f(x)=xcosx Find the Maclaurin series for the function.

Maclaurin series  is a special case of Taylor series which is centered at a=0. We follow the formula:
f(x) =sum_(n=0)^oo (f^n(0))/(n!)x^n
or
f(x) = f(0) + (f'(0))/(1!)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +(f^5(0))/(5!)x^5 +...
To list of f^n(x)  up to n=9 , we may apply the Product rule for differentiation: d/(dx) (u*v) = u'*v +u*v' .
f(x) = xcos(x)
f'(x) = cos(x)-xsin(x)
f''(x) = -xcos(x)-2sin(x)
f'''(x) = xsin(x)-3cos(x)
f^4(x) = xcos(x)+4sin(x)
f^5(x) = 5cos(x)-xsin(x)
f^6(x) = -xcos(x)-6sin(x)
f^7(x) = xsin(x)-7cos(x)
f^8(x) = xcos(x)+8sin(x)
f^9(x) = 9cos(x)-xsin(x)
Note: d/(dx)x=1 ,  d/(dx) cos(x) =-sin(x) , and d/(dx) sin(x)=cos(x).
Plug-in x =0 , we get:
f(0) = 0*cos(0)
        = 0*1
        =0
f'(0) = cos(0)-0*sin(0)
         = 1 -0*0
         =1
f''(0) = -0*cos(0)-2sin(0)
          =-0*1 - 2*0
          =0
f'''(0) = 0*sin(0)-3cos(0)
             =0*0 - 3*1
            =-3
f^4(0) = 0*cos(0)+4sin(0)
           =0*1 +4*0
            =0
f^5(0) = 5cos(0)-0*sin(0)
          =5*1 -0*0
          =5
f^6(0) = -0*cos(0)-6sin(0)
          =-0*1 -6*0
          =0
f^7(0) = 0*sin(0)-7cos(0)
           = 0*0-7*1
           =-7
f^8(0) = 0*cos(0)+8sin(0)
           =0*1+8*0
           =0
f^9(0) = 9cos(0)-0*sin(0)
           =9*1 -0*0
           =9
Note: cos(0)=1 and sin(0)=0 .
Plug-in the values in the formula, we get:
f(x) = 0 + 1/(1!)x+0/(2!)x^2+(-3)/(3!)x^3+0/(4!)x^4+5/(5!)x^5+0/(6!)x^6+ (-7)/(7!)x^7+0/(8!)x^8+9/(9!)x^9
=0 + 1/(1)x+0/(1*2)x^2-3/(1*2*3)x^3+ 0/(1*2*3*4)x^4 + 5/(1*2*3*4*5)x^5+ 0/(1*2*3*4*5*6)x^6
-7/(1*2*3*4*5*6*7)x^7 + 0/(1*2*3*4*5*6*7*8)x^8 + 9/(1*2*3*4*5*6*7*8*9)x^9+...

=0 + x+0/2x^2-3/6x^3+ 0/24x^4 + 5/120x^5 + 0/720x^6 -7/5040x^7 + 0/40320x^8 + 9/362880x^9+...
=0 + x+0-1/2x^3 + 0+ 1/24x^5 + 0 -1/720x^7 + 0+ 9/40320x^9+...
= x-1/2x^3 +1/24x^5 -1/720x^7 + 9/40320x^9 +...
Therefore, the Maclaurin series for the function f(x) =xcos(x) can be expressed as:
xcos(x)= x-1/2x^3 +1/24x^5 -1/720x^7 + 9/40320x^9 +...