经济数学y=1除以x-2 ln(4-x)-搜答案-搜狗做题网

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

y'=1-2x/(1+x²)=(1+x²-2x)/(1+x²)=(x-1)²/(1+x²)显然y'>0所以y单调增加

分子有理化,分子分母同乘以-x-√(x²-a²)结果是2lna-ln(-x-√(x²-a²))

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

由题意得：y的方程式可以写为y=[1/根号下（1-x）]*[ln根号下(1+x)]求导y'=[1/根号下（1-x）]'*[ln根号下(1+x)]+[1/根号下（1-x）]*[ln根号下(1+x)]=[

两边相加都是0,没啥意义啊,我有一种方法

y=ln(1+x^2),求y

y'=[1/(1+x^2)]*(1+x^2)'=[1/(1+x^2)]*2x=2x/(1+x^2)

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=(1+x)⁻¹ln(1-x)y'=(-1)(1+x)⁻²ln(1-x)+(1+x)⁻¹(1-x)⁻¹(-1)

ln(x-y)/ln(x+y)=ln(x-y-x-y)=ln(-2y)

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(1-x^2) 求y''

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

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(2x^-1)求导

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

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）