y=ln(x∧2-3x)
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y'=(lnlnx)'/lnlnx=(lnx)'/lnxlnlnx=1/xlnxlnlnx
y=x^5+ln^3xy'=(x^5)’+(ln^3x)‘=5x^4+3(lnx)²/X
chainruley=f(g(x))y'=g'(x)f'(g(x))
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(
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'=3(lnx)^2·(lnx)'=3/x·(lnx)^2y''=-3/x^2·(lnx)^2+3/x·2lnx·1/x=3/x^2·lnx·(2-lnx)再问:再问:能不能发图给我再答:
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'=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)
y=xln³x所以y'=x'*ln³x+x*(ln³x)'=1*ln³x+x*3ln²x*(lnx)'=ln³x+x*3ln²x*
y=e^c·x^(-1/3)
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=ln(3^x)=xln3所以y'=ln3
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)