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Derivative of $e^x$

In this article, we are going to prove that the derivative of $e^x$ is equal to $\boxed{e^x}$. To find the derivative of $e^x$, we must first define $e$. The formal definition of $e$ is:

\[\begin{align*} e = \lim_{n\to\infty}(1+\dfrac1n)^n \end{align*}\]

$e$ is approximately equal to $2.71828$. This value does not need to me memorized, as it can be found on a calculator, but a general idea of its scale should be known. This can alternatively be written as

\[\begin{align*} e = \lim_{n\to0}(1+\dfrac1n)^{\dfrac{1}{n}} \end{align*}\]

We will now use the limit definition of a derivative to find the derivative of $e^x$. For a refresher on the limit of a derivative, check out our blog on the limit definition of a derivative. We write the derivative of $e^x$ as:

\[\begin{align*} \dfrac{d}{dx}(e^x)= \lim_{h\to0} \dfrac{e^{x+h} – e^x}{h}\longrightarrow \dfrac{d}{dx}(e^x)= \lim_{h\to0} \dfrac{e^x\cdot e^h- e^x}{h} \end{align*}\]

We will now factor $e^x$ out of the limit. Because the limit is taken with respect to $h$ and not $x$, $e^x$ will be constant when the limit is evaluated. The new derivative is:

\[\begin{align*} \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{h\to0} \dfrac{e^h – 1}{h} \end{align*}\]

We will now define a new variable, $n$, such that:

\[\begin{align*} n = e^h – 1,\\ n + 1 = e^h, \end{align*}\]

By taking the natural logarithm of both sides, we get:

\[\begin{align*} \ln(n+1) = h \end{align*}\]

We will substitute $n$ into the numerator and denominator of the original equation. We will also change the variable in the limit. Since $n$ approaches $0$ as $h$ approaches $0$, we substitute appropriately:

\[\begin{align*} \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{n\to0} \dfrac{n}{\ln(n+1)} \end{align*}\]

We will multiply the numerator and denominator of the limit by $\dfrac1n$ and move the $\dfrac1n$ in the denominator inside of the $\ln(n+1)$ function using our logarithm rules:

\[\begin{align*} \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{n\to0} \dfrac{1}{\dfrac1n\cdot\ln(n+1)}\longrightarrow \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{n\to0} \dfrac{1}{\cdot\ln((n+1)^{\dfrac1n})} \end{align*}\]

Since we’ve established that:

\[\begin{align*} e = \lim_{n\to0}(1+\dfrac1n)^{\dfrac{1}{n}}, \end{align*}\]

We can substitute $e$ into the equation like so:

\[\begin{align*} \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{n\to0} \dfrac{1}{\cdot\ln(e)}\\ \end{align*}\]

Since $ln(e)$ = 1:

\[\begin{align*} \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{n\to0} \dfrac{1}{1}\\ \dfrac{d}{dx}(e^x)= e^x \end{align*}\]

Thus, we can conclude that the derivative of $e^x$ is $\boxed{e^x}$. While sometimes $e^x$ is defined as a function whose derivative equals itself, it is useful to know the proof behind the property.

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