f(x)=sin(3x) ,c=0 Use the definition of Taylor series to find the Taylor series, centered at c for the function.

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 apply the definition of Taylor series for the given function f(x) = sin(3x) , we list f^n(x) using  the derivative formula for trigonometric function: d/(dx) sin(u) = cos(u) *(du)/(dx) and d/(dx) cos(u)= -sin(u)*(du)/(dx) .
Let u = 3x then (du)/(dx) =3 .
f(x) =sin(3x)
f'(x) = d/(dx) sin(3x)
           = cos(3x)*3
           =3cos(3x)
f^2(x) = d/(dx) 3cos(3x)
            =3 d/(dx) cos(3x)
            =3*( -sin(3x)*3)
            =-9sin(3x)
f^3(x) = d/(dx)-9sin(3x)
            = -9 d/(dx)sin(3x)
             =-9 * cos(3x)*3
           = -27cos(3x)
f^4(x) = d/(dx) -27cos(3x)
            =-27*d/(dx) cos(3x)
            = -27 * (-sin(3x)*3)
            =81 sin(3x)
 f^5(x) = d/(dx) 81sin(3x)
            =81*d/(dx) sin(3x)
            = 81* (cos(3x)*3)
            =243cos(3x)
 Plug-in x=0 on each f^n(x) , we get:
 f(0) =sin(3*0)
         =sin(0)
         =0
 f'(0)= 3cos(3*0)
           =3cos(0)
           = 3*1
           =3  
 f^2(0)= -9sin(3*0)
           =-9sin(0)
            =-9 *0
            =0
 f^3(0)= -27cos(3*0)
            =-27 cos(0)
             =-27*1
            =-27
 f^4(0)= 81sin(3*0)
            =81sin(0)
            =81*0
             =0
 f^5(0)= 243cos(3*0)
            =243cos(0)
             =243*1
             =243
 Plug-in the values on the formula for Taylor series, we get:
 sin(3x) = sum_(n=0)^oo (f^n(0))/(n!) (x-0)^n
 =sum_(n=0)^oo (f^n(0))/(n!) x^n
 =f(0)+f'(0)x +(f'^2(0))/(2!)x^2 +(f^3(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +(f^4(0))/(4!)x^4 +...
=0+3x +0/(2!)x^2 +(-27)/(3!)x^3 + 0/(4!)x^4 +243/(5!)x^5+...
=0+3x +0/2x^2 +(-27)/6x^3 + 0/24x^4 +243/120x^5+...
=0+3x +0 -9/2x^3 + 0 +81/40x^5+...
=3x -9/2x^3 +81/40x^5+...
The Taylor series for the given function f(x)=sin(3x) centered at c=0 will be:
sin(3x) =3x -9/2x^3 +81/40x^5+...