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      <text>11.7.3 Second order imp diff</text>
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      <text><![CDATA[<h4>Second order implicit differentiation</h4>
We sometimes need to obtain the second derivative \( \dfrac{d^2y}{dx^2}\) for a function defined implicitly.
<hr>
<h4 class="HELM_example">Example</h4>
<p>Obtain \( \dfrac{dy}{dx} \) and \( \dfrac{d^2y}{dx^2}\) at the point \( (4, 2)\) on the curve defined by the equation 
\[ x^2 - xy - y^2 - 2y = 0\]</p>
<h4 class="HELM_solution">Solution</h4>
Firstly we obtain \( \dfrac{dy}{dx} \) by differentiating the equation implicitly and then evaluate it at \( (4, 2)\).
<p></p>
We have
\[ 2x -x\dfrac{dy}{dx} - y -2y\dfrac{dy}{dx} -2\dfrac{dy}{dx} = 0 \hspace{4cm} (1)\]
from which
\[ \dfrac{dy}{dx} = \dfrac{2x - y}{x + 2y + 2} \hspace{5cm} (2) \]
so at \( (4,2)\),
\[ \dfrac{dy}{dx} = \dfrac{6}{10} = \dfrac{3}{5}. \]
to obtain the second derivative  \( \dfrac{d^2y}{dx^2}\) it is easier to use \( (1) \) than \( (2) \) becuase the latter is a quotient. We simplify \( (1)\) first:
\[ 2x - y - (x + 2y + 2)\dfrac{dy}{dx} = 0. \hspace{5cm} (3)\]
We will have to use the product rule to differentiate the third term here.
<p></p>
Hence differentiating \( (3) \) with respect to \(x\):
\[ 2 - \dfrac{dy}{dx} - (x + 2y + 2) \dfrac{d^2y}{dx^2} - (1 + 2\dfrac{dy}{dx})\dfrac{dy}{dx} = 0\]
or
\[ 2 - 2\dfrac{dy}{dx} - 2\left( \dfrac{dy}{dx} \right)^2 - (x + 2y + 2)  \dfrac{d^2y}{dx^2} = 0. \hspace{4cm} (4)\]
Note carefully that the third term here, \( \left( \dfrac{dy}{dx} \right)^2\), is the square of the first derivative. It should not be confused with the second derivative denoted by \( \dfrac{d^2y}{dx^2}\).
<p></p>
Finally, at \( (4,2)\), where \(\dfrac{dy}{dx} = \dfrac{3}{5}\) we obtain from \( (4) \):
\[ 2 - 2\left(\dfrac{3}{5}\right) - 2\left(\dfrac{9}{25}\right) - (4+4+2)\dfrac{d^2y}{dx^2} = 0\]
from which
\[ \dfrac{d^2y}{dx^2} = \dfrac{1}{125}\]
at \( (4,2)\).]]></text>
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