y=ln(x sqrt(4 x^2))的反函数-搜答案-智慧作业帮

设y=ln ln ln x,求y’

y'=(lnlnx)'/lnlnx=(lnx)'/lnxlnlnx=1/xlnxlnlnx

求导y=ln ln ln(x^2+1)

y=ln(1-x^2)

chainruley=f(g(x))y'=g'(x)f'(g(x))

y=ln[ln(ln x)] 求导

复合函数f(x)=lnxg(x)=ln[ln(x)]r(x)=ln{lnln(x)]}r'(x)=[1/lnln(x)]g'(x)=[1/lnln(x)][1/ln(x)]f'(x)=[1/lnln(

y=ln^2x求导

1/x再问：求写一下过程拍照再答：再问：不是是ln二次方x再答：再答：懂了么再答：再问：懂了再答：别忘了采纳最佳答案

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

y=ln^2(1-x)求导

Y=[LN(1-X)]^2?Y'=2LN|1-X|/(1-X)(-1)=-2LN|1-X|/(1-X)

y=ln(2-x) 值域

由y=ln（2-x）定义域：2-x＞0,∴x＜2,值域：y∈R.

求导 y=ln(-x)

答：y=ln(-x)y'(x)=[1/(-x)]*(-1)=1/x所以：y=ln(-x)的导数为y'(x)=1/x再问：非常感谢，，，那y=ln（3x+2）的导和y=ln（4x+2）的导分别怎么算呢？

求导 y=ln(tan(x/2))

y'=1/(tan(x/2))*(tan(x/2))'=1/(tan(x/2))*(sec^2(x/2))*(x/2)'=1/(2sin(x/2)*cos(x/2))=1/sin(x)=csc(x)

z可以分为两部分f(x)=√(4x-y^2)和g(x)=ln(x+y-1)z=f(x)+g(x)分布求定义域再求交集4x-y^2≥0x+y-1>0y^2≤4x且x>1-y

1)恩对,考虑错了,a

1,y=ln(1-x)y'=1/(1-x)*(1-x)'=1/(1-x)*(-1)=1/(x-1);2,y=ln[1/√(1-x)]=-ln√(1-x)y'=-1/√(1-x)*[√(1-x)]'=-

y=ln(1+x^2)求导

2x/(1+x^2)

y=ln(x^2+e^x) 求Y'X

如果是求导数的话,y'=(2x+e^x)/(x^2+e^x)

y=ln(2x^-1)求导

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

y=ln(x^2+sinx)求导

复合函数求导,应用链式法则y'=dy/dx=[dy/d(x^2+sinx)]*[d(x^2+sinx)/dx]=[1/(x^2+sinx)]*(2x+cosx)故y'=(2x+cosx)/(x^2+s

y=ln(3x-1)3x-1>0x>1/3即定义域为(1/3,+∞)y=log2/(4x-3)2/(4x-3)>04x-3>0x>3/4即定义域为(3/4,+∞)

y=ln(x+√x^2+1),求y

x≤0时√x^2=-x所以y=0x>0时√x^2=x所以y=ln（2x+1）