方程y=xliny确定y是x的函数,求y的导-搜答案-灵鹊做题网

xy+e^y=y+1（1）求d^2y/dx^2在x=0处的值：（1）两边分别对x求导：y+xy'+e^yy'=y'y/y'+x+e^y=1(2)(2)两边对x再求导一次：(y'y'-yy'')/y'^

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

1）y|x=o当x=0时sin(0)-1/y-0=1得：y|x=0=-1(2)y'|x=osin(xy)-1/y-x=1两边对x求导：cos(xy)(y+xy')+y'/y^2-1=0当x=0时y=-

y=sin(x+y),y'=cos(x+y)*(1+y'),y'=cos(x+y)/(1-cos(x+y))=dy/dx

两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)

这个题目要用到微分的形式不变性e^y*dy+d(xy)=0e^y*dy+xdy+ydx=0-ydx=(x+e^y)dydy=-y*dx/(x+e^y)

两边对x求导得y+xy'=(1+y')/(x+y)y(x+y)+x(x+y)y'=1+y'y'[x(x+y)-1]=1-y(x+y)y'=[1-y(x+y)]/[x(x+y)-1]dy=[1-y(x+

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)]

ln(x²+y²)=x+y-1两边对x求导得：(2x+2yy')/(x²+y²)=1+y'整理得：y'=(2x-x²-y²)/(x²

看不懂求什么?用一下ln吧,不知道你要干什么,可能有用

隐函数微分法两边同时取自然对数sinx*lny=y*lnsinx,两边同时对x的微商cosxlny+(y'sinx)/y=y'lnsinx+ycosx/sinx写不清楚了,格式太多,整

y'=-2sin2(x+y)-2y'sin2(x+y)(1+2sin2(x+y))y'=-2sin2(x+y)y'=-2sin2(x+y)/(1+2sin2(x+y))

两边求导e^y×y'=xy'+yy'=y/(e^y-x)dy/dx=y/(e^y-x)

再问：你是用对数求导法么？第一步右边那个怎么写得出来的再答：

方程y=sin(x+y)两边对x求导数有：y'=cos(x+y)(x+y)'=cos(x+y)(1+y')移项整理得：[1-cos(x+y)]y'=cos(x+y)因此：y'=cos(x+y)/[1-

把它看成关于y的一元二次方程,整理得x²y²＋y－1＝0解得y＝－1＋√（1＋4x²）／2x²＞0或者y＝－1－√（1＋4x²）／2x²＜0

主要利用复合函数的求导：z=f(y),y=g(x),则z对x求导dz/dx=f'(y)*(dy/dx).等式左边对x求导过程：d(lny)/dx=(1/y)y',等式右边对x求导过程：d(x-y)/d

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]