f(x)=2/x n=3,c=1 Find the n'th Taylor Polynomial centered at c

Taylor series is an example of infinite series derived from the expansion of f(x) about a single point. It is represented by infinite sum of f^n(x) centered at x=c. The general formula for Taylor series is:
f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n
or
f(x) =f(c)+f'(c)(x-c) +(f^2(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f^4(c))/(4!)(x-c)^4 +...
To determine the Taylor polynomial of degree n=3 from the given function f(x)=2/x centered at x=1 , we may apply the definition of Taylor series.
To determine the list f^n(x) up to n=3 , we may apply Law of Exponent: 1/x^n = x^-n  and  Power rule for derivative: d/(dx) x^n= n *x^(n-1) .
f(x) = 2/x or 2x^(-1)
f'(x) = d/(dx) 2/x
            = d/(dx) 2x^(-1)
           = 2*d/(dx) x^(-1)
           =2*(-1 *x^(-1-1))
           =-2x^(-2) or -2/x^2
f^2(x)= d/(dx) -2x^(-2)
            =-2 *d/(dx) x^(-2)
           =-2 *(-2x^(-2-1))
           =4x^(-3) or 4/x^3
f^3(x)= d/(dx) 4x^(-3)
           =4 *d/(dx) x^(-3)
          =4 *(-3x^(-3-1))
          =-12x^(-4) or -12/x^4
Plug-in x=1 , we get:
f(2)=2/1 =2
f'(2)=-2/1^2 = -2
f^2(2)=4/1^3 =4
f^3(2)=-12/1^4 = -12
Applying the formula for Taylor series, we get:
sum_(n=0)^3 (f^n(1))/(n!) (x-1)^n
=f(1)+f'(1)(x-1) +(f^2(1))/(2!)(x-1)^2 +(f^3(1))/(3!)(x-1)^3
=2+(-2)(x-1) +4/(2!)(x-1)^2 +(-12)/(3!)(x-1)^3
=2-2(x-1) +4/2(x-1)^2 -12/6(x-1)^3
=2-2(x-1) +2(x-1)^2 -2(x-1)^3
The Taylor polynomial of degree n=3  for the given function f(x)=2/x centered at x=1 will be:
P_3(x)=2-2(x-1) +2(x-1)^2 -2(x-1)^3