Calculus help!!!! Whats the first derivative of the function [x^5/2]lnx.

For what x-values does it have horizontal tangents?

Product rule:

(5/2)(x^3/2)lnx + (x^5/2)/x

= (5/2)(x^3/2)lnx + x^3/2

Forizontal tangent when f' = 0

0 = 5/2 (x^3/2) lnx + x^3/2

0 = x^3/2 ((5/2)lnx + 1)

By zero property law,
x^3/2 = 0
so x = 0
But 0 is not in domain

and (5/2)lnx + 1 = 0
5/2 lnx = -1
lnx = 2/5

x = e^2/5

y = [x^5/2]lnx
Diff. both sides w.r.t x
[Use chain rule]
dy/dx = [x^5/2]*(1/x) + (lnx)[{(2*5*x^4)-0}/4]
= x^4/2 + {5x^4/2}lnx
= 1/2{(5x^4ln(x)+x^4)}

It is x^3/2 + (5/2)x^(3/2)lnx.
As regards horizontal, you did not mention direction in horizontal plane!

d/dx (x^(5/2) ln x) = 5/2 x^(3/2) ln x + x^(3/2) = 0

x^(3/2) (ln x + 1) = 0
x^(3/2) = 0 or ln x + 1 = 0
x = 0 (reject)......ln x = -1
..........................x = 1/e