The Maclaurin series for #f(x)# is
#sum_(n=0)^oo (f^('n)(0))/(n!) * (x)^n#
I assume the major difficulty here is evaluating
#f^('n)(0)# for successive terms of the series
Before getting into individual derivatives, let's first note:
#e^0 = 1#
#cos(0) = 1#
#sin(0) = 0#
#(d e^x)/(dx) = e^x#
#(d sin(x)) = cos(x)# and
#(d cos(x)) = - sin(x)#
If #f(x) = e^x sin(x)#, then
#f'(x) = (d e^x) / (dx) * sin(x) + e^x*(d sin(x))/(dx)#
#=e^x sin(x) + e^x cos(x)#
so #f'(0) = 1#
#f''(x) = ((d (e^x sin(x)))/(dx)) + ((d (e^x cos(x)))/(dx))#
#= (e^x sin(x) + e^x cos(x)) + (e^x cos(x) + e^x (-sin(x)) )#
#=2 e^x cos(x)#
and #f''(0) = 2#
continuing on in the same manner we can evaluate subsequent derivatives.
Simply insert these values into the definition of the Maclaurin series.
