f(x) = xe^x , n=4 Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at c=0 . The expansion of the function about 0 follows the formula:
f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n
 or
f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...
To determine the Maclaurin polynomial of degree n=4 for the given function f(x)=xe^x , we may apply the formula for Maclaurin series.
To list f^n(x) up to n=4 , we may apply the Product rule for differentiation: d/(dx) (u*v) = u' *v +u*v' and derivative property: d/(dx) (f+g) = d/(dx) f +d/(dx) g .
f(x)=xe^x
Let: u =x then u' = 1
      v = e^x  then v' = e^x
d/(dx) (xe^x) =(1*e^x) + (x*e^x)
                  =e^x +xe^x
f'(x)=d/(dx) (xe^x)
            = e^x +xe^x
f^2(x) = d/(dx) (e^x +xe^x)
           =d/(dx) e^x + d/(dx) xe^x
           = e^x + (e^x+xe^x)
           = 2e^x+xe^x
f^3(x) = d/(dx) ( 2e^x +xe^x)
            =d/(dx) 2e^x + d/(dx) xe^x
            = 2e^x + (e^x+xe^x)
            = 3e^x+xe^x
f^4(x) = d/(dx) ( 3e^x +xe^x)
            =d/(dx) 3e^x + d/(dx) xe^x
            = 3e^x + (e^x+xe^x)
            = 4e^x+xe^x
Plug-in x=0 for each f^n(x) , we get:
f(0)=0*e^0
         =0*1
          =0
f'(0)=e^0+0*e^0
            =1 +0*1
           =1
f^2(0)=2e^0+0*e^0
            =2*1 +0*1
            =2
f^3(0)=3e^0+0*e^0
            =3*1 +0*1
            =3
f^4(0)=4e^0+0*e^0
            =4*1 +0*1
            =4
Plug-in the values on the formula for Maclaurin series, we get:
sum_(n=0)^4 (f^n(0))/(n!) x^n
       = f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4
       = 0+1/(1!)x+2/(2!)x^2+3/(3!)x^3+4/(4!)x^4
       = 0+1/1x+2/2x^2+3/6x^3+4/24x^4
       = 0+x+x^2+1/2x^3+1/6x^4
       = x+x^2+1/2x^3+1/6x^4
The Maclaurin polynomial of degree n=4 for the given function f(x)=xe^x will be:
P(x)=x+x^2+1/2x^3+1/6x^4