f(x^2)的导数等于1/x（其中x>0）且f(1)=2,则f(x)=?
问题描述：
f(x^2)的导数等于1/x（其中x>0）且f(1)=2,则f(x)=?
请写下详细解答过程，谢谢！
可是答案是：2X^(1/2)
答
d/dx {f(x^2) } = 1/x
2x f'(x^2) = 1/x
f'(x^2) = 1/(2x^2)
f'(x) = 1/(2x)
f(x) = (1/2) lnx + C
f(1) = C=2
f(x) = (1/2)lnx + 2
那是
f'(x^2) = 1/x
f'(x) = x^(-1/2)
f(x) = 2x^(1/2) + C
f(1) = 2 + C =2 =>C = 0
f(x) = 2x^(1/2)