Derivation of Exponential Functions
Let’s consider an exponential function:
\[y(x) = b^x\]We want to measure the rate of change of this function with respect to \(x\).
Let’s pull a page from the Limits and Derivatives and apply the core logic to get the derivative:
\[\begin{aligned} y'(x) &= \lim_{h \to 0} \frac{y(x+h) - y(x)}{h} \\[1em] &= \lim_{h \to 0} \frac{b^{(x+h)} - b^{(x)}}{h} \end{aligned}\]We cannot apply the Binomial Theorem, because we have a varying power, not a varying base.
(A refresher on why this is different from the Power Rule):
If we have \(y(x) = x^n\) (like \(x^2\)), the base \(x\) is the variable.
\[\begin{aligned} y'(x) &= \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \\[1em] &= \lim_{h \to 0} \frac{\cancel{x^2}+2xh+h^2 - \cancel{x^2}}{h} \\[1em] &= \lim_{h \to 0} \frac{\cancel{h}(2x+h)}{\cancel{h}} \\[1em] &= \lim_{h \to 0} (2x+\overbrace{h}^{\to 0}) = 2x \end{aligned}\]This works because we can use the Binomial Theorem on \((x+h)^2\) and it fails for \(b^{x+h}\)
No worries, we just use a different perspective for \(b^{x+h}\):
\[\begin{aligned} y'(x) &= \lim_{h \to 0} \frac{b^{(x+h)} - b^{x}}{h} \\[1em] &= \lim_{h \to 0} \frac{b^{x} \cdot b^{h} - b^{x}}{h} \\[1em] &= \lim_{h \to 0} \frac{b^{x}(b^{h} - 1)}{h} \end{aligned}\]As we can see, the limit \(\lim_{h \to 0}\) only affects \(h\). The term \(b^x\) has no \(h\), so we can pull it out as a constant:
\[y'(x) = b^{x} \cdot \underbrace{\left( \lim_{h \to 0} \frac{b^h - 1}{h} \right)}_{C_b}\]Let’s assign \(C_b\) for this constant limit.
\[\frac{d}{dx}(b^x) = y'(x) = b^{x} \cdot C_b\]This result is quite remarkable: the derivative of an exponential function (its rate of change) is the function’s value at that point, times a constant value \(C_b\) that depends only on the base \(b\).
Since the derivative gives us the slope, we can say the slope of the function at any point \(x\) is \(b^x \cdot C_b\). To better understand what this constant \(C_b\) represents, let’s evaluate the slope at the specific point where \(x=0\):
\[\text{Slope at } x=0 = y'(0) = b^0 \cdot C_b = 1 \cdot C_b = C_b\]So, \(C_b\) is the slope of the function at \(x=0\).
We can conclude that this function behaves in the way that no matter the \(x\), the derivative of this function is simply the evaluated current value (\(b^x\)) times its starting slope (\(C_b\)).
This constant \(C_b\) is different for every base. Let’s look at the term once again:
\[C_b = \lim_{h \to 0} \frac{b^h - 1}{h}\]Let’s test this with a small value for \(h\), like \(h=0.0001\):
For \(b=2\): \(C_2 \approx \frac{2^{0.0001} - 1}{0.0001} \approx \frac{1.000069317 - 1}{0.0001} \approx 0.693 < 1\)
For \(b=3\): \(C_3 \approx \frac{3^{0.0001} - 1}{0.0001} \approx \frac{1.000109867 - 1}{0.0001} \approx 1.098 > 1\)
hmmm… so there must exist a base somewhere between 2 and 3 where the slope \(C_b\) is exactly 1.
This special number is \(e \approx 2.718...\). (Learn on how this value came to be here).
To summarize:
- For any base \(b\): \(\frac{d}{dx}(b^x) = C_b \cdot b^x\) (where \(C_b = \lim_{h \to 0} \frac{b^h - 1}{h}\))
- For the special base \(e\): \(\frac{d}{dx}(e^x) = 1 \cdot e^x = e^x\)