Question

[ln(x+√x²+1)]`

如题如图，请问这一步是怎么来的？

Answer 1

{ln[x+√(x^2+1)]}'
={1/[x+√(x^2+1)]}·[x+√(x^2+1)]'
={1/[x+√(x^2+1)]}·[1+1/2(x^2+1)^(-1/2)·(x^2+1)']
={1/[x+√(x^2+1)]}·[1+1/2(x^2+1)^(-1/2)·2x]
={1/[x+√(x^2+1)]}·[1+x·(x^2+1)^(-1/2)]
={1/[x+√(x^2+1)]}·{1+x/[√(x^2+1)]}
={1/[x+√(x^2+1)]}·{[√(x^2+1)+x]/[√(x^2+1)]}
=1/[√(x^2+1)]

Answer 2

=ln[(x+√(x^2+1))(-x+√(x^2+1))]=ln[(x^2+1)-x^2]=ln1 =0 所以f(-x)=-f(x)所以f(x)是奇函数。函数的单调性：设函数f（x）的定义域为D，区间I包含于D。如果对于区间上任意两点x1及x2，当...

Answer 3

=ln[(x+√(x^2+1))(-x+√(x^2+1))]=ln[(x^2+1)-x^2]=ln1 =0 所以f(-x)=-f(x)所以f(x)是奇函数。函数的单调性：设函数f（x）的定义域为D，区间I包含于D。如果对于区间上任意两点x1及x2，当...