求一阶容易,求二阶更容易：y = ln[x+√(a+x)] dy/dx = 1/[x+√(a+x)] * [1+2x / 2√(a+x)] =1/[x+√(a+x)] * [√(a+x)+x]/√(a+x) =1/√(a+x) dy/dx = [-2x / 2√(a+x)] / (a+x),依照d/dx 1/f(x) = -f'(x)/[f(x)] =-x/(a+x)^(3/2)
y=ln[x+√(a^2+x^2)],求导数y',和二阶导数y''
求一阶容易,求二阶更容易：y = ln[x+√(a+x)] dy/dx = 1/[x+√(a+x)] * [1+2x / 2√(a+x)] =1/[x+√(a+x)] * [√(a+x)+x]/√(a+x) =1/√(a+x) dy/dx = [-2x / 2√(a+x)] / (a+x),依照d/dx 1/f(x) = -f'(x)/[f(x)] =-x/(a+x)^(3/2)