How do you find the Maclaurian series for ${e^x}$?

Hint:
We will first write the general Maclaurian series for any function f(x) and then we will just plug in the function f (x) equal to ${e^x}$ in that series to get the required answer.

Complete step by step solution:
We are given that we are required to find the Maclaurian series for ${e^x}$.
Since, we know that the Maclaurian series for any function f (x) is given by the following expression:-
$ \Rightarrow f(x) = f(0) + xf'(0) + \dfrac{{{x^2}}}{{2!}}f''(0) + \dfrac{{{x^3}}}{{3!}}f'''(0) + ........... + \dfrac{{{x^n}}}{{n!}}{f^n}(0) + ........$
Now, we just need to plug in $f(x) = {e^x}$.
Now, since we know that the derivative of f (x) is ${e^x}$, no matter which derivative is it.
Therefore, we have $f(0) = {e^0} = 1,f'(0) = {e^0} = 1$ and so on.
Therefore, we have ${f^n}(0) = 1$ for all the values of n.
Now, putting it in the above mentioned expression, we will then obtain the following expression:-
$ \Rightarrow {e^x} = 1 + x(1) + \dfrac{{{x^2}}}{{2!}}(1) + \dfrac{{{x^3}}}{{3!}}(1) + ........... + \dfrac{{{x^n}}}{{n!}}(1) + ........$
Simplifying it, we will then obtain the following expression:-
$ \Rightarrow {e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + ........... + \dfrac{{{x^n}}}{{n!}} + ........$
Thus, we have the required answer.

Note:
The students must note that “Maclaurin Series” is just “Taylor Series” about the point 0.
The taylor series of any function f (x) about the point x = a is given by the following expression:-
$ \Rightarrow f(x) = f(a) + (x - a)f'(a) + \dfrac{{{{(x - a)}^2}}}{{2!}}f''(a) + \dfrac{{{{(x - a)}^3}}}{{3!}}f'''(a) + ........... + \dfrac{{{{(x - a)}^n}}}{{n!}}{f^n}(a) + ........$
If we put in a = 0, we will then obtain the following expression with us:-
$ \Rightarrow f(x) = f(0) + (x - 0)f'(0) + \dfrac{{{{(x - 0)}^2}}}{{2!}}f''(0) + \dfrac{{{{(x - 0)}^3}}}{{3!}}f'''(0) + ........... + \dfrac{{{{(x - 0)}^n}}}{{n!}}{f^n}(0) + ........$
Simplifying the values in the above expression, we will then obtain the following expression:-
$ \Rightarrow f(x) = f(0) + xf'(0) + \dfrac{{{x^2}}}{{2!}}f''(0) + \dfrac{{{x^3}}}{{3!}}f'''(0) + ........... + \dfrac{{{x^n}}}{{n!}}{f^n}(0) + ........$
This is exactly the Maclaurin Series as we mentioned in the above solution.
The students must also know that, to find the Taylor series and the Maclaurian series, the function must be infinitely times differentiable and continuous, so that we have defined series and all the values that we are required to plug in.
We use Taylor series so that we have an idea and an approximation of the values, which we cannot exactly find and thus we can use the approximate values.