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    <name>
      <text>10.1.14 (Derivatives of exp and log)</text>
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      <text><![CDATA[<h3>Derivatives of exponentials</h3>

<p>Suppose \(f(x) = a^x\) for some constant \(a&gt;0\).&nbsp; Now the variable \(x\) is in the power, not the base, so our usual formula for differentiating powers of \(x\) does not apply: the functions \(x^2\) and \(2^x\) are two very different functions.&nbsp; How then can we differentiate this function \(f\)?&nbsp; Let us consider the definition of the derivative.&nbsp; Using the laws of indices, we have that 
\[
\begin{align*}
f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} 
&amp;= \lim_{h \to 0}\frac{a^{x+h} - a^x}{h}\\
&amp;= \lim_{h \to 0}\frac{a^x a^h - a^x}{h} \\
&amp;= \lim_{h \to 0}\frac{a^x ( a^h - 1)}{h} \\
&amp;= a^x \lim_{h \to 0}\frac{a^h - 1}{h}.
\end{align*}
\]
Here we have been able to move the \(a^x\) outside the limit, because it does not depend on \(h\), which is the variable in which the limit is taken.&nbsp; The point is that the value of this limit of difference quotients depends on the value of the limit 
\[
\lim_{h \to 0}\frac{a^h - 1}{h},
\] 
which we notice is exactly the definition of the value of the derivative of \(f\) at \(0\):
\[
f'(0) = \lim_{h \to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h \to 0}\frac{a^{0+h} - a^0}{h} = \lim_{h \to 0}\frac{a^h - 1}{h}.
\]
So in general, for \(f(x)=a^x\) we have that
\[
f'(x) = a^x f'(0) = f(x)f'(0),
\] 
where \(f'(0)\) is a fixed value that depends only on the value of our base \(a\).&nbsp; This in particular gives a rigorous demonstration of a characteristic feature of exponential growth: that the rate of change (the derivative) depends on the value of the function itself.&nbsp; Thus (unregulated) population size typically behaves exponentially, since the larger a population, the faster it grows, since there are more individuals reproducing.</p>
<br/>
<p>The number \(e \approx 2.71828\) which met met in detail in week 4, satisfies 
\[
\lim_{h \to 0}\frac{e^h - 1}{h} = 1.
\]
In fact this is one way to define the value \(e\).&nbsp; So for the natural exponential function \(y = e^x\), we have \(\dfrac{dy}{dx} = e^x\).&nbsp; In other words, the derivative is equal to the function itself.</p>
\[ 
\mbox{Suppose } f(x) = e^x. \mbox{ Then } f'(x) = e^x = f(x).
\]
</p>
<br/>
<h3>Derivative of the logarithm</h3>

<p>The natural logarithm \(f(x) = \ln (x)\) is defined to be the inverse of the exponential function \(e^x\).&nbsp; Since \(e^x\) only produces positive numbers, \(\ln(x)\) is only defined for positive numbers, and is defined such that the equation 
\[
e^{\ln(x)} = x
\] 
holds for all \(x&gt; 0\).
<p>
\[
\mbox{Suppose } f(x) = \ln(x). \mbox{ Then } f'(x) = 1/x.
\]</p>]]></text>
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