题目

求复合函数的导数y=ln[x+√(x^2-a^2)]
答案=1/√(x^2-a^2)

提问时间：2020-07-24

答案

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)
这其实和+号是一样的