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<p><img alt="" src="../../files/differentiation/implicit-differentiation/main1.jpg" style="float: left; width: 100px; height: 100px;"></p> <p>This page is all about implicit equations and how we find the gradient of them. Implicit equations are equations which are not written explicitly. Examination questions typically ask you to find the equation of a tangent or a normal to the graphs. This requires you to be able to find the gradient and we often have to use the product and chain rule. Before we get on to that, it is a good idea to understand what implicit equations are and why we use them.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <p>On this page, you should learn about</p> <ul> <li>explicit and implicit equations</li> <li>implicit differentiation</li> </ul> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <h3>Definition of Implicit Equations</h3> <p>You may be more familiar with<strong> explicit</strong> functions like y = x² and y = sinx. They are written explicitly since ‘y’ is the subject of these equations. You should know how to find the gradient function of these!</p> <p>x² + y² = 1is an <strong>implicit </strong>equation, since ‘y’ is not the subject of this equation. It is difficult, and sometimes impossible, to write some relations like this explicitly. Many implicit equations have really cool and interesting graphs. If you want to play around with some of these graphs, get some background knowledge and understand why implicit differentiation is important visit to <a href="../565/graphs-of-implicit-equations.html" target="_blank" title="Graphs of Implicit Equations">Graphs of Implicit Equations</a>.</p> <hr class="hidden-separator"> <p>The following videos will help you understand all the concepts from this page</p> <div class="panel panel-yellow panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>What and How</p> </div> </div> <div class="panel-body"> <div class="smart-object center" data-id="508"> <p>In this section, we'll look at what implicit functions are, what implicit differentiation is used for and how to do implicit differentiation.</p> <p>y=x² and f(x) = sin(x) are both <strong>explicit</strong> functions.</p> <p><strong>Implicit</strong> equations are equations that are <strong>not</strong> expressed explicitly, for example <em><strong>y</strong></em> is not given explicitly as a function of <em><strong>x</strong>. </em> The following are two examples of implicit equations:<em> x + y = 1 </em>and <em>x² + y² = 4 . </em>We use implicit differentiation to find the gradientfunctions of these.</p> <p>The following video explains why we might need to use implicit differentiation and how we use the <a href="../562/chain-rule.html" title="Chain Rule">Chain Rule</a> to do it. We'll see how we can find the gradient of the circle x² + y² = 4</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/216308832"></iframe></div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>The Technique</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="509"> <p>Before you attempt exam-style questions, you really need to be confident with the basics of implicit differentiation. In the following video, we'll be using implicit differentiation along with the <a href="../563/product-and-quotient-rule.html" title="Product and Quotient Rule">Product Rule </a>and the <a href="../562/chain-rule.html" title="Chain Rule">Chain Rule</a> to find the following:</p> <p style="text-align: center;"><span class="math-tex">\(\frac{d}{dx}(3y^{4})\qquad,\qquad\frac{d}{dx}(3x+2y^{2})\qquad,\qquad\frac{d}{dx}(xy^{2})\qquad,\qquad\frac{d}{dx}(sin(xy)) \)</span></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/216459440"></iframe></div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-expandable panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>A Typical Question</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="510"> <p>Now you are ready to try some typical exam-style questions. Often in the examination, you are required to find the <a href="../564/equation-of-tangent-and-normal.html" title="Equation of Tangent and Normal">Equation of a Tangent or a Normal</a>. In the following video we'll answer the question</p> <p>Find the equation of the tangent to the curve <span class="math-tex">\(sin(x+y)−cos(xy)− \frac{1}{2}=0\)</span> at the point <span class="math-tex">\((\frac{\pi}{6},0)\)</span></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/216499185"></iframe></div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <p> </p> <p> </p> <p> </p> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander pull-right" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <p>Here is a quiz that will get you to practise deriving simple expressions using implicit differentiation</p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#f6ed9f3a"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="f6ed9f3a"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Implicit Differentiation <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="11-155-741" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>Find <span class="math-tex">\(\frac{d}{dx}(3y^{3})\)</span></p></div><div class="q-answer"><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\((9y^{2})\frac{dy}{dx}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\((9x^{2})\frac{dx}{dy}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(9y^{2}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\((9x^{2})\frac{dy}{dx}\)</span></label></p></div><div class="q-explanation"></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Find <span class="math-tex">\(\frac{d}{dx}(e^y)\)</span></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(e^y\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\((e^y)\frac{dx}{dy}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\((e^x)\frac{dy}{dx}\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\((e^y)\frac{dy}{dx}\)</span></label></p></div><div class="q-explanation"></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Find <span class="math-tex">\(\frac{d}{dx}(siny)\)</span></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\((siny)\frac{dy}{dx}\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\((cosy)\frac{dy}{dx}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(cosy\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\((-cosy)\frac{dy}{dx}\)</span></label></p></div><div class="q-explanation"></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Find <span class="math-tex">\(\frac{d}{dx}(x^{2}y)\)</span></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(2xy+x^2\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(2xy+x^{2}\frac{dy}{dx}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(2x\frac{dy}{dx}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(2x+\frac{dy}{dx}\)</span></label></p></div><div class="q-explanation"><p><span class="math-tex">\(x^2y\)</span> is the product of two functions. Use the product rule!</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Find <span class="math-tex">\(\frac{d}{dx}(e^{xy})\)</span></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(e^{x\frac{dy}{dx}+y}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(\frac{dy}{dx}e^{xy}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\((x+y)\frac{dy}{dx}e^{xy}\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\((x\frac{dy}{dx}+y)e^{xy}\)</span></label></p></div><div class="q-explanation"><p>You need to use the chain rule: <span class="math-tex">\(\frac{d}{dx}(e^{f(x)}) = f'(x)e^{f(x)}\)</span> and the product rule to differentiate the function <span class="math-tex">\(xy\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Find <span class="math-tex">\(\frac{d}{dx}(sin(x+y))\)</span></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\((1+\frac{dy}{dx})sin(x+y)\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(\frac{dy}{dx}cos(x+y)\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\((1+\frac{dy}{dx})cos(x+y)\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(cos(1+\frac{dy}{dx})\)</span></label></p></div><div class="q-explanation"><p>You need to use the chain rule: <span class="math-tex">\(\frac{d}{dx}(sin[{f(x)}]) = f'(x)cos[f(x)]\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next <i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> </div> <div class="panel-footer"> <div> <p>text</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Exam-style Questions</p> </div> </div> <div class="panel-body"> <div class="panel panel-has-colored-body panel-default panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 1</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="511"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>Find the equation of the tangent to the curve <span class="math-tex">\(x^3 + y^3 -6xy = 0\)</span> at the point (3,3)</p> <p><img alt="" src="../../files/differentiation/implicit-differentiation/qu1.png" style="width: 300px; height: 284px;"></p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>This is a question about implicit differentiation.</p> <p>Differentiate <span class="math-tex">\(x^3 + y^3 -6xy = 0\)</span> with respect to x and find the gradient function</p> <content> </content></section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><span></span><span class="math-tex">\(3x^2 +3y^2 \frac{dy}{dx} – (6y + 6x \frac{dy}{dx}) = 0\)</span><span></span></p> <p>Substitute x=3 and y=3 to find the gradient of the tangent</p> <p><span class="math-tex">\(3\times3^2 +3\times3^2 \frac{dy}{dx} – (6 \times3+ 6\times3 \frac{dy}{dx} ) = 0 \)</span></p> <p><span class="math-tex">\(27 +27 \frac{dy}{dx} – (18+ 18 \frac{dy}{dx}) = 0\)</span></p> <p><span class="math-tex">\(9 \frac{dy}{dx} +9 = 0\)</span></p> <p>Our question becomes: find the equation of the tangent with gradient -1 which passes through the point (3,3)</p> <p><span class="math-tex">\(-1=\frac{y-3}{x-3}\)</span></p> <p><span class="math-tex">\(-x+3=y-3\)</span></p> <p><span class="math-tex">\(y=-x+6\)</span></p> </section> <h4> </h4> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 2</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="512"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>Find the gradient of the curve <span class="math-tex">\(x^2+2e^{(x+2y)}=3\)</span> at the point when x=-1</p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>This is a question about implicit differentiation.</p> <p><span calibri="" style="font-size:11.0pt;line-height:107%; font-family:">You will need to find the y coordinate when x=-1 and then find </span><span class="math-tex">\(\frac{dy}{dx} \)</span> at this point.</p> <p>To find <span class="math-tex">\(\frac{dy}{dx} \)</span> <span calibri="" style="font-size:11.0pt;line-height:107%; font-family:">, differentiate <span class="math-tex">\(x^2+2e^{(x+2y)}=3\)</span> with respect to x.</span></p> <p style="margin:0cm;margin-bottom:.0001pt">From the Chain Rule, we know that <span class="math-tex">\(\frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)}\)</span></p> <content> </content></section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>Substitute x=-1 into equation</p> <span class="math-tex">\((-1)^2+2e^{(-1+2y)}=3\\ 1+2e^{(-1+2y)}=3\\ 2e^{(-1+2y)}=2\\ e^{(-1+2y)}=1\\ -1+2y=0\\ y=0.5\\ \)</span> <p>Find <span class="math-tex">\(\frac{dy}{dx}\)</span> by differentiating <span class="math-tex">\( x^2+2e^{(x+2y)}=3\)</span> with respect to x</p> <p>From the Chain Rule, we know that <span class="math-tex">\(\frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)}\)</span></p> <p><span class="math-tex">\(2x+2(1+2\frac{dy}{dx})e^{(x+2y)}=0\)</span></p> <p>Substitute x=-1, y=0.5</p> <p><span class="math-tex">\(2(-1)+2(1+2\frac{dy}{dx})e^{(-1+2(0.5)}=0\\ -2+2(1+2\frac{dy}{dx})e^{(0)}=0\\ 2(1+2\frac{dy}{dx})=2\\ 1+2\frac{dy}{dx}=1\\ 2\frac{dy}{dx}=0\\ \frac{dy}{dx}=0 \)</span></p> </section> <h4> </h4> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-expandable panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 3</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="513"> <p><img class="sibico" src="../../../img/sibico/hl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL difficult"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>A curve is defined by <span class="math-tex">\(x\arcsin y=e^{2y}\)</span></p> <p>Show that <span class="math-tex">\(\frac{dy}{dx}=\frac{\sqrt{1-y^{2}}\arcsin y}{2e^{2y}\sqrt{1-y^{2}}-x}\)</span></p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"><content> </content> <p>This is a question about implicit differentiation.</p> <p>Differentiate <span class="math-tex">\(x\arcsin y=e^{2y}\)</span> with respect to x and find the gradient function.</p> <p>You might need to look up <span class="math-tex">\(\frac{d}{dx}( arcsin x)\)</span> in the information booklet.</p> <p>You will need to use the Product Rule, since <span class="math-tex">\(x\arcsin y\)</span> is the product of two functions.</p> <p>Careful rearrangement will be required!</p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>Note that <span class="math-tex">\(\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^{2}}}\)</span></p> <p>Therefore, <span class="math-tex">\(\frac{d}{dx}(\arcsin y)=\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}\)</span></p> <hr class="hidden-separator"> <p>Differentiate <span class="math-tex">\(x\arcsin y=e^{2y}\)</span> with respect to x</p> <p><span class="math-tex">\(1\arcsin y+x\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}=2e^{2y}\frac{dy}{dx}\)</span></p> <p>Rearrange the equation to collect terms with <span class="math-tex">\(\frac{dy}{dx}\)</span> on one side</p> <p><span class="math-tex">\(x\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}-2e^{2y}\frac{dy}{dx}=-\arcsin y\)</span></p> <p>Multiply both sides by -1</p> <p><span class="math-tex">\(2e^{2y}\frac{dy}{dx}-x\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}=\arcsin y\)</span></p> <p>Factorise</p> <p><span class="math-tex">\((2e^{2y}-\frac{x}{\sqrt{1-y^{2}}})\frac{dy}{dx}=\arcsin y\)</span></p> <p>Divide both sides through by <span class="math-tex">\(2e^{2y}-\frac{x}{\sqrt{1-y^{2}}}\)</span></p> <p><span class="math-tex">\(\frac{dy}{dx}=\frac{\arcsin y}{2e^{2y}-\frac{x}{\sqrt{1-y^{2}}}}\)</span></p> <p>Multiply numerator and denominator by <span class="math-tex">\({\sqrt{1-y^{2}}}\)</span></p> <p><span class="math-tex">\(\frac{dy}{dx}=\frac{\sqrt{1-y^{2}}\arcsin y}{2e^{2y}\sqrt{1-y^{2}}-x}\)</span></p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 4</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="514"> <p><img class="sibico" src="../../../img/sibico/hl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL difficult"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <ol style="list-style-type:lower-alpha;"> <li>Find the coordinates of the stationary point on the graph <span class="math-tex">\(x^2y+2x=1\)</span></li> <li value="2">Show that <span class="math-tex">\(2y + 4x\frac{dy}{dx}+ x^2\frac{d^2y}{dx^2}=0\)</span> and hence justify that the stationary point is a local minima.</li> </ol> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>This is a question about implicit differentiation. If you need a reminder about stationary points visit <em>stationary points page</em></p> <p>a) Differentiate <span class="math-tex">\(x^2y+2x=1\)</span> with respect to x. The stationary point exists where <span class="math-tex">\( \frac{dy}{dx}=0\)</span>. You can find an equation for y in terms of x by substituting this into your equation. Can you use this and the initial equation to find the coordinates?</p> <p>b) Differentiate the equation for a second time. Remember that local minima occurs when <span class="math-tex">\(\frac{d^2y}{dx^2}>0 \)</span></p> <content> </content></section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p style="margin-left:18.0pt;">a)</p> <p style="margin-left:18.0pt;">Differentiate <span class="math-tex">\(x^2y+2x=1\)</span> with respect to x.</p> <p style="margin-left:18.0pt;"><span class="math-tex">\(2xy + x^2\frac{dy}{dx}+2=0\)</span></p> <p style="margin-left:18.0pt;">Substitute <span class="math-tex">\(\frac{dy}{dx}=0 \)</span></p> <p style="margin-left:18.0pt"><span class="math-tex">\(2xy + 0 + 2 = 0\)</span></p> <p style="margin-left:18.0pt"><span class="math-tex">\(xy = -1\)</span></p> <p style="margin-left:18.0pt"><span class="math-tex">\(y = \frac{-1}{x}\)</span></p> <p style="margin-left:18.0pt">Substitute <span class="math-tex">\(y = \frac{-1}{x}\)</span> into initial implicit equation.</p> <p style="margin-left:18.0pt"><span class="math-tex">\(x^2(\frac{-1}{x})+2x=1\\ -x + 2x = 1\\ x=1\\ y=-1 \)</span></p> <p style="margin-left:18.0pt">Hence, a stationary point exists at (1, -1)</p> <p style="margin-left:18.0pt">b)</p> <p><span class="math-tex">\(2xy + x^2\frac{dy}{dx}+2=0\)</span></p> <p>Differentiate <span class="math-tex">\(2xy + x^2\frac{dy}{dx}+2=0\)</span> with respect to x</p> <p><span class="math-tex">\(2y + 2x\frac{dy}{dx}+ 2x\frac{dy}{dx}+ x^2\frac{d^2y}{dx^2}=0\)</span></p> <p>Hence</p> <p><span class="math-tex">\(2y + 4x\frac{dy}{dx}+ x^2\frac{d^2y}{dx^2}=0\)</span></p> <p>Substitute x=1 , y= -1, <span class="math-tex">\(\frac{dy}{dx}=0\)</span></p> <p><span class="math-tex">\(-2 + 0 + \frac{d^2y}{dx^2}=0\\ \frac{d^2y}{dx^2}=2 \)</span></p> <p>Since <span class="math-tex">\(\frac{d^2y}{dx^2}>0\)</span> , stationary point at (1, -1) is a local minima</p> </section> </div> </div> </div> <div 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