**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`

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