哈哈
把x看成常量
1/n看作x1,1/(n+1)看作x2
n^2 [x^(1/n)-x^(1/n+1)]=[(n+1)/n][1/n-1/(n+1)]*[x^(1/n)-x^(1/n+1)]
[1/n-1/(n+1)]*[x^(1/n)-x^(1/n+1)]为a^x在0处的导数
即得答案是lnx
哈哈
把x看成常量
1/n看作x1,1/(n+1)看作x2
n^2 [x^(1/n)-x^(1/n+1)]=[(n+1)/n][1/n-1/(n+1)]*[x^(1/n)-x^(1/n+1)]
[1/n-1/(n+1)]*[x^(1/n)-x^(1/n+1)]为a^x在0处的导数
即得答案是lnx