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    <name>
      <text>18.2.1.2 First partial derivatives - y derivative</text>
    </name>
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      <text><![CDATA[<p></p>
<p><strong>The \(y\) partial derivative</strong></p>
<p>For functions of two variables \(f(x,y)\) the \(x\) and \(y\) variables are on the same footing, so what we have done for the \(x\)-variable we can do for the \(y\)-variable. We can thus imagine keeping the \(x\)-variable fixed and determining the rate
    of change of \(f\) as \(y\) changes. This rate of change is denoted by \(\dfrac{\partial f}{\partial y}\).</p>

<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p><strong>The Partial Derivative of \(f\) with respect to \(y\)</strong></p>
    <p> For a function of two variables \(z=f(x,y)\) the partial derivative of \(f\) with respect to \(y\) is denoted by \(\dfrac{\partial f}{\partial y}\) and is obtained by differentiating \(f(x,y)\) with respect to \(y\) in the usual way but treating the
        \(x\)-variable as if it were a constant.</p>
    <p>Alternative notations for \(\dfrac{\partial f}{\partial y}\) are \(f_y(x,y)\) or \(f_y\) or \(\dfrac{\partial z}{\partial y}\).
    </p>
</div>
<p><br></p>
<p>Returning to \(f(x,y)=x^3+2x^2y+y^2+2x+1\) once again, we therefore obtain:</p>
\[\begin{aligned} \frac{\partial f}{\partial y} &amp;= 0+2x^2 \cdot 1+2y+0+0 \\ &amp;= 2x^2+2y \end{aligned}\]

<hr>
<h4 class="HELM_example">Example</h4>
<ol class="HELM_parts_inline">
    <p>Find \(\dfrac{\partial f}{\partial y}\) for</p>
    <li> \(f(x,y)=x^3+x+y^2+y\), </li>
    <li> \(f(x,y)=x^2y+xy^3\).</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>\(\dfrac{\partial f}{\partial y}=0+0+2y+1=2y+1\)</li>
    <li>\(\dfrac{\partial f}{\partial y}=x^2\cdot 1+x\cdot 3y^2=x^2+3xy^2\)</li>
</ol>
<hr>

<p>We can calculate the partial derivative of \(f\) with respect to \(x\) and the value of \(\dfrac{\partial f}{\partial x}\) at a specific point e.g. \(x=1, y=-2\).</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>[Remember \(f_x\) means \(\dfrac{\partial f}{\partial x}\) and \(f_y\) means \(\dfrac{\partial f}{\partial y}\).]
</p>
<p>Find \(f_x(1,-2)\) and \(f_y(-3,2)\) for \(f(x,y)=x^2+y^3+2xy\)</p>

<h4 class="HELM_solution">Solution</h4>
<p>\(f_x(x,y)=2x+2y\), so \(f_x(1,-2)=2-4=-2\);</p>
<p>\(f_y(x,y)=3y^2+2x,\) so \(f_y(-3,2)=12-6=6\).</p>

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