求复合函数的导数y=ln[x+√(x^2-a^2)]答案=1/√(x^2-a^2)
问题描述：
求复合函数的导数y=ln[x+√(x^2-a^2)]
答案=1/√(x^2-a^2)
答
y对[x+√(x^2-a^2)]求导=1/[x+√(x^2-a^2)]
[x+√(x^2-a^2)]对x求导=1+√(x^2-a^2)对x求导
=1+x/√(x^2-a^2)=[x+√(x^2-a^2)]/√(x^2-a^2)
由链式法则，y对x的导数=上边两个式子乘积=1/√(x^2-a^2)
答
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