Question

已知函数f(x)的导函数为f'(x)且满足f(x)=2xf'(1)+lnx则f'(1)=?

Answer 1

f(x)=2xf'(1)+lnx
两边求导得
f'(x)=2f'(1)+1/x
令x=1
则f'(1)=2f'(1)+1
f'(1)=-1

Answer 2

f(x)=2xf'(1)+lnx,,f'(x)=2f'(1)+1/X,于是f'(1)=2f'(1)+1/1，得f'(1)=-1 【注意f'(1)是个常数哦，相当于f（x）=2ax+lnx，f'(1)=a】