Question

ln根号(1+t^2)的导数是多少?

Answer 1

[ln√(1+t²)]'

=1/√(1+t²)*[√(1+t²)]'

=1/√(1+t²)*1/2√(1+t²)*(1+t²)'

=1/[2(1+t²)]*2t

=t/(1+t²)