求xcosy=sin(x y)的导数-搜答案-搜狗做题网

将y看成是关于x的函数即y=f(x)我们在求导的同时要记得y也要对x求导即dy/dx我们两边分别对x求导得e^x+e^y*dy/dx=cos(xy)*(y+x*dy/dx)移项e^x-y*cos(xy

3f(x)+f(-1/x)=2x-x(1)令x=-1/x则3f(-1/x)+f(x)=2/x+1/x(2)(1)×3-(2)8f(x)=6x-3x-2/x+1/x所以f(x)

等式两边对x求偏导,cosy+z'(x)*(-sinxy)*y=0,z'(x)=cosy/y*sinxyz''(xy)=-(cosy/y)*(1/(sinxy)^2)*cosxy*y原式两边对y求偏导

一阶dz/dx=ycosxydz/dy=xcosxy二阶d^2z/dx^2=y^2cosxyd^2z/dy^2=x^2cosxy还有混合导数相等就写一个了=cosxy-xcosy

再答：隐函数高阶求导。再答：

e^(xy)+sin(xy)=y(y+xy')e^(xy)+(y+xy')cos(xy)=y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))

将原方程两边微分得d[xe^y+sin(xy)]=0→e^ydx+xe^ydy+cos(xy)(ydx+xdy)=0→移项[xe^y+xcos(xy)]dy=-[e^y+ycos(xy)]dx整理→d

如下图：

方程两边对变量x求导有d[sin(xy)]/dx=dx/dxcos(xy)*d(xy)/dx=1cos(xy)*(dx*y+x*dy)/dx=1cos(xy)*[y+x*(dy/dx)]=1所以：dy

(sinydx+xcosydy)+(y^2sinxdx-2ycosx)dy=0[sinydx+xd(siny)]+[y^2d(-cosx)-cosx(dy^2)]=0d(xsiny)+d(-y^2co

cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]

Zx=ycos(xy)-2ycos(xy)sin(xy)=ycos(xy)-ysin(2xy)Zy=xcos(xy)-xsin(2xy)

dy/dx=1/(xcosy+sin2y)=1/(xcosy+2sinycosy)所以cosydy/dx=1/(x+2siny)所以dsiny/dx=1/(x+2siny)所以dx/dsiny=x+2

y+xy'-cos(πy²)2πyy'=0y=[2πycos(πy²)-x]y'y'=y/[2πycos(πy²)-x]即：dy/dx=y/[2πycos(πy²

sin(xy)=x 求dx/dy

x/[sec(xy)-y]dx/dy.

答案是y'+cosy-xsiny*y'=2x

补上线段y=0则令P=e^xsiny-y,dP/dy=e^xcosy-1Q=e^xcosy-1,dQ/dx=e^xcosy∫_L(e^xsiny-y)dx+(e^xcosy-1)dy=∫∫_D[(e^

dx/dy=xcosy+sin2yx'-cosyx=sin2yx的一阶微分方程注意是x=x(y)两边同乘e^(-siny)[e^(-siny)*x]'=sin2y*e^(-siny)e^(-siny)

两边对X求导得：cosy-x(siny)y'=cos(x+y)(1+y')化得：y'=[cosy-cos(x+y)]/[cos(x+y)+xsiny]再问：没搞懂啊！确信这个对吗？我都好久不学这个了，

设dz=(2siny)dx+(2xcosy+1)dy那么∂z/∂x=2siny于是：z=2xsiny+g(y)∂z/∂y=2xcosy+g'(y),而已