# Exponential function

• Apr 10th 2008, 09:20 AM
matty888
Exponential function
Define the exponential function in terms of the natural logarithm and show that the derivitave of e^x is e^x
• Apr 10th 2008, 09:52 AM
ThePerfectHacker
Quote:

Originally Posted by matty888
Define the exponential function in terms of the natural logarithm and show that the derivitave of e^x is e^x

Define $\ln x = \int_1^x \frac{d\mu}{\mu}$, for $x>0$. This function is differenciable by the fundamental theorem for any $x>0$ and $\ln x = \frac{1}{x}$. Since the derivative is positive the function is increasing, so $\ln x$ is a one-to-one function on $\mathbb{R}^+$. Let $\exp (x)$ be its inverse function on [tex]D[tex] (where $D$ is the range of $\ln x$, it happens to be that $D=\mathbb{R}$ but it does not matter here). Thus, $\ln (\exp (x)) = x$. Using the chain rule we get $\left( \exp x \right) ' \frac{1}{\exp x} = 1 \implies \left( \exp x \right) ' = \exp x$.

We can go further and define some additional properties. We know that there is a unique number $e$ such that $\ln e = 1$ by intermediate value theorem. Let $q$ be any rational number. We can show that $\exp ( q ) = e^q$ where $e^q$ is ordinary exponentiation of rational exponents (try provong this). Thus, it is reasonable to define $e^r = \exp (r)$ for any real number $r$.