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)).
