**It is somewhat difficult to read the question as written. Thus, to prevent confusion, it will be listed that the answer provided below will help to find f'(x) if #f(x) = 2xsqrtx - 3sqrtx +1/sqrtx#. **

For the sake of simplicity, one will wish to recall that #x = x^1, sqrt x = x^(1/2)#, #1/x = x^(-1)# and thus that #1/sqrtx = x^(-1/2)#. Further, when multiplying together two expressions which have the same base but different exponents, one simply adds the exponents. (For example, #x^1 x^(1/2) = x^((1/2 + 1)) =x^(3/2)#). Thus...

#f(x) = 2xsqrtx - 3sqrtx +1/sqrtx = 2x^(3/2) - 3x^(1/2) +x^(-1/2)#.

Via use of the power rule (stating that, for #g(x) = x^n, g'(x) = nx^(n-1)#)...

#f'(x) = (3/2)2x^(1/2) - (1/2)3x^(-1/2) - (1/2)x^(-3/2) = 3x^(1/2) -3/2x^(-1/2) -1/2x^(-3/2)#

Returning to the initial notation, which utilized #sqrt x#...

#f'(x) = 3sqrtx -3/(2sqrtx) - 1/(2xsqrtx) #