x^y 5^y=6确定了隐函数y=y(x)

最后乘以dy/dx实际上是对Iny中的y求导,因为Iny是复合函数（y是关于x的函数）,所以(Iny）'=1/y*y'=1/y*dy/dx

第一问设yˆ5=k3xˆ4因为x=1时,y=2所以2ˆ5=k3*1ˆ464=12kk=16/3所以函数的表达式为yˆ5=(16/3)3xˆ4

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

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]

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))

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)]

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)

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)(

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))

将x=0代入方程得：lny=1,得y=e方程两边对x求导：y+xy'+e^xlny+y'e^x/y=0代入x=0,y=e得：e+lne+y'/e=0,得y'=-e(e+1)即y'(0)=-e(e+1)

两边求导得e^(2y)-arcsinx=y+xy'解出来y'就可以了再问：为什么是e^(2y)而不是e^(y^2)?再答：因为你的被积分函数是e^(2t)，不是e^(t^2)

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'(

两边对x求导：2y'-1=y'cosy得：y'=1/(2-cosy)因此dy=dx/(2-cosy)

等式两边求导y'=e^y*y'y'=1/(e^y-1)(y'=dy/dx)

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

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)]=[