设可导函数y=y(x)由方程
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函数y=arctane^x求dyy'=e^x/(1+e^2x)dy=e^xdx/(1+e^2x)函数y=y(x)由方程x-y-e^y=0确定,求y'(0)两边对x求导:1-y'-y'e^y=0y'=1
两边对【x】求导,注意,y是x的函数,利用复合函数求导1/[1+(y/x)^2]×(y/x)'=1/2×1/(x^2+y^2)×(x^2+y^2)',也就是:x^2/(x^2+y^2)×(xy'-y)
lny+x/y=0等式两边求导:y'*1/y+1/y+x*y'(-1/y²)=0(1/y-x/y²)y'=-1/y∴y'=(-1/y)/(1/y-x/y²)=-y/(y-
由隐函数微分法可得:-sin(x+y)(1+y′)+y′=0-sin(x+y)+[1-sin(x+y)]y′=0∴y′=sin(x+y)/[1-sin(x+y)].
直接在等式中零,x=0,y=y(0),可得关于y(0)的方程解出y(0)即可.具体:e^0*y(0)+lny(0)/1=0即-y(0)=lny(0)作图y1=-x,y2=ln(x),两者的交点的横坐标
两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)
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'=cos(x+y)(1+y')y'=cos(x+y)/(1-cos(x+y))
取对数xlny=ylnx求导lny+x*1/y*y'=y'*lnx+y*1/x(x/y-lnx)y'=y/x-lny所以dy/dx=(y/x-lny)/(x/y-lnx)
分别对y求导,求左边为1+【e^(x+y)×(dx/dy+1)】右边为2×dx/dy推的dx/dy:自己算下,没得草稿纸.
y'=-2sin2(x+y)-2y'sin2(x+y)(1+2sin2(x+y))y'=-2sin2(x+y)y'=-2sin2(x+y)/(1+2sin2(x+y))
xy+e^y=1e^y(0)=1y(0)=0xy'+y+e^yy'=00+y(0)+y'(0)=0y'(0)=0xy''+y'+y'+e^yy''+(y')^2e^y=00+2y'(0)+y''(0)
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,e^y-e^0=0,则e^y=1,则y=0所以y'(
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-
ln(x+y)=x·lny(1+y‘)/(x+y)=lny+x/y·y‘y+y·y‘=y(x+y)lny+x(x+y)·y‘y‘=【y(x+x)lny-y】/【y-x(x+y)】再问:лл����
两边对x求导数,得y'*e^y+y+xy'=0,在原方程中令x=0可得y=1,因此,将x=0,y=1代入上式可得y'+1=0,即y'(0)=-1.再问:对x求导时y可以当成一个常数吗?为什么要用公式(
/>e^y+xy+e^x=0两边同时对x求导得:e^y·y'+y+xy'+e^x=0得y'=-(y+e^x)/(x+e^y)y''=-[(y'+e^x)(x+e^y)-(y+e^x)(1+e^y·y'
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)
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)]=[