## Content

### Derivatives of general exponential functions

Since we can now differentiate \(e^x\), using our knowledge of differentiation we can also differentiate other functions.

In particular, we can now differentiate functions of the form \(f(x) = e^{kx}\), where \(k\) is a real constant. From the chain rule, we obtain

\[ f'(x) = k \, e^{kx}. \]We saw in the previous section, when differentiating \(2^x\), that it can be written as \(e^{\log_e 2 \cdot x}\), which is of the form \(e^{kx}\). The same technique can be used to differentiate any function \(a^x\), where \(a\) is a positive real number. A function \(a^x\) is just a function of the form \(e^{kx}\) in disguise.

Writing \(a\) as \(e^{\log_e a}\), we can rewrite \(f(x) = a^x\), using the index laws, as

\[ f(x) = a^x = \bigl( e^{\log_e a} \bigr)^x = e^{x \, \log_e a}. \]The function is then in the form \(e^{kx}\) (with \(k=\log_e a\)) and differentiating gives

\[ f'(x) = \log_e a \cdot a^x. \]So the derivative of \(a^x\) is a constant times itself, and that constant is \(\log_e a\).

Exercise 8

Use what we've done so far to explain why, for any \(a>0\),

\[ \lim_{h \to 0} \dfrac{a^h - 1}{h} = \log_e a. \]Exercise 9

Let \(f(x) = x^x\), for \(x>0\). Differentiate \(f(x)\) and find its stationary points.

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