2yy'*f(x)+y^2f'(x)+f(x)+xf'(x)=2x得:y'=[2x-xf'(x)-y^2f'(x)]/(2yf(x)]dy=[2x-xf'(x)-y^2f'(x)]/(
若z=f(x,y)由方程F(x,y,z)=0确定,则将F(x,y,z)=0两边对x,y求导(x,y视为独立变量,z视为x,y的函数)这个是没有问题的,但此处x,y为两个独立的变量;题1.设y=f(x,
F(x,y)=x^2+y^2-ln(x+2y)Fx=2x-1/(x+2y)Fy=2y-2/(x+2y)F(x)=-Fx/Fy=-[2x(x+2y)-1]/[2y(x+2y)-2]
两边对x求导:1+y'=y'e^y得dy/dx=y'=1/(e^y-1)