ln(x^2 y^2)^(1 2)对于X求导
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分子分母同乘以√x^2+1-x再问:哪里来的分子分母?我问的是第一步是怎么来的?再答:把x+√x^2+1看成(x+√x^2+1)/1,分母看成1
分子有理化,分子分母同乘以-x-√(x²-a²)结果是2lna-ln(-x-√(x²-a²))
y'=[1/(1+x^2)]*(1+x^2)'=[1/(1+x^2)]*2x=2x/(1+x^2)
chainruley=f(g(x))y'=g'(x)f'(g(x))
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
1.求导:y=ln(3-2x-x²)dy/dx=(-2-2x)/(3-2x-x²)=-2(1+x)/(1-x)(3+x)2.设y=lncosx,求dy/dx是多少?dy/dx=-s
Y=[LN(1-X)]^2?Y'=2LN|1-X|/(1-X)(-1)=-2LN|1-X|/(1-X)
由y=ln(2-x)定义域:2-x>0,∴x<2,值域:y∈R.
y=ln(x^2+2)是复合函数所以y'=[ln(x^2+2)]'[x^2+2]'=[1/(x^2+2)][2x]=2x/(x^2+2)
y=ln(1-x^2)y'=(1-x^2)'/(1-x^2)=-2x/(1-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=ln[x+a^(-y^2)],以下'表示对y求偏导,z'=[a^(-y^2)]'/[x+a^(-y^2)]=(-y^2)'a^(-y^2)lna/[x+a^(-y^2)],z'=-2ya^(-y^
2x/(1+x^2)
如果是求导数的话,y'=(2x+e^x)/(x^2+e^x)
y'=ln(2x^-1)'=(x/2)*2*(-1)/x^2=-1/x
y′=(3x-2)′/(3x-2)+e^(2x)·(2x)′=3/(3x-2)+2e^(2x).
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
x≤0时√x^2=-x所以y=0x>0时√x^2=x所以y=ln(2x+1)