What is the derivative of $b^x$ ?

29 March 2018 in MAE 113: Calculus for the Liberal Arts | No comments

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 $b = e \approx 2.718$ .