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      <text>10.1.1 (Derivatives from first principals)</text>
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      <text><![CDATA[<h1>Derivatives from first principals</h1>
<p>In the previous section, we have seen how we can calculate instantaneous rates of change. We have seen that calculating the gradient of the tangent at a point on the curve corresponds to the instantaneous rate of change at that point. This is actually the physical interpretation of the definition of the derivative. In this section we will focus on deriving the derivative in a more formal manner from first principals.</p>
<p>The derivative of a function at a point \(x\) is found by looking at how the values of the function change at that point. However, what does it mean for something to change at a single point? This is an issue which vexed some of the greatest mathematicians for a very long time. This issue was not resolved until the nineteenth century and the development of a rigorous theory of analysis.  We shall not pursue this here, as it is beyond the scope of this course.</p>
<p>We have seen that the rate of change of a linear function \(f(x) = mx + c\) is the role of the gradient \(m\).  If we change my \(x\) value by \(\delta x\) units, then the \(y\) value changes by \(m\delta x\) units. Let us fix an increment \(\delta x\) by which we change our \(x\) value.  Then the value \(\delta y\) by which the \(y\) value changes is given by 
\[ 
\delta y = m \delta x. 
\] 
Rearranging this we see that the gradient \(m\) of a linear function satisfies the formula \[ 
m = \frac{\delta y }{\delta x}. 
\]
</p>
<p>Our approach to interpreting the rate of a change of a nonlinear function, is inspired by this approach to linear functions.  Let us consider evaluating the derivative at a point \(x\) where the value of the function \(f\) is \(f(x) = y\).  We again fix a certain value \(\delta x\) by which we increment our \(x\) value.  This determines a change \(\delta y\) in our \(y\) value: \(f(x + \delta x) = f(x) + \delta y = y + \delta y\).  If this were a linear function, then the gradient would be given by 
\[ 
\frac{\delta y}{\delta x} = \frac{f(x + \delta x) - f(x)}{\delta x}. 
\] 
But our function is not linear. The trick is to observe that, for some functions at least, if we zoom in on the graph about \(x\) it will look more and more linear. So as we make \(\delta x\) smaller, the value \(\dfrac{f(x + \delta x) - f(x)}{\delta x}\) will become a better approximation for the rate of change of \(f\) at \(x\). The tangent line to the graph of \(f\) at the point \((x, f(x))\) is the graph of the linear function which best approximates \(f\) at the point \(x\), and its gradient is the limiting value of \(\dfrac{f(x + \delta x) - f(x)}{\delta x}\) as \(\delta x \) goes to \(0\). This value is a natural interpretation of the rate of change of \(f\) at the point \(x\).  </p>
<p>The following demonstrates this idea. (note that we we write, as is common, \(h\) instead of \(\delta x\)). As \(h\) gets smaller, the line through \(P\) and \(Q\) better approximates the tangent to the curve at \(P\).  Let \(h\) get closer to \(0\) to see this.  You can also change the point \(x_0\), and the definition of the function \(f\) itself.</p>
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