Question

第三题求导数?

详细解答

Answer 1

解答如下图：

Answer 2

y=√[(lnx)^2 +1]
y'
=(1/2) {1/√[(lnx)^2 +1] } . [(lnx)^2 +1]'
=(1/2) {1/√[(lnx)^2 +1] } . (2lnx) .(lnx)'
=(1/2) {1/√[(lnx)^2 +1] } . (2lnx) .(1/x)
=lnx/ {x.√[(lnx)^2 +1] }