Recall that the derivative of f(x) = e^x is f '(x) = e^x. In fact, all the derivatives are e^x.

f '(0) = e⁰ = 1

f ''(0) = e⁰ = 1

f '''(0) = e⁰ = 1

We see that all the derivatives, when evaluated at x = 0, give us the value 1.

Also, f(0) = 1, so we have:

The Maclaurin Series expansion will be simply: