求复合函数的导数y=ln[x+√(x^2+a^2)]

2025-01-19 08:23:51
推荐回答(1个)
回答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)