Question

y=ln[x+(x^2+a^2)^(1/2)]的导数怎么求.

Answer 1

y'=1/[x+√(x²+a²)]*[x+√(x²+a²)]

=1/[x+√(x²+a²)]*[1+1/2√(x²+a²)*(x²+a²)']

=1/[x+√(x²+a²)]*[1+2x/2[x+√(x²+a²)]

=1/[x+√(x²+a²)]*[x+√(x²+a²)]/√(x²+a²)

=1/√(x²+a²)