已知方程y=sin(y x)确定了y=y(x)求dy
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方程x2+y2-4x+1=0表示以点(2,0)为圆心,以3为半径的圆.设yx=k,即y=kx,由圆心(2,0)到y=kx的距离为半径时直线与圆相切,斜率取得最大、最小值,由|2k−0|k2+1=3,解
2x+yx2-2xy+y2•(x-y)=2x+y(x-y)2•(x-y)(2分)=2x+yx-y;(4分)当x-3y=0时,x=3y;(6分)原式=6y+y3y-y=7y2y=72.(8分)
y=sin(x+y),y'=cos(x+y)*(1+y'),y'=cos(x+y)/(1-cos(x+y))=dy/dx
答案是(ycosxy-1)/(1-xcosxy).亲、加油哦.
两边求导得:cos(xy)*(y+xy')+1+y'=0y'[xcos(xy)+1]=-ycos(xy)-1所以,y'=-[ycos(xy)+1]/[xcos(xy)+1]
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))
y'=cos(x+y)(1+y')y'=cos(x+y)/(1-cos(x+y))
e^y=sin(x+y)两边求导得e^y*y'=cos(x+y)(x+y)'=cos(x+y)(1+y')=cos(x+y)+y'cos(x+y)[e^y-cos(x+y)]y'=cos(x+y)y'
y=sin(x+y).两边对x求导得:y’=cos(x+y)(1+y')y'=cos(x+y)/(1-cos(x+y))所以:dy=[cos(x+y)/(1-cos(x+y))]dx再问:y'=cos
dy=dsin(x+y)dy=cos(x+y)d(x+y)dy=cos(x+y)(dx+dy)dy=cos(x+y)dx+cos(x+y)dy所以dy/dx=cos(x+y)/[1-cos(x+y)]
对x求导2cos(x+2y-3z)乘以(1-3Fx)=1+3Fx对y求导2cos(x+2y-3z)乘以(2-3Fy)=2+3Fy整理可得,再问:juti具体点吧咯咯咯再答:隐函数求导,Fx就是Z对x求
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将方程x2+y2-4x+1=0化简得,(x-2)2+y2=3,∴方程表示以点(2,0)为圆心,以r=3为半径的圆yx+1表示两点(x,y),(-1,0)的斜率设k=yx+1,即kx-y+k=0当直线与
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代入y=5/4原方程可化为:x/4+5/4=5/4(x-2)等号两端同乘以4x+5=5x-104x=15x=15/4再问:若三分之x+2=x则x-()=2再答:移项,x-x/3=2所以()中填x/3
dy/dx=-fx/fy,你自己可以算吧
化为:e^(ylnx)-e^y=sin(xy)两边对x求导:e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')y'[lnxe^(ylnx)-e^y-xcos(xy)]=[