f(x) = x/(x+1) ,n=4 Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at a=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)=x/(x+1) , we may apply the formula for Maclaurin series.
The list f^n(x) up to n=4 will be:
f(x)=x/(x+1)
Apply the Quotient rule for differentiation: d/(dx) (u/v) = (u' *v -u*v' )/v^2
Let u = x then u'=1
     v = x+1 then v' =1  and v^2= (x+1)^2
f'(x) = d/(dx) (x/(x+1))
           =(1 *(x+1) -x*1)/(x+1)^2
           =((x+1) -x)/(x+1)^2
           =(x+1 -x)/(x+1)^2
           =1/(x+1)^2
Apply Law of Exponent: 1/x^n = x^(-n) and Power Rule for differentiation: d/(dx) u^n= n* u^(n-1) *(du)/(dx).
Let:  u =x+1 then (du)/(dx) = 1
d/(dx) c*(x+1)^n = c *d/(dx) (x+1)^n
                          = c *(n* (x+1)^(n-1)*1)
                          = cn(x+1)^(n-1)
 
f^2(x)= d/(dx) (1/(x+1)^2)
           = d/(dx) (1)(x+1)^(-2)
           =1*(-2)(x+1)^(-2-1)
           =-2(x+1)^(-3) or -2/(x+1)^3
f^3(x)= d/(dx) [-2(x+1)^(-3)]
            =(-2)*(-3)(x+1)^(-3-1)
            =6(x+1)^(-4) or 6/(x+1)^4
f^4(x)= d/(dx) [6(x+1)^(-4)]
             =6*(-4)(x+1)^(-4-1)
             =-24(x+1)^(-5) or -24/(x+1)^5
Plug-in x=0 for each f^n(x) , we get:
f(0)=0/(0+1) =0
f'(0)=1/(0+1)^2 = 1
f^2(0)=-2/(0+1)^3 = -2
f^3(0)=6/(0+1)^4 = 6
f^4(0)=-24/(0+1)^5 = -24
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+6/(3!)x^3+(-24)/(4!)x^4
       = 0+1/1x-2/2x^2+6/6x^3-24/24x^4
       = 0+x-x^2+x^3-x^4
      =x-x^2+x^3-x^4
The Maclaurin polynomial of degree n=4 for the given function f(x)=x/(x+1) will be:
P(x)=x-x^2+x^3-x^4