求微分y＝ln²(1-x)-搜答案-搜狗做题网

Z=（1/2）ln(1+x²+y²)dz=(1/2)2x/(1+x²+y²)dx+(1/2)2y/(1+x²+y²)dy=x/(1+x&su

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

(1)y=3x^2-ln1/x=3x^2+lnxdy=6xdx+(1/x)dx=(6x+1/x)dx(2)y=e^(-x)cosxdy=-e^(-x)cosxdx-e^(-x)sinxdx=-e^(-

-((2x)/(1-x^2))dx;(-E^-x-Sin[3+x])dx;2Cos[2x]dx

等式两边同时求导得:2y*y'+y'/y=4*x^3-->y'=4y*x^3/(2y^2+1)y'=dy/dx-->dy=y'*dx=dx*4y*x^3/(2y^2+1)

F(x,y)=x+lny-y=0dF(x,y)=0=(∂F(x,y)dx/∂x)+(∂F(x,y)dy/∂y)dy/dx=-(∂F(x,y)

symsx>>y=log(x+sqrt(1+x^2));>>simple(diff(y)ans=1/(1+x^2)^(1/2)>>y=log(2*x+sqrt(1+x^2));>>simple(dif

y＝ln[x+√(1+x²)]∴y'＝[x+√(1+x²)]'/[x+√(1+x²)]＝[1+x/√(1+x²)]/[x+√(1+x²)]＝[x+√(

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这是复合函数的求导：y=√u,u=lnv,v=3x^2则y'=1/(2√u)*u'=1/(2√u)*1/v*v'=1/(2√u)*1/v*6x=1/(2√u)*1/(3x^2)*6x=1/(x√u)=

dy/dt=2t/(1+t²)dx/dt=1-[1/(1+t²)]=t²/(1+t²)dy/dx=(dy/dt)/(dx/dt)=2/t

解y=ln²(1-2x)y'=dy/dx=[ln²(1-2x)]'=2ln(1-2x)[ln(1-2x)]'(1-2x)'=2ln(1-2x)[1/(1-2x)(-2)=[-4ln

y=[ln(1-x)^2]^2y'=2[ln(1-x)^2]*[ln(1-x)^2]'=2[ln(1-x)^2]*[2ln(1-x)]'=2[ln(1-x)^2]*2*1/(1-x)=4*[ln(1-

(e^-x)=-e^(-x)arcsinx^2=1/√(1-x^4)*(x²)'=2x/√(1-x^4)ln(sinx)=1/sinx*cosx=cotx所以dy=[-(e^-x)arcsi

dy=dx/(√(1+x^2))不好意思,我没办法将过程打出来

u'x=2x/(x^2+y^2+z^2)u'y=2y/(x^2+y^2+z^2)u'z=2z/(x^2+y^2+z^2)du=2xdx/(x^2+y^2+z^2)+2ydy/(x^2+y^2+z^2)

z=1/2*ln(x^2+y^2+4)Z'x=1/2*1/(x^2+y^2+4)*(2x)=x/(x^2+y^2+4)Z'y=1/2*1/(x^2+y^2+4)*(2y)=y/(x^2+y^2+4)所

对等式两边求全微分du=【1/(2x+3y+4z^2)】【2dx+3dy+8zdz】

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