Question

函数y=

1nx

x2+1

的导数是______．

Answer 1

因为y=

1nx

x2+1

，
所以y′=

(lnx)′•(x2+1)-lnx•(x2+1)′

(x2+1)2

=

1 x (x2+1)-2xlnx

(x2+1)2

=

x2+1-2x2lnx

x(x2+1)2

．