y=ln根号x 的微分
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两边对x求导得1/[1+(y/x)^2]*(y/x)'=1/[ln(x^2+y^2)]*[ln(x^2+y^2)]'1/[1+(y/x)^2]*(y'x-y)/x^2=1/[2ln(x^2+y^2)]
两边相加都是0,没啥意义啊,我有一种方法
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)
dz=[-3ysin3xy+1/(1+x+y)]dx+[-3xsin3xy+1/(1+x+y)]dy
z偏x=-sin3xy*3y+1/(x+y+1)z偏y=-sin3xy*3x+1/(x+y+1)dz=[-sin3xy*3y+1/(x+y+1)]dx+[sin3xy*3x+1/(x+y+1)]dy
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=In√x=In(x)^1/2=1/2*Inxx=e^(2y).反函数为y=e^(2x)没理解错吧?
z=arctanx/y+ln√(x^2+y^2)编微分的符号打不出来,只有用d代替了dz/dx=1/(1+(x/y)^2)*1/y+1/√(x^2+y^2)*1/2√(x^2+y^2)*2x=y/(x
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=1/2ln(1+sinx)-1/2ln(1-sinx)y'=1/2×1/(1+sinx)×cosx-1/2×1/(1-sinx)×(-cosx)=1/2×[cosx/(1+sinx)+cosx/(
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-
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】