# `f(x) = 1/(x+1) , n=5` Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at `x=0` . The expansion of the function `f(x)` 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 +(f^5(0))/(5!)x^5+...`

To determine the Maclaurin polynomial of degree `n=5` from the given function `f(x)=1/(x+1)` , we may apply Law of Exponent: `1/x^n = x^(-n)` . The function becomes:

`f(x) = (x+1)^(-1)`

Apply Power rule for differentiation: `d/(dx) u^n = n * u^(n-1)* (du)/(dx)` to list the derivative of `f(x)` .

Let `u =1+x` then `(du)/(dx) = 1` for each derivatives.

`f'(x) = d/(dx)(x+1)^(-1)`

`= (-1) *(x+1)^(-1-1)*1`

`= -(x+1)^(-2) or -1/(x+1)^2`

`f^2(x)= d/(dx) -(x+1)^(-2)`

`=-1*(-2 *(x+1)^(-2-1))*1`

`=2(x+1)^(-3) or 2/(x+1)^3`

`f^3(x)= d/(dx) 2(x+1)^(-3)`

`=2*d/(dx) (x+1)^(-3)`

`=2* (-3*(x+1)^(-3-1))*1`

`=-6(x+1)^(-4) or -6/(x+1)^4`

`f^4(x)= d/(dx) -6(x+1)^(-4)`

`=-6*d/(dx) (x+1)^(-4)`

`=-6* (-4*(x+1)^(-4-1))*1`

`=24(x+1)^(-5) or 24/(x+1)^5`

`f^5(x)= d/(dx) 24(x+1)^(-5)`

`=24*d/(dx) (x+1)^(-5)`

`=24* (-5*(x+1)^(-5-1))*1`

`=-120(x+1)^(-6) or -120/(x+1)^6`

Plug-in `x=0` , we get:

`f(0) = 1/(0+1)`

`=1/1`

`=1`

`f'(0) = -1/(0+1)^2`

`=-1/1^2`

`=-1`

`f^2(0) =2/(0+1)^3`

`=2/1^3`

`=2`

`f^3(0) =-6/(0+1)^4`

`=-6/1^4`

`=-6`

`f^4(0) =24/(0+1)^5`

`=24/1^5`

` =24`

`f^5(0) =-120/(0+1)^6`

`=-120/1^6`

`=-120`

Plug-in the values on the formula for Maclaurin series.

`sum_(n=0)^5 (f^n(0))/(n!) x^n`

`= 1+(-1)/(1!)x+2/(2!)x^2+(-6)/(3!)x^3+24/(4!)x^4+ (-120)/(5!)x^5`

` = 1-1/1x+2/2x^2-6/6x^3+24/24x^4-120/120x^5`

`= 1-x+x^2-x^3+x^4-x^5`

The Maclaurin polynomial of degree ` n=5` for the given function` f(x)= 1/(x+1)` will be:

`P_5(x)=1-x+x^2-x^3+x^4-x^5`

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