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335 lines
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<h1 class="center">Implicit Differentiation</h1>
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<p class="center"><i>Finding the derivative when you can’t solve for y<br>
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</i></p>
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<p class="center"><img src="images/imply.jpg" alt="What are you trying to imply?" height="266" width="509"></p>
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<p>You may like to read <a href="derivatives-introduction.html">Introduction to Derivatives</a> and <a href="derivatives-rules.html">Derivative Rules</a> first.</p>
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<h2>Implicit vs Explicit</h2>
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<p>A function can be explicit or implicit:</p>
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<p><b>Explicit</b>: "y = some function of x". <i>When we know x we can calculate y directly.</i></p>
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<p><b>Implicit</b>: "some function of y and x equals something else". <i>Knowing x does not lead directly to y.</i></p>
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<div class="example">
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<h3>Example: A Circle</h3>
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<table align="center" cellpadding="5">
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<tbody>
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<tr style="text-align:center;">
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<td><b>Explicit Form</b></td>
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<td style="width:20px;"> </td>
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<td><b>Implicit Form</b></td>
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</tr>
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<tr style="text-align:center;">
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<td class="large" style="white-space:nowrap">y = ± √ (r<sup>2</sup> − x<sup>2</sup>)</td>
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<td> </td>
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<td class="large" style="white-space:nowrap">x<sup>2</sup> + y<sup>2</sup> = r<sup>2</sup></td>
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</tr>
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<tr style="text-align:center;">
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<td>In this form, y is expressed<br>
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as a function of x.</td>
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<td> </td>
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<td>In this form, the function is<br>
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expressed in terms of both y and x.</td>
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</tr>
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</tbody></table>
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<p class="center"><img src="images/graph-x2-y2-9.gif" alt="graph x^2 + y^2 = 9, a circle" height="194" width="223"><br>
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The <a href="../data/grapher-equation1bdb.html?func1=x%5E2+y%5E2=9&xmin=-5.000&xmax=5.000&ymin=-3.750&ymax=3.750">graph</a> of x<sup>2</sup> + y<sup>2</sup> = 3<sup>2</sup></p>
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</div>
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<h2>How to do Implicit Differentiation</h2>
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<ul>
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<li>Differentiate with respect to x</li>
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<li>Collect all the <span class="intbl"> <em>dy</em> <strong>dx</strong> </span> on one side</li>
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<li>Solve for <span class="intbl"> <em>dy</em> <strong>dx</strong> </span></li>
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</ul>
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<div class="example">
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<h3>Example: x<sup>2</sup> + y<sup>2</sup> = r<sup>2</sup></h3>
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<p>Differentiate with respect to x:</p>
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<p class="center large"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(x<sup>2</sup>) + <span class="intbl"> <em>d</em> <strong>dx</strong> </span>(y<sup>2</sup>) = <span class="intbl"> <em>d</em> <strong>dx</strong> </span>(r<sup>2</sup>)</p>
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<p>Let's solve each term:</p>
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<div class="tbl">
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<div class="row"><span class="left">Use the <a href="derivatives-rules.html">Power Rule</a>:</span><span class="right"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(x<sup>2</sup>) = 2x</span></div>
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<div class="row"><span class="left">Use the Chain Rule (explained below):</span><span class="right"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(y<sup>2</sup>) = 2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span></span></div>
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<div class="row"><span class="left">r<sup>2</sup> is a constant, so its derivative is 0:</span><span class="right"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(r<sup>2</sup>) = 0</span></div>
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</div>
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<p>Which gives us:</p>
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<p class="center large">2x + 2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = 0</p>
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<p>Collect all the <span class="intbl"> <em>dy</em> <strong>dx</strong> </span> on one side</p>
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<p class="center large">y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = −x</p>
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<p>Solve for <span class="intbl"> <em>dy</em> <strong>dx</strong> </span>:</p>
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<p class="center larger"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = <span class="intbl"> <em>−x</em> <strong>y</strong> </span></p>
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</div>
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<div class="def">
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<h3>The Chain Rule Using <span class="intbl"> <em>dy</em> <strong>dx</strong> </span></h3>
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<p>Let's look more closely at how <span class="large"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(y<sup>2</sup>)</span> becomes <span class="large">2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span></span></p>
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<p>The Chain Rule says:</p>
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<p class="center large"><span class="intbl"> <em>du</em> <strong>dx</strong> </span> = <span class="intbl"> <em>du</em> <strong>dy</strong> </span><span class="intbl"> <em>dy</em> <strong>dx</strong> </span></p>
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<p>Substitute in <span class="large">u = y<sup>2</sup></span>:</p>
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<p class="center large"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(y<sup>2</sup>) = <span class="intbl"> <em>d</em> <strong>dy</strong> </span>(y<sup>2</sup>)<span class="intbl"> <em>dy</em> <strong>dx</strong> </span></p>
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<p>And then:</p>
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<p class="center large"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(y<sup>2</sup>) = 2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span></p>
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<h3>Basically, all we did was differentiate with respect to y and multiply by <span class="intbl"> <em>dy</em> <strong>dx</strong> </span></h3>
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</div>
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<p>Another common notation is to use <span class="hilite">’</span> to mean <span class="intbl"> <em>d</em> <strong>dx</strong> </span></p>
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<div class="def">
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<h3>The Chain Rule Using <span class="center large">’</span></h3>
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<p>The Chain Rule can also be written using <span class="center large">’</span> notation:</p>
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<p class="center large">f(g(x))’ = f’(g(x))g’(x)</p>
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<p>g(x) is our function "y", so:</p>
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<p class="center large">f(y)’ = f’(y)y’</p>
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<p>f(y) = <span class="large"> y<sup>2</sup></span>, so f<span class="center large">’</span>(y) = 2y:</p>
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<p class="center large">f(y)’ = 2yy’</p>
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<p class="center large">or alternatively: f(y)’ = 2y <span class="intbl"> <em>dy</em> <strong>dx</strong> </span></p>
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<h3>Again, all we did was differentiate with respect to y and multiply by <span class="intbl"> <em>dy</em> <strong>dx</strong> </span></h3>
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</div>
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<h2>Explicit</h2>
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<p>Let's also find the derivative using the <b>explicit</b> form of the equation.</p>
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<ul>
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<li>To solve this explicitly, we can solve the equation for y</li>
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<li>Then differentiate</li>
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<li>Then substitute the equation for y again</li>
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</ul>
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<div class="example">
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<h3>Example: x<sup>2</sup> + y<sup>2</sup> = r<sup>2</sup></h3>
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<div class="tbl">
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<div class="row"><span class="left">Subtract x<sup>2</sup> from both sides:</span><span class="right">y<sup>2</sup> = r<sup>2</sup> − x<sup>2</sup></span></div>
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<div class="row"><span class="left">Square root:</span><span class="right">y = ±√(r<sup>2</sup> − x<sup>2</sup>) </span></div>
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<div class="row"><span class="left">Let's do <b>just the positive</b>: </span><span class="right">y = √(r<sup>2</sup> − x<sup>2</sup>) </span></div>
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<div class="row"><span class="left">As a power: </span><span class="right">y = (r<sup>2</sup> − x<sup>2</sup>)<sup>½</sup></span></div>
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<div class="row"><span class="left"><b>Derivative (Chain Rule)</b>:</span><span class="right">y<span class="center large">’</span> =½(r<sup>2</sup> − x<sup>2</sup>)<sup>−½</sup>(<span class="center large">−</span>2x)</span></div>
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<div class="row"><span class="left">Simplify:</span><span class="right">y<span class="center large">’</span> = −x(r<sup>2</sup> − x<sup>2</sup>)<sup>−½</sup></span></div>
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<div class="row"><span class="left">Simplify more:</span><span class="right">y<span class="center large">’</span> = <span class="center large"><span class="intbl"> <em>−x</em> <strong>(r<sup>2</sup> − x<sup>2</sup>)<sup>½</sup></strong> </span></span></span></div>
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<div class="row"><span class="left">Now, because <b>y = (r<sup>2</sup> − x<sup>2</sup>)<sup>½</sup></b>: </span><span class="right">y<span class="center large">’</span> = −x/y</span></div>
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</div>
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<p>We get the same result this way!</p>
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<p>You can try taking the derivative of the negative term yourself.</p>
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</div>
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<div class="def">
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<h3>Chain Rule Again!</h3>
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<p>Yes, we used the Chain Rule again. Like this (note different letters, but same rule):</p>
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<p class="center large"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = <span class="intbl"> <em>dy</em> <strong>df</strong> </span><span class="intbl"> <em>df</em> <strong>dx</strong> </span></p>
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<p>Substitute in <span class="large">f = (r<sup>2</sup> − x<sup>2</sup>)</span>:</p>
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<p class="center large"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(f<sup>½</sup>) = <span class="intbl"> <em>d</em> <strong>df</strong> </span>(f<sup>½</sup>)<span class="intbl"> <em>d</em> <strong>dx</strong> </span><span class="intbl"> </span><span class="large">(r<sup>2</sup> − x<sup>2</sup>)</span></p>
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<p>Derivatives:</p>
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<p class="center large"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(f<sup>½</sup>) = ½(f<sup>−½</sup>) <span class="large">(−2x)</span></p>
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<p>And substitute back <span class="large">f = (r<sup>2</sup> − x<sup>2</sup>)</span>:</p>
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<p class="center large"><span class="intbl"> <em>d</em> <strong>dx</strong> </span><span class="large">(r<sup>2</sup> − x<sup>2</sup>)</span><sup>½</sup> = ½(<span class="large">(r<sup>2</sup> − x<sup>2</sup>)</span><sup>−½</sup>) (−2<span class="large">x)</span></p>
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<p>And we simplified from there.</p>
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</div>
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<h2>Using The Derivative</h2>
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<p>OK, so why find the derivative <span class="large">y’ = −x/y </span>?</p>
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<p>Well, for example, we can find the slope of a tangent line.</p>
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<div class="example">
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<h3>Example: what is the slope of a circle centered at the origin with a radius of 5 at the point (3, 4)?</h3>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/graph-x2-y2-25.gif" alt="graph x^2 + y^2 = 25 with tangent line" height="319" width="261"></p>
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<p>No problem, just substitute it into our equation:</p>
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<p class="center"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = −x/y</p>
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<p class="center"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = −3/4</p>
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<p>And for bonus, the equation for the tangent line is:</p>
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<p class="center">y = −3/4 x + 25/4</p>
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</div>
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<h2>Another Example</h2>
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<p>Sometimes the <b>implicit</b> way works where the explicit way is hard or impossible.</p>
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<div class="example">
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<h3>Example: 10x<sup>4</sup> − 18xy<sup>2</sup> + 10y<sup>3</sup> = 48</h3>
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<p>How do we solve for y? We don't have to!</p>
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<ul>
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<li>First, differentiate with respect to x (use the Product Rule for the xy<sup>2</sup> term).</li>
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<li>Then move all dy/dx terms to the left side.</li>
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<li>Solve for dy/dx</li>
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</ul>
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<p>Like this:</p>
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<div class="tbl">
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<div class="row"><span class="left">Start with:</span><span class="right">10x<sup>4</sup> − 18xy<sup>2</sup> + 10y<sup>3</sup> = 48</span></div>
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<div class="row"><span class="left"><b>Derivative</b>:</span><span class="right">10 (4x<sup>3</sup>) − 18(x(2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span>) + y<sup>2</sup>) + 10(3y<sup>2</sup><span class="intbl"> <em>dy</em> <strong>dx</strong> </span>) = 0</span></div>
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<div class="row"><span class="left"></span><span class="right">
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<p>(the middle term is explained<br>
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in "Product Rule" below)</p></span></div>
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<div class="row"><span class="left">Simplify:</span><span class="right">40x<sup>3</sup> − 36xy<span class="intbl"> <em>dy</em> <strong>dx</strong> </span> − 18y<sup>2</sup> + 30y<sup>2</sup><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = 0</span></div>
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<div class="row"><span class="left"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> on left:</span><span class="right">−36xy<span class="intbl"> <em>dy</em> <strong>dx</strong></span> + 30y<sup>2</sup><span class="intbl"><em>dy</em> <strong>dx</strong> </span> = −40x<sup>3</sup> + 18y<sup>2</sup></span></div>
|
||
<div class="row"><span class="left">Simplify :</span><span class="right">(30y<sup>2</sup>−36xy)<span class="intbl"> <em>dy</em> <strong>dx</strong> </span>= 18y<sup>2</sup> − 40x<sup>3</sup></span></div>
|
||
<div class="row"><span class="left">Simplify :</span><span class="right">3(5y<sup>2</sup>−6xy)<span class="intbl"> <em>dy</em> <strong>dx</strong> </span>= 9y<sup>2</sup> − 20x<sup>3</sup></span></div>
|
||
</div>
|
||
<p>And we get:</p>
|
||
<div class="center large"><span class="intbl"><em>dy</em><strong>dx</strong></span> = <span class="intbl"><em>9y<sup>2</sup> − 20x<sup>3</sup></em><strong>3(5y<sup>2</sup> − 6xy)</strong></span></div>
|
||
<!-- dy/dx = 9y^2~-20x^3/3(5y^2~-6xy) -->
|
||
<p> </p>
|
||
|
||
<h3>Product Rule</h3>
|
||
<p>For the middle term we used the Product Rule: (fg)’ = f g’ + f’ g</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">(xy<sup>2</sup>)’ </span><span class="right">= x(y<sup>2</sup>)’ + (x)’y<sup>2</sup></span></div>
|
||
<div class="row"><span class="left"> </span><span class="right">= x(2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span>) + y<sup>2</sup></span></div>
|
||
</div>
|
||
<p>Because (y<sup>2</sup>)’ = 2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span>(we worked that out in a previous example)</p>
|
||
<p>Oh, and <span class="intbl"><em>dx</em><strong>dx</strong></span> = 1, in other words x’ = 1</p>
|
||
</div>
|
||
|
||
|
||
<h2>Inverse Functions</h2>
|
||
|
||
<p>Implicit differentiation can help us solve inverse functions.</p>
|
||
<p>The general pattern is:</p>
|
||
<ul>
|
||
<li>Start with the inverse equation in explicit form. Example: y = sin<sup>−1</sup>(x)</li>
|
||
<li>Rewrite it in non-inverse mode: Example: x = sin(y)</li>
|
||
<li>Differentiate this function with respect to x on both sides.</li>
|
||
<li>Solve for dy/dx</li>
|
||
</ul>
|
||
<p>As a final step we can try to simplify more by substituting the original equation.</p>
|
||
<p>An example will help:</p>
|
||
<div class="example">
|
||
|
||
<h3>Example: the inverse sine function y = sin<sup>−1</sup>(x)</h3>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Start with:</span><span class="right">y = sin<sup>−1</sup>(x)</span></div>
|
||
<div class="row"><span class="left">In non−inverse mode:</span><span class="right">x = sin(y)</span></div>
|
||
<div class="row"><span class="left"><b>Derivative</b>:</span><span class="right"><span class="intbl"> <em>d</em> <strong>dx</strong> </span>(x) = <span class="intbl"> <em>d</em> <strong>dx</strong> </span>sin(y)</span></div>
|
||
<div class="row"><span class="left"> </span><span class="right">1 = cos(y) <span class="intbl"> <em>dy</em> <strong>dx</strong> </span></span></div>
|
||
<div class="row"><span class="left">Put <span class="intbl"> <em>dy</em> <strong>dx</strong> </span> on left:</span><span class="right"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span>= <span class="intbl"> <em>1</em> <strong>cos(y)</strong> </span></span></div>
|
||
</div>
|
||
<p>We can also go one step further using the Pythagorean identity:</p>
|
||
<p class="center">sin<sup>2</sup> y + cos<sup>2</sup> y = 1</p>
|
||
<p class="center">cos y = √(1 − sin<sup>2</sup> y )</p>
|
||
<p>And, because sin(y) = x (from above!), we get:</p>
|
||
<p class="center">cos y = √(1 − x<sup>2</sup>)</p>
|
||
<p>Which leads to:</p>
|
||
<p class="center larger"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span>= <span class="intbl"> <em>1</em> <strong>√(1 − x<sup>2</sup>)</strong> </span></p>
|
||
</div>
|
||
<div class="example">
|
||
|
||
<h3>Example: the derivative of square root √x</h3>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Start with:</span><span class="right">y = √x</span></div>
|
||
<div class="row"><span class="left">So:</span><span class="right">y<sup>2</sup> = x</span></div>
|
||
<div class="row"><span class="left"><b>Derivative</b>:</span><span class="right">2y<span class="intbl"> <em>dy</em> <strong>dx</strong> </span>= 1</span></div>
|
||
<div class="row"><span class="left">Simplify:</span><span class="right"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = <span class="intbl"> <em>1</em> <strong>2y</strong> </span></span></div>
|
||
<div class="row"><span class="left">Because y = √x:</span><span class="right"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = <span class="intbl"> <em>1</em> <strong>2√x</strong> </span></span></div>
|
||
</div>
|
||
<p>Note: this is the same answer we get using the Power Rule:</p>
|
||
<div class="tbl">
|
||
<div class="row"><span class="left">Start with:</span><span class="right">y = √x</span></div>
|
||
<div class="row"><span class="left">As a power:</span><span class="right">y = x<sup>½</sup></span></div>
|
||
<div class="row"><span class="left">Power Rule <span class="intbl"> <em>d</em> <strong>dx</strong> </span>x<sup>n</sup> = nx<sup>n−1</sup>:</span><span class="right"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = (½)x<sup>−½</sup></span></div>
|
||
<div class="row"><span class="left">Simplify:</span><span class="right"><span class="intbl"> <em>dy</em> <strong>dx</strong> </span> = <span class="intbl"> <em>1</em> <strong>2√x</strong> </span></span></div>
|
||
</div>
|
||
</div>
|
||
|
||
|
||
<h2>Summary</h2>
|
||
|
||
<ul class="bigul">
|
||
<li>To Implicitly derive a function (useful when a function can't easily be solved for y)
|
||
|
||
|
||
<ul>
|
||
<li>Differentiate with respect to x</li>
|
||
<li>Collect all the dy/dx on one side</li>
|
||
<li>Solve for dy/dx</li>
|
||
</ul></li>
|
||
<li>To derive an inverse function, restate it without the inverse then use Implicit differentiation</li>
|
||
</ul>
|
||
<p> </p>
|
||
<div class="questions">11312, 11313, 11314, 11315, 11316, 11317, 11318, 11319, 11320, 11321</div>
|
||
|
||
<div class="related">
|
||
<a href="derivatives-introduction.html">Introduction to Derivatives</a>
|
||
<a href="derivatives-rules.html">Derivative Rules</a>
|
||
<a href="index.html">Calculus Index</a>
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