Question

这个函数求二阶导，怎样简洁运算

如图所示

Answer 1

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