How do you find the maclaurin series expansion of f(x) = e^(3x)f(x)=e3x?

1 Answer
Jun 18, 2018

e^(3x) = sum_(n=0)^oo (3^nx^n)/(n!) e3x=n=03nxnn!

Explanation:

As:

d/dx (e^(3x)) = 3e^(3x)ddx(e3x)=3e3x

and thus:

(d^n)/dt^n (e^(3x)) = 3^n e^(3x)dndtn(e3x)=3ne3x

the MacLaurin expansion is:

e^(3x) = sum_(n=0)^oo [(d^n)/dt^n (e^(3x))]_(x=0) x^n/(n!) e3x=n=0[dndtn(e3x)]x=0xnn!

e^(3x) = sum_(n=0)^oo (3^nx^n)/(n!) e3x=n=03nxnn!