先将x+√(1+x^2)看做整体,求ln的导数,再乘以x+√(1+x^2)的导数,而x+√(1+x^2)的导数为x导数加√(1 x^2)的导数,所以
ln(x+√(1+x²))'
=[1/(x+√(1+x²))]*[x+√(1+x²)]'
=[1/(x+√(1+x²))]*[x'+[(1+x²)½]']
=[1/(x+√(1+x²))]*[1+[(1/2)(1+x²)^(-1/2)]*(1+x²)']
=[1/(x+√(1+x²))]*[1+[(1/2)(1+x²)^(-1/2)]*2x]
=x / {[x+√(1+x²)]*[(√(1+x²)]}
最终结果分子为x,分母为[x+√(1+x²)]*[(√(1+x²)]
不知道那个二次方你手机能否显示,我没有用^2,