求y=ln^22x的微分-搜答案-智慧作业帮

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

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)

求x=y^y的微分

如果对x求导,则ln|x|=yln|y|,1/x=y'/y+yy'/y=y'/y+y',.对数求导法.如果对y求导,则ln|x|=yln|y|,x'/x=ln|y|+y/y,x'=y^y(1+ln|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)=

y=(ln(ln(x))'/ln(ln(x))=(ln(x))'/(ln(x)(ln(ln(x)))=1/(xln(x)ln(ln(x)))

分解为y=5^u,u=lnv,v=tanx,用复合函数的求导法则；dy/dx=dy/du×du/dv×dv/dx用求导公式,dy/du=5^u×ln5,du/dv=1/v,dv/dx=(secx)^2

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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