write both #x#-terms as numerical exponents of #x:#
#sqrt(x) = x^(1/2)#
#root3(x) = x^(1/3)#
#1/(root3(x)) = x^(-1/3)#
#(sqrt(x) + 1/(root3(x)))^2 = (x^(1/2) + x^(-1/3))^2#
using FOIL:
#(x^(1/2) + x^(-1/3))(x^(1/2) + x^(-1/3)) = x^1 + x^(1/6) + x^(1/6) + x^(-2/3)#
then you can differentiate these values separately, using the power rule
#f(x) = nx^(n-1)#.
when #x^n = x^1#, #nx^(n-1)# is #x^0#, or #1#.
when #x^n = x^(1/6)#, #nx^(n-1)# is #1/6x^(-5/6)#.
when #x^n = x^(-2/3)#, #nx^(n-1)# is #-2/3x^(-5/3)#.
hence, when #x^1 + x^(1/6) + x^(1/6) + x^(-2/3)# is differentiated, the result is
#1 + 1/6x^(-5/6) + 1/6x^(-5/6) + -2/3x^(-5/3)#.
this can be simplified to #1 + 1/3x^(-5/6) -2/3x^(-5/3)#.
or even #(3 + x^(-5/6) - 2x^(-5/3))/3#.
#v'(x) = 1 + 1/3x^(-5/6) -2/3x^(-5/3)#.
