y=1=ln(2x-3)的反函数-搜答案-智慧作业帮

y=ln√[(x+1)(x-2)/(x+3)]y=(1/2)[In(x+1)+In(x-2)-In(x+3)]y'=(1/2)[1/(x+1)+1/(x-2)-1/(x+3)]

y'=[ln(x+√(1+x²))]'=1/(x+√(1+x²))*[x+√(1+x²)]'=1/(x+√(1+x²))*[1+2x/2√(1+x²)

两边相加都是0,没啥意义啊,我有一种方法

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[1/(2x+1)]=-ln(2x+1)y'=-2/(2x+1)若有不懂请追问,如果解决问题请点下面的“选为满意答案”.

表示以e为底的对数函数符号

y=ln[x(2x+1)]=ln(2x^2+x)所以：y'=[1/(2x^2+x)]*(2x^2+x)'=[1/(2x^2+x)]*(4x+1)=(4x+1)/(2x^2+x).如果是：y=lnx*(

y=ln(3-2x-x^2)的导数

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

x-1>0x>1x^2-4x+3>0(x-3)(x-1)>0x>3orx3

1)这两个函数对所有实数有定义；2）ln[-x+根号下(x^2+1)]=ln[1/(x+根号下(x^2+1))]=-ln[x+根号下(x^2+1)]

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=e^c·x^(-1/3)

y=ln(1+x^2)求导

2x/(1+x^2)

y=ln(2x^-1)求导

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

y=ln(x+√(1+x^2))y'=1/[x+√(1+x^2)]*[x+√(1+x^2)]'又∵[x+√(1+x^2)]'=1+(1/2)(1+x²)^(-1/2)*2x=1-x*(1+x

看一下图片够详细了没有~~~

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-