The #nth# term of the Taylor series of #f(x)# centered at #a# is given by
#f^((n))(a)(x-a)^n/(n!)#
For our case, #a=0,# so the #nth# term is given by
#f^((n))(0)x^n/(n!)#
So, to find the first #4# terms using this formula, we'll need the function itself evaluated at #a=0#, as well as the first three derivatives (as the #0th# derivative is the function itself).
#f(0)=0(e^0)=0#
#f'(x)=e^x+xe^x, f'(0)=e^0=1#
#f''(x)=2e^x+xe^x, f''(0)=2e^0=2#
#f'''(x)=3e^x+xe^x, f'''(0)=3e^0=3#
#0th# term: #0(x^0)/(0!)=0, 0! =1# is a convention we'll adopt here.
#1st# term: #x/(1!)=x#
#2nd# term: #2x^2/(2!)=x^2#
#3rd# term #3x^3/(3!)=x^3/2#:
As the #0th# term is #0,# and we're generally interested in non-zero terms , we can take the fourth derivative and find the next term (this is, I presume, why the first four derivatives were taken, as you need four derivatives to get four non-zero terms with this function):
#f^((4))(x)=4e^x+xe^x, f^((4))(0)=4e^0=4#
#4th# (non-zero) term: #4x^4/(4!)=x^4/6#
Alternatively, we can use known Taylor series, which is easier with this function: (this method requires no derivatives)
#e^x=sum_(n=0)^oox^n/(n!)#
#xe^x=xsum_(n=0)^oox^n/(n!)#
#xe^x=sum_(n=0)^oox^(n+1)/(n!)#
Evaluate from #n=0# to #n=3#:
#sum_(n=0)^3x^(n+1)/(n!)=x+x^2+x^3/2+x^4/6#
