x^y-y^x=1确定函数y=f(x),求dy dx

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)

第一题,这是个隐函数,两边对x求导得：2y'-1=(1-y')*ln(x-y)+(x-y)*(1-y')/(x-y)=(1-y')*ln(x-y)+(1-y')所以[3+ln(x-y)]y'=ln(x

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]

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

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.

函数f（x）在（-∞，0）上递增；证明：设x1＜x2＜0，则f（x1）-f（x2）=x1-1x1-x2+1x2=（x1-x2）+（1x2-1x1）=（x1-x2）+x1−x2x1x2=(x1−x2)(

d(e^x+e^y)=dyde^x+de^y=dye^xdx+e^ydy=dy(1-e^y)dy=e^xdxdy/dx=e^x/(1-e^y)

y^(1/x)=x^(1/y)就是y^y=x^x两边取对数就是ylny=xlnx两边求一阶倒数就是y'lny+y/y=x'lnx+x/x即y'lny+1=lnx+1就是y'lny=lnx解得y'=ln

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)

.y/x=ty=txy=xtdy/dx=t+t'xdy=(t+t'x)dxy^2(x-y)=x^2t^2(x-tx)=1x=1/[t^2(1-t)]y=1/[t(1-t)]1/y^2=t^2(1-t)

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����

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

e^(x+y)+sin(xy)=1e^(x+y)*(1+y')+cos(xy)(y+xy')=0y'*[e*(x+y)+xcos(xy)]=-[ycos(xy)+e^(x+y)]y'=-[ycos(x

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