The importance of the exponential function \( e^x \) in mathematics cannot be overstated. The derivative of the exponential is fundamental to calculus. This article demonstrates a proof of this derivative from first principles.
There are several ways to define \( e \). The textbook definition as a limit is the one I learned first and the one I want to use for the proof.
$$ e = \lim_{N \to \infty}\left(1 + \frac{1}{N}\right)^N $$
Let's define \( e^x \) using the definition provided above. We can start by exponentiating the limit.
$$ e^x = \left( \lim_{N \to \infty}\left(1 + \frac{1}{N}\right)^N \right)^x $$
Since exponentiation is a continuous function, we can bring the exponent \( x \) inside of the limit and combine it with the \( N \) following exponent rules.
$$ e^x = \lim_{N \to \infty}\left(1 + \frac{1}{N}\right)^{Nx} $$
We will now make a substitution. Let \( M = Nx \). We will assume \( x \) is positive; otherwise we can consider \( e^{-x} \) to make the exponent positive. In the case of \( x = 0 \), we know the value of \( e^0 = 1 \) which handles that case. Since \( x \) is positive, then \( Mx \to \infty \) as \( N \to \infty \). We can write this as simply \( M \to \infty \) in the limit, which is equivalent. Also we know that \( \frac{1}{N} = \frac{x}{M} \) by rearranging \( M = Nx \). After making these substitutions, we arrive at the following.
$$ e^x = \lim_{M \to \infty}\left(1 + \frac{x}{M}\right)^M $$
The choice of limiting variable \( M \) above is arbitrary; we can reset it back to \( N \). This gives us the representation of \( e^x \) we will use going forward.
$$ \tag{$\star$} e^x = \lim_{N \to \infty}\left(1 + \frac{x}{N}\right)^N $$
This equation is important enough that I've labeled it with a \( (\star) \). I'll refer back to it more throughout this article.
What about negative values of the exponent? We can approach that by still assuming \( x \) is positive and analyzing \( e^{-x} \). By exponent rules, this is equal to \( \frac{1}{e^x} \). We can substitute our definition of \( e^x \).
$$ e^{-x} = \frac{1}{e^x} = \frac{1}{\lim_{N \to \infty}\left(1 + \frac{x}{N}\right)^N} $$
We are taking a limit on an interval bounded away from \( 0 \). Keep in mind we handled the case of \( e^0 \) separately earlier. The reciprocal function is continuous on any strictly positive interval, and the reciprocal can therefore be absorbed into the limit.
$$ e^{-x} = \lim_{N \to \infty}\frac{1}{\left(1 + \frac{x}{N}\right)^N} $$
We will work on modifying the fraction inside of the limit. First, we can bring the \( N \) as the exponent of the entire fraction. Second we can write \( 1 + \frac{x}{N} \) as \( \frac{N + x}{N} \).
$$ e^{-x} = \lim_{N \to \infty}\left(\frac{1}{1 + \frac{x}{N}}\right)^N = \lim_{N \to \infty}\left(\frac{1}{\frac{N + x}{N}}\right)^N $$
Now, we can take the reciprocal inside of the parentheses.
$$ e^{-x} = \lim_{N \to \infty}\left(\frac{N}{N + x}\right)^N $$
The next step is a favorite sleight of hand by mathematicians. We will add and subtract \( x \) from the numerator and then split the fraction in two and simplify.
$$ e^{-x} = \lim_{N \to \infty}\left(\frac{N + x - x}{N + x}\right)^N = \lim_{N \to \infty}\left(\frac{N + x}{N + x} - \frac{x}{N + x}\right)^N $$
Simplifying the \( \frac{N + x}{N + x} \) to \( 1 \), we obtain the following.
$$ e^{-x} = \lim_{N \to \infty}\left(1 - \frac{x}{N + x}\right)^N $$
Next is another substitution. Consider mapping \( N \mapsto N - x \). This does not change the limiting value, as \( N - x \) approaches \( \infty \) as \( N \) does (remember, \( x \) is fixed and finite).
$$ e^{-x} = \lim_{N \to \infty}\left(1 - \frac{x}{N}\right)^{N -x} $$
The exponent \( N - x \) can be split into separate factors using exponent rules. Since limit is a product is the product of limits as long as both limits exist, then we can split the limit.
$$ e^{-x} = \lim_{N \to \infty}\left[\left(1 - \frac{x}{N}\right)^{N} \left(1 - \frac{x}{N}\right)^{-x}\right] = \lim_{N \to \infty}\left(1 - \frac{x}{N}\right)^N \lim_{N \to \infty}\left(1 - \frac{x}{N}\right)^{-x} $$
The second limit evaluates to \( 1 \), yielding the final expression for \( e^{-x} \).
$$ e^{-x} = \lim_{N \to \infty}\left(1 - \frac{x}{N}\right)^N $$
Notice that this is equivalent to the earlier equation \( (\star) \), where \( x \) has been replaced with \( -x \). It also holds true for \( x = 0 \). Therefore, equation \( (\star) \) is valid for all real values of \( x \).
We can now explore the derivative of \( e^x \) using the equation \( (\star) \) we built here.
As this is a proof from first principles, we will use the definition of the derivative.
$$ \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$
Setting \( f(x) = e^x \) in the definition gives:
$$ \frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x + h} - e^x}{h} $$
Using exponential rules, we can factor a \( e^x \) from the numerator as follows. Since the \( e^x \) does not depend on \( h \), it can be pulled in front of the limit.
$$ \frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h - 1}{h} $$
The next step is to analyze this remaining limit \( \lim_{h \to 0} \frac{e^h - 1}{h} \) more closely.
Starting from the previous limit, we will plug in the definition of \( e^h \) using equation \( (\star) \).
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{1}{h} \lim_{N \to \infty}\left[\left( 1 + \frac{h}{N} \right)^N - 1 \right] $$
We will assume \( N \) is a positive integer. Since the limit of \( e \) exists, we can assume the limit of \( N \) to \( \infty \) ascends on integer values. This allows us to apply the binomial expansion.
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{1}{h} \lim_{N \to \infty}\left[\sum_{k=0}^{\infty}\binom{N}{k}\left(\frac{h}{N}\right)^k - 1 \right] $$
Notice when \( k = 0 \), the term within the sum is \( 1 \). We can cancel this with the \( -1 \) outside the sum by indexing the sum to start at \( k = 1 \).
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{1}{h} \lim_{N \to \infty}\left[\sum_{k=1}^{\infty}\binom{N}{k}\left(\frac{h}{N}\right)^k \right] $$
Each term within the sum is a rational polynomial in \( N \). As \( N \to \infty \), we need to look at the coefficient of the largest power of \( N \) in both the numerator and the denominator of the fraction to determine the limiting value.
We can write \( \binom{N}{k} \) as \( \frac{(N)(N-1)\cdots(N-k+2)(N-k+1)}{k!} \). Notice there are exactly \( k \) terms in the numerator. The highest power of \( N \) in the numerator is \( N^k \). We can also see that the highest power of \( N \) in the denominator is also \( N^k \). In the limit, we need to take the ratio of the coefficients of these terms.
The \( N^k \) in the numerator has a coefficient of \( h^k \), which is provided by the term \( \left(\frac{h}{N}\right)^k \) in the sum. The \( N^k \) in the denominator has a coefficient of \( k! \), which we derived in the previous paragraph from \( \binom{N}{k} \). Therefore, in the limit of \( N \to \infty \), we get the value \( \frac{h^k}{k!} \). Let's plug this back into our equation.
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{1}{h} \sum_{k=1}^{\infty} \frac{h^k}{k!} $$
We are almost done. We will bring the \( \frac{1}{h} \) into the sum by reducing the power of \( h \) in the numerator.
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \sum_{k=1}^{\infty} \frac{h^{k-1}}{k!} $$
In this sum, as \( h \to 0 \), all the powers of \( h \) will tend toward zero, except whenever the exponent of \( h \) is exactly \( 0 \). This only occurs when \( k = 1 \). The infinite sum collapses to a single term, which evaluates to \( 1 \).
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{h \to 0} \frac{h^0}{1!} = 1 $$
Plugging this value in back into our previous equation for the derivative of \( e^x \) brings us to our final result.
$$ \frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x $$
We have proven the derivative of \( e^x \) is \( e^x \), what I consider to be one of the most fundamental results in all of calculus. \( \square \)