Before turning to logs, let's take a closer look at the function $f(x)=e^x$, where $e \approx 2.718281828$. In the last learning module, we saw that the rate of change of any exponential function $f(x)=a^x$ is proportional to $a^x$, and that the rate of change of $e^x$ is exactly $e^x$. This is the only function (up to multiplication by a constant) whose derivative is itself. That is, whenever you see a rate equation $f'(x) = f(x)$, you should think $f = A e^x$.
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$\displaystyle\frac{d}{dx}(e^x)=e^x$. The
derivative of $e^x$ is $e^x$. |
So why is that true? First, you can check (graphically, or numerically) that $\lim_{h \to 0} (2^h -1)/h$ is less than 1 (it's around 0.69), and that $\lim_{h \to 0} (3^h-1)/h$ is bigger than 1 (it's around 1.1). This means that there has to be a number $a$, somewhere between 2 and 3, such that $\lim_{h \to 0}(a^h - 1)/h = 1$. We call this magic number "$e$", and it's around 2.718281828.
Now, if $f(x)=e^x$, then \begin{eqnarray*} f'(x) & = & \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \cr & = & \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} \cr & = & \lim_{h \to 0} \frac{e^xe^h - e^x}{h} \cr & = & e^x \cdot\left(\lim_{h \to 0} \frac{e^h - 1}{h}\right) \cr & = & e^x, \end{eqnarray*} since we already determined that $\lim_{h \to 0}(e^h-1)/h = 1$. Thus, the derivative of $e^x$ is $e^x$.If any of those steps are unclear to you, the following video explains them.