Derivative of Exponential Functions

Basic exponential functions have the form `f(x) = b^x` where `b` is some positive number. In this applet we see what `f(x) = b^x` looks like for various values of `b`, and we see how `f'(x)` is related to `f(x)`. In particular, it turns out that

If `f(x) = b^x`, then `f'(x)` is proportional to `f(x)`. That is, [`f'(x) = m_b f(x)`] for some number `m_b` that depends on `b`.

In fact,

`m_b` is the slope of the line tangent to `y = b^x` through the point `(0, 1)`.

And that fact, coupled with the definition for `e` (see below) implies

If `f(x) = e^x`, then `f'(x) = e^x`.

Explore

Use the slider to change the value of `b` and observe how the graph of `f(x) = b^x` changes.
Every exponential function goes through the point `(0,1)`, right? Why is this?
Click the checkbox to see `f'(x)`, and verify that the derivative looks like what you would expect (the value of the derivative at `x = c` look like the slope of the exponential function at `x = c`).
Now click the checkbox to show the line tanget to `f(x)`. How is the slope of that line related to the relationship between `f(x)` and `f'(x)`?
Let `m_b` denote the slope of the tangent line. It turns out (proof below) that `f'(x)` is proportional to `f(x)` by a factor of `m_b`, that is, `f'(x) = m_b f(x)`. Confirm that when `m_b = 0.5`, then `f'(x)` is half as big as `f(x)`, when `m_b = 1/3`, then `f'(x)` is a third as big as `f(x)`, and so forth.
We define `e` to be that number which, when used as the base of an exponential function, causes the tangent line through `(0,1)` to have slope 1.
Use the slider to find an approximation for `e`. If `f(x) = e^x`, then what is `f'(x)`?

Proof

Let `f(x) = b^x`, and let `m_b` denote the slope of the line tangent to the graph of `f(x)` at `x = 0` (in other words, `m_b = f'(0)`).
We prove that `f'(x) = m_b f(x)`.

`f'(x) = lim_(h-->0) (f(x + h) - f(x))/h = lim_(h-->0) (b^(x+h) - b^x)/h = lim_(h-->0) (b^x b^h - b^x)/h = lim_(h-->0) b^x (b^h - 1)/h = b^x lim_(h-->0) (b^h - b^0)/h`
At this point notice that `lim_(h-->0) (b^h - b^0)/h = lim_(h-->0) (f(0 + h) - f(0))/h = f'(0)`, which is, by definition, `m_b`.
Thus we have `f'(x) = m_b f(x)` as needed.
Finally, if `b = e`, then `m_b = m_e = 1` and `f'(x) = f(x)`. The derivative of `e^x` is just itself, `e^x`!