How do you find the maclaurin series expansion of #f(x) = e^(3x)#?

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Answer 1 · 2018-06-18T11:28:00

#e^(3x) = sum_(n=0)^oo (3^nx^n)/(n!) #

As:

#d/dx (e^(3x)) = 3e^(3x)#

and thus:

#(d^n)/dt^n (e^(3x)) = 3^n e^(3x)#

the MacLaurin expansion is:

#e^(3x) = sum_(n=0)^oo [(d^n)/dt^n (e^(3x))]_(x=0) x^n/(n!) #

#e^(3x) = sum_(n=0)^oo (3^nx^n)/(n!) #