Question

求复合函数的导数y=ln[x+√(x^2+a^2)]

x^2+a^2都在根号内
答案=1/√(x^2-a^2).同样的,x^2-a^2都在根号内
他舅，请教：
你的第一步根号里面为什么都是+号，题目是－号呀，怎么变成+号的呢？
y'={1/[x+√(x²+a²)]}*[x+√(x²+a²)]'

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√(x²+a²)]
={1/[x+√(x²+a²)]}*{[x+√(x²+a²)]/√(x²+a²)}
=1/√(x²+a²)