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      <text>11.5.2 The derivative of a function of a function</text>
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      <text><![CDATA[<h4>The derivative of a function of a function</h4>
<p></p>
To differentiate a function of a function we use the following Key Point:

<div class="HELM_keypoint">
<h4>Key Point</h4>
<b>The Chain rule</b>
<p></p>
If \(y = f(g(x))\), that is, a function of a function, then 
\[ \dfrac{dy}{dx} = \dfrac{df}{dg} \times \dfrac{dg}{dx}\]
This is called the <b>chain rule</b>.
</div>

<hr>
<h4 class="HELM_example">Example</h4>
<p> Find the derivatives of the following composite functions using the chain rule and
check the result using other methods  </p>
<ol class="HELM_parts_inline">
<li>\( (2x^2 - 1)^2 \)</li>
<li>\( \ln e^x\)</li>
</ol>


<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
  <li>Here \(y = f(g(x))\) where \(f(g) = g^2\) and \(g(x) = 2x^2 - 1\). Thus
\[\dfrac{df}{dg} = 2g \text{   and   }  \dfrac{dg}{dx} = 4x\] 
\[\therefore  \dfrac{dy}{dx} = 2g(4x) = 2(2x^2 - 1)(4x) = 8x(2x^2 - 1)\]
<p></p>
This result is easily checked by using the rule for differentiating products:
<p></p>
\(y = (2x^2-1)(2x^2-1)\) so \(\dfrac{dy}{dx} = 4x(2x^2-1)+(2x^2-1)(4x) = 8x(2x^2-1)\) as obtained above.
</li>


  <li>Here \(y = f(g(x))\) where \(f(g) = \ln g\) and \(g(x) = e^x\). Thus
\[ \dfrac{df}{dg} = \dfrac{1}{g} \text{   and   } \dfrac{dg}{dx} = e^x\]
\[ \therefore \dfrac{dy}{dx} = \dfrac{1}{g}&nbsp; e^x = \dfrac{1}{e^x}&nbsp; e^x = 1\]
<p></p>
This is easily checked since, of course, \(y = \ln e^x = x\) and so, obviously \( \dfrac{dy}{dx} = 1\) as obtained above.</li>
</ol>]]></text>
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