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Subsection 2.10.1 Back to \(\mathbf{\diff{}{x} a^x}\)

We can also now finally get around to computing the derivative of \(a^x\) (which we started to do back in Section 2.7).

\begin{align*} f(x) &= a^x & \text{take log of both sides}\\ \log f(x) &= x \log a & \text{exponentiate both sides base $e$}\\ f(x) &= e^{x \log a} & \text{chain rule}\\ f'(x) &= e^{x \log a} \cdot \log a\\ &= a^x \cdot \log a \end{align*}

Notice that we could have also done the following:

\begin{align*} f(x) &= a^x & \text{take log of both sides}\\ \log f(x) &= x \log a & \text{differentiate both sides}\\ \diff{}{x} \left( \log f(x) \right) &= \log a\\ \end{align*}

We then process the left-hand side using the chain rule

\begin{align*} f'(x) \cdot \frac{1}{f(x)} &= \log a\\ f'(x) &= f(x) \cdot \log a = a^x \cdot \log a \end{align*}

We will see \(\diff{}{x} \log f(x)\) more below in the subsection on “logarithmic differentiation”.

To summarise the results above:

Recall that we need the caveat \(a \neq 1\) because the logarithm base 1 is not well defined. This is because \(1^x = 1\) for any \(x\text{.}\) We do not need a similar caveat for the derivative of the exponential because we know (recall Example 2.7.1)

\begin{align*} \diff{}{x} 1^x &= \diff{}{x} 1= 0 &\text{while the above corollary tells us}\\ &= \log 1 \cdot 1^x = 0 \cdot 1 = 0. \end{align*}