Proof for the differentiation of e^x

To prove that $\frac{d}{d x} e^{x} = e^{x}$ , start with

\frac{d y}{d x} = y

So the search is for a function that is equal to it's derivative. Using a polynomial, a well known function, and starting with the simplest,

y = a_{0} + a_{1} x

The derivative of is

\frac{d y}{d x} = a_{1}

Not particularly similar to the original function. Now try a higher order polynomial

y = a_{0} + a_{1} x + a_{2} x^{2}

The derivative of is

\frac{d y}{d x} = a_{1} + 2 a_{2} x

This has some of the structure of the original function. Now for a higher order

y = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}

The derivative of is

\frac{d y}{d x} = a_{1} + 2 a_{2} x + 3 a_{3} x^{2}

Here we have even more similarity to the orginal function and suggest taking a higher order.

y = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + ... + a_{(n - 2)} x^{(n - 2)} + a_{(n - 1)} x^{(n - 1)} + a_{n} x^{n}

the derivative of is

\frac{d y}{d x} = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} + ... + (n - 2) a_{(n - 2)} x^{(n - 3)} + (n - 1) a_{(n - 1)} x^{(n - 2)} + n a_{n} x^{(n - 1)}

$y$ and it's derivative differ by only one term, so by comparing coefficients,

For the constant term, $a_{1} = a_{0}$

For the $x$ term, $2 a_{2} = a_{1}$ . So, $a_{2} = \frac{a_{0}}{2}$

For the $x^{2}$ term, $3 a_{3} = a_{2}$ . So, $a_{3} = \frac{a_{0}}{2 \cdot 3}$

For the $x^{3}$ term, $4 a_{4} = a_{3}$ . So, $a_{4} = \frac{a_{0}}{2 \cdot 3 \cdot 4}$

... and so on ...

For the $x^{(n - 1)}$ term, $n a_{n} = a_{(n - 1)}$ . So, $a_{n} = \frac{a_{0}}{n!}$

So the function can be rewritten,

y = a_{0} + a_{0} x + \frac{a_{0} x^{2}}{2!} + \frac{a_{0} x^{3}}{3!} + ... + \frac{a_{0} x^{(n - 2)}}{(n - 2)!} + \frac{a_{0} x^{(n - 1)}}{(n - 1)!} + \frac{a_{0} x^{n}}{n!}

= a_{0} (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!})

So the proposed function has the above form. Plugging this into the differential equation to be solved,

\frac{d}{d x} (a_{0} (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!})) = a_{0} (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!})

a_{0} \frac{d}{d x} (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!}) = a_{0} (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!})

\frac{d}{d x} (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!}) = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!}

So, the function is of the form,

y = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ... + \frac{x^{(n - 2)}}{(n - 2)!} + \frac{x^{(n - 1)}}{(n - 1)!} + \frac{x^{n}}{n!}

Each derivative will be one term short of the original function. So the function becomes more true as $n \to ∞$ . Therefore,

y = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + ...

Many will recognise this infinite series, but let's investigate this functions nature. This function has the value of one when $x$ is zero. Taking ratios of equally spaced values of $x$ in the function then a constant is found.

\frac{y (2)}{y (1)} = \frac{1 + 2 + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + ...}{1 + 1 + \frac{1^{2}}{2!} + \frac{1^{3}}{3!} + ...} = 2.718281828...

also

\frac{y (10)}{y (9)} = 2.718281828...

All of these properties are of an exponential function and the base here is $2.718281828... = e$ . So

y = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + ... = e^{x}

and

\frac{d}{d x} e^{x} = e^{x}

As required.

Proof for the differentiation of ex

Proof for the differentiation of $e^{x}$