Question

求y=ln[x+根号下（1+x的平方）]的二阶导数

Answer 1

y'=1/[x+√(1+x²)]*[x+√(1+x²)]'

=1/[x+√(1+x²)]*[1+2x/2√(1+x²)]

=1/[x+√(1+x²)]*[x+√(1+x²)]/√(1+x²)

=1/√(1+x²)

=(1+x²)^(-1/2)

所以y''=(-1/2)(1+x²)^(-3/2)*(1+x²)'

=-1/[(1+x²)√(1+x²)]