ln(1 e^2)的微分
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ln(1/e^2)=log(e)1-1og(e)e^2=0-2*log(e)e=0-2*1=-2分析:ln是以e为底的对数又因为真数相除等于对数相减即真数1除以e的平方等于log以e为底1的对数减去l
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
(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
1、y=2x/(x+1),求y'(0)y'=2/(x+1)-2x/(x+1)^2=2/(x+1)^2y'(0)=22、y=ln(x^2+3),求dyy'=1/(x^2+3)*2x=2x/(x^2+3)
(e^e^x)'=(e^e^x)*(e^x)'=(e^e^x)*(e^x)(ln3(x+1)^2)'=1/3(x+1)^2*(3(x+1)^2)'=(1/3(x+1)^2)*(6(x+1))=2/(x
全微分,分别对x和y求微分,再综合起来
dz=[-3ysin3xy+1/(1+x+y)]dx+[-3xsin3xy+1/(1+x+y)]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
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-
(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
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
1.e^(e^x+x)2.2/(x+1)3.-2/(x^2-1)都是复合函数求导再问:可以给我一下过程么。。
对等式两边求全微分du=【1/(2x+3y+4z^2)】【2dx+3dy+8zdz】
两个都是求不出来的,只能求近似值.这是我用计算器算的,都逃不开这个Li2函数.12那个ln(1-e^(-kx))的积分,也是求不出来的.我是用级数来求得.因为对于|x-1|<1, ln