y = lg2x,则dy=

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y = lg2x,则dy=
设y=x+lnx,则dx/dy=

dx/dy=x'(y)y=x+lnx,对等号两边求导∴1=x'+x'/x∴x'(1+1/x)=1∴dx/dy=x'=x/(x+1)

设y=f(-x^2) 则dy=

--2xdx,负2xdx再问:过程呢再答:这是个非常简单的微分运算,如果你是自学高等数学你可以上网搜‘’‘’‘’微分;;;;过程就是根据求导法则,

微积分.若x^y=y^x,则dy/dx.

两边取对数得:ylnx=xlny两边对x求导得:(dy/dx)lnx+y/x=lny+(x/y)dy/dx解得:dy/dx=[x^2-xylnx]/[y^2-xylny]

(x^2)dy+(y^2)dx=dx-dy

(x^2+1)dy=(1-y^2)dxdy/(1-y)(1+y)=dx/(x^2+1)1/2lnl(y-1)/(y+1)l=arctanx+c再问:在帮我一个,我给再加五分,y′=y,y(0)=1.谢

设函数y=xe^y,则dy/dx=?

x=y*e^(-y)故dx/dy=e^(-y)+y*(-e^(-y))=(1-y)*e^(-y)故dy/dx=e^y/(1-y)再问:是吧dy/dx看成分数的是吧?

微分方程(x+y)(dx-dy)=dx+dy的通解

两边同除以dx,整理后得到dy/dx=(x+y-1)/(x+y+1),然后转化一下,d(x+y)/dx=2(x+y)/(x+y+1).设u=x+y,得到du/dx=2u/(u+1).以下略.结果:x-

y=ln(sinx)求y",dy

y=ln(sinx)y'=cosx/sinx=cotxy''=-1/sin²x∴y''=-1/sin²xdy=cotxdx

为什么y'=dy/dx?

是定义的.

y=[sin(x^4)]^2,则dy/dx=?,dy^2/dx^2=?,dy/d(x^2)=?

dy/dx=2sin(x^4)cos(x^4)*4x^3复合函数求导dy^2/dx^2=[8x^3sin(x^4)cos(x^4)]^2dy/d(x^2)=2sin(x^4)cos(x^4)*2x^2

dy/dx-y=cosx

y'-y=cosx为一阶线性微分方程通解为y=C*e^[∫-P(x)dx]+e^[∫-P(x)dx]*∫e^[∫P(x)dx]*q(x)dx=Ce^x+e^x*∫cosx*e^(-x)dx①其中:∫e

已知x>0,y>0,lg2x+lg8y=lg2,则1x+13y的最小值是(  )

∵lg2x+lg8y=lg2,∴lg(2x•8y)=lg2,∴2x+3y=2,∴x+3y=1.∵x>0,y>0,∴1x+13y=(x+3y)(1x+13y)=2+3yx+x3y≥2+23yx•x3y=

设y=tanx 则dy=

求导即可因为(tanx)'=sec^2x所以dy=sec^2xdx

y=1/x^2,则dy

y=x^-2dy=-2x^-3dx=-2/x^3dx

dx= dy/(y+k)

x=ln(y+k)+c或者y=e^(x-c)-k其中c是任意常数

∫y √{y+5} dy=?

求不定积分∫[y√(y+5)]dy令√(y+5)=u,则y+5=u²,y=u²-5,dy=2udu,代入原式得:原式=2∫[(u²-5)u²du=2∫(u

y=x*e^y,则dy=

y=x*e^y,则:y'=e^y+x*e^y*y',所以:y'=e^y/(1-xe^y)=e^y/(1-y)所以:dy={e^y/(1-y)}dx

dy/dx=x+y

线性一阶微分方程,公式解:利用积分因子法,可得到积分因子为:e^(-x)结果为:y=C*e^x-(x+1)C为任意常数