f(x) = e^(x/3) , 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)x)/(1!)+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...
We may apply the formula for Maclaurin series to determine the Maclaurin polynomial of degree n=4 for the given function f(x)=e^(x/3) .
Apply derivative formula for exponential function: d/(dx) e^u = e^u * (du)/(dx) to list f^n(x) as:
Let u =x/3 then (du)/(dx)= 1/3
Applying the values on the derivative formula for exponential function, we get:
d/(dx) e^(x/3) = e^(x/3) *(1/3)
         = e^(x/3)/3 or 1/3e^(x/3)
Applying d/(dx) e^(x/3)= 1/3e^(x/3)  for each f^n(x) , we get:
f'(x) = d/(dx) e^(x/3)
          =1/3e^(x/3)
f^2(x) = d/(dx) (1/3e^(x/3))
          =1/3 *d/(dx)e^(x/3)
          =1/3 *(1/3e^(x/3))
          =1/9e^(x/3)
f^3(x) = d/(dx) (1/9e^(x/3))
           =1/9 *d/(dx) e^(x/3)
          =1/9 *(1/3e^(x/3))
          =1/27e^(x/3)
f^4(x) = d/(dx) (1/27e^(x/3))
           =1/27 *d/(dx) e^(x/3)
           =1/27 *(1/3e^(x/3))
           =1/81e^(x/3)
Plug-in x=0 on each f^n(x) , we get:
f(0)=e^(0/3) = 1
f'(0)=1/3e^(0/3) = 1/3
f^2(0)=1/9e^(0/3)=1/9
f^3(0)=1/27e^(0/3)=1/27
f^4(0)=1/81e^(0/3)=1/81
Note: e ^(0/3) = e^0 =1.
Plug-in the values on the formula for Maclaurin series, we get:
f(x)=sum_(n=0)^4 (f^n(0))/(n!) x^n
    = 1+(1/3)/(1!)x+(1/9)/(2!)x^2+(1/27)/(3!)x^3+(1/81)/(4!)x^4
   =1+1/3x+1/18x^2+1/162x^3+1/1944x^4
The Maclaurin polynomial of degree n=4 for the given function f(x)=e^(x/3) will be:
P_4(x)=1+1/3x+1/18x^2+1/162x^3+1/1944x^4