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		<title>Wie findet man # dy / dx # durch implizite Differenzierung von # y = sin (xy) #?</title>
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		<dc:creator><![CDATA[Lynea]]></dc:creator>
		<pubDate>Mon, 17 Feb 2020 18:52:35 +0000</pubDate>
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					<description><![CDATA[Wie findet man # dy / dx # durch implizite Differenzierung von # y = sin (xy) #? Antworten: # dy/dx={ycos(xy)}/ {1-xcos(xy)},# ,ODER, #dy/dx={y^2sqrt(1-y^2)}/{y-sqrt(1-y^2)arc siny}.# Erläuterung: #y=sin(xy).# #:. dy/dx," using the Chain Rule,"# #=d/dx(sin(xy))={cos(xy)}{d/dx(xy)}," &#38;, using the Product Rule,"# #={x*d/dx(y)+y*d/dx(x)}cos(xy),# #:. dy/dx=xcos(xy)dy/dx+ycos(xy),# #rArr {1-xcos(xy)}dy/dx=ycos(xy).# #:. dy/dx={ycos(xy)}/ {1-xcos(xy)}.# Andernfalls #y=sin(xy) rArr arc siny=xy, or, x=(arc siny)/y.# ... <a title="Wie findet man # dy / dx # durch implizite Differenzierung von # y = sin (xy) #?" class="read-more" href="https://dieklugeeule.com/wie-findet-man-dy-dx-durch-implizite-differenzierung-von-y-sin-xy/" aria-label="Mehr dazu unter Wie findet man # dy / dx # durch implizite Differenzierung von # y = sin (xy) #?">Weiterlesen</a>]]></description>
										<content:encoded><![CDATA[<h1 class="questionTitle">Wie findet man # dy / dx # durch implizite Differenzierung von # y = sin (xy) #?</h1>
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<h4 class="answerHeader">Antworten:</h4>
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<p># dy/dx={ycos(xy)}/ {1-xcos(xy)},#</p>
<p class="gt-block"><strong>,ODER,</strong> </p>
<p>#dy/dx={y^2sqrt(1-y^2)}/{y-sqrt(1-y^2)arc siny}.#</p>
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<h4 class="answerHeader">Erläuterung:</h4>
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<p>#y=sin(xy).#</p>
<p>#:. dy/dx," using the Chain Rule,"#</p>
<p>#=d/dx(sin(xy))={cos(xy)}{d/dx(xy)}," &amp;, using the Product Rule,"#</p>
<p>#={x*d/dx(y)+y*d/dx(x)}cos(xy),#</p>
<p>#:. dy/dx=xcos(xy)dy/dx+ycos(xy),#</p>
<p>#rArr {1-xcos(xy)}dy/dx=ycos(xy).#</p>
<p>#:. dy/dx={ycos(xy)}/ {1-xcos(xy)}.#</p>
<p class="gt-block"><strong>Andernfalls</strong> #y=sin(xy) rArr arc siny=xy, or, x=(arc siny)/y.#</p>
<p class="gt-block">Daher unterscheiden sich beide Seiten bzgl #y,# wir haben durch die <strong><a href="https://socratic.org/calculus/basic-differentiation-rules/quotient-rule">Quotientenregel</a>,</strong></p>
<p>#dx/dy={y*d/dy(arc siny)-(arc siny)*d/dy(y)}/y^2,#</p>
<p>#={y*(1/sqrt(1-y^2))-(arc siny)*1}/y^2,#</p>
<p>#={y-sqrt(1-y^2)arc siny}/{y^2*sqrt(1-y^2)},#</p>
<p>Deswegen, #dy/dx={y^2sqrt(1-y^2)}/{y-sqrt(1-y^2)arc siny}.#</p>
<p class="gt-block"><strong>Ich überlasse es dem Fragesteller, zu zeigen, dass beide Antworten übereinstimmen.</strong></p>
<p class="gt-block"><strong>Viel Spaß beim Rechnen!</strong></p>
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