What is the derivative of b^x?

If f(x) = b^x, then f'(x) = b^x \ln x.

Let f(x) = b^x. Using the definition of the derivative, we have

    \begin{align*} f'(x) &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}\\ &= \lim_{\Delta x \to 0} \frac{b^{x + \Delta x} - b^x}{\Delta x}\\ &= \lim_{\Delta x \to 0} \frac{b^x b^{\Delta x} - b^x}{\Delta x}\\ &= \lim_{\Delta x \to 0} \frac{b^x (b^{\Delta x} - 1)}{\Delta x}\\ &= \lim_{\Delta x \to 0} b^x \underbrace{\lim_{\Delta x \to 0} \frac{b^{\Delta x} - 1}{\Delta x}}_{\ln x}\\ &= b^x \ln b. \end{align*}

It would be nice if \displaystyle\lim_{\Delta x \to 0} \frac{b^{\Delta x} - 1}{\Delta x} = 1. Well, this occurs for a special value of b; we let b = e \approx 2.718.

This leads to the following:

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


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