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Derivative of $e^x$

To find the derivative of $e^x$, we must first define $e$. We define $e$ as:
\[ \begin{align*} e = \lim_{n\to\infty}(1+\dfrac1n)^n \end{align*} \]
This can alternatively be written as:
\[ \begin{align*} e = \lim_{n\to0}(1+\dfrac1n)^\dfrac1n \end{align*} \]
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 (insert link). We write the derivative of $e^x$ as:
\[ \begin{align*} \dfrac{d}{dx}(e^x)= \lim_{h\to0} \dfrac{e^{x+h} – e^x}{h} \dfrac{d}{dx}(e^x)= \lim_{h\to0} \dfrac{e^x\cdot e^h- e^x}{h} \end{align*} \]
We will now factor $e^x$ out of the limit. We can do this because the limit is with respect to $h$, not $x$. 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:
\[ \begin{align*} n = e^h – 1\\ n + 1 = e^h\\ \ln(n+1) = h \end{align*} \]
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 $ln$ function:
\[ \begin{align*} \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{n\to0} \dfrac{1}{\dfrac1n\cdot\ln(n+1)}\\ \dfrac{d}{dx}(e^x)= e^x \cdot \lim_{n\to0} \dfrac{1}{\cdot\ln((n+1)^{\dfrac1n})}\\ \end{align*} \]
Since we’ve established that $e = \lim_{n\to0}(1+\dfrac1n)^\dfrac1n$, 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, the derivative of $e^x$ is $e^x$.

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