Note that we have several functions in #y = x(lnx)^(1/2)#: there is the function #f(x) = x#, which multiplies the composite function #g(x) = (lnx)^(1/2)#. #g(x)# is nothing more thant the composite function of #h(x) = x^(1/2)# and #j(x) = ln(x)#. Therefore, to find the derivative of #y#, we will need to use the product and chain rule.

#dy/(dx) = d/(dx)[x(lnx)^(1/2)]#;

#dy/(dx) = d/(dx)(x) . (lnx)^(1/2) + x.d/dx[(lnx)^(1/2)].#

Since # d/(dx)(x) = 1#, then:

#dy/(dx) = (lnx)^(1/2) + x.d/dx[(lnx)^(1/2)].#

Now, the derivative #d/dx[(lnx)^(1/2)]# is where we apply the chain rule. If we take #u = ln(x)#, then:

#d/dx[(lnx)^(1/2)] = d/(du)(u^(1/2)).(du)/(dx)#;

#d/dx[(lnx)^(1/2)] = 1/2 u^(-1/2).(1/x)#.

Now we return to our original variable, #x#, since we know the relation between #u# and #x#:

#d/dx[(lnx)^(1/2)] = 1/2 [ln(x)]^(-1/2).(1/x)#.

Then:

#dy/(dx) = (lnx)^(1/2) + x.1/2 [ln(x)]^(-1/2).(1/x)#;

#dy/(dx) = (lnx)^(1/2) + 1/2 [ln(x)]^(-1/2)#.

If you want to, you can also rewrite this expression:

#dy/(dx) = sqrt(ln(x)) + 1/(2sqrt(ln(x))#.

#dy/(dx) = (2ln(x) + 1)/(2sqrt(ln(x))#.