Question

求导：f(X)=ln(根号下x^2+1)

Answer 1

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