求由方程cos(xy)=x^2y^2所确定的函数的微分
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隐函数求导设z=x²y²-cos(xy)dy/dx=-(δz/δx)/(δz/δy)=-(2xy²+ysin(xy))/(2x²y+xsin(xy))=-y/x
cos(xy)=x+y两边微分,得dx+dy-sin(xy)*(x*dy+y*dx)=0dx(1-ysin(xy))+dy(1-xsin(xy))=0dy/dx=(ysin(xy)-1)/(1-xsi
-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy-2x^2*ydy-sin(xy)xdy=2xy^
cos(xy)-x^2·y=1两边对x求导-sin(xy)*(y+xy')-2xy-x^2y'=0===>x=1,y=0,y'=0-cos(xy)(y+xy')^2-(y'+y'+xy")-2y-2x
两边对x求导:-(y+xy')sin(xy)=2xy^2+2x^2yy'解得:y'=-[ysin(xy)+2xy^2]/[2x^2y+xsin(xy)]所以dy=-[ysin(xy)+2xy^2]/[
令F(x,y)=cos(xy)-x-yF'(x,y)x=-ysin(xy)-1对x求偏导F'(x,y)y=-xsin(xy)-1对y求偏导切线方程为:(x-0)/F'(x,y)=(y-1)/F'(x,
ysinx=cos(xy)两边分别求导y'sinx+ycosx=-sin(xy)(y+xy')y'=-y(sin(xy)+cosx)/(sinx+xsin(xy))
对方程两边同时求导得,﹣﹙y+xy′﹚sin﹙xy﹚+e^y+﹙x+1﹚y′e^y=0令x=0则方程cos(xy)+(x+1)*e^y=2为1+e^y=2,得y=0,即切点坐标为﹙0,0﹚将﹙0,0﹚
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
直接求导,用xy表示导数【欢迎追问,
=-[ysin(xy)+2e^(2x+y)]/[ysin(xy)+e^(2x+y)]*(dx)再问:麻烦给我写出解的过程。。再答:等式两边取对数,得:d[e^(2x+y)]-d[cos(xy)]=0(
因为x、y都为自变量,不是宗量,故此题没有全微分,应只有偏微分.详解如下:对方程两边微分:左边:de^z=e^z*dz右边d[xyz+cos(xy)]=xydz+yzdx+xzdy-(sinxy)*(
网上有很多高数课后习题答案,你可以下载一个参考~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,原式
你指的是ezplot函数吧.ezplot('sin(x^2+932/1000*y^2)=cos(x*y)')
对两边取对数:xy+3lny=lncos(x-y)两边同时对x求导:y+xy'+y'*3/y=-tan(x-y)*(1-y')整理得:y'=tan(x-y)+y/tan(x-y)-x-3/y不知道对不
x=0时,代入方程得:1+1=y,得:y=2对x求导:(y+xy')e^xy-sin(xy)*(y+xy')=y'将x=0,y=2代入得:2=y'故dy(0)=2dx
dy/dx=-fx/fy,你自己可以算吧
两边对x求导xy^2+sinx=e^yy^2+2xyy'+cosx=e^y*y'y'(e^y-2xy)=y^2+cosxy'=(y^2+cosx)/(e^y-2xy)
在方程ex+y+cos(xy)=0左右两边同时对x求导,得:ex+y(1+y′)-sin(xy)•(y+xy′)=0,化简求得:y′=dydx=ysin(xy)−ex+yex+y−xsin(xy).