Calculus of a Single Variable, Chapter 9, 9.7, Section 9.7, Problem 21

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