## 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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