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	<title>Aida &#8211; Die Kluge Eule</title>
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		<title>Wie finden Sie die Ableitung von # 2e ^ -x #?</title>
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		<pubDate>Tue, 04 Feb 2020 18:48:40 +0000</pubDate>
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					<description><![CDATA[Wie finden Sie die Ableitung von # 2e ^ -x #? Antworten: #(dy)/(dx)=-2e^-x# Erläuterung: Erinnere dich daran #d/dxe^x=e^x# Mit Kettenregel, #(dy)/(dx)=(dy)/(du)*(du)/(dx)#, Lassen #u=-x# #(dy)/(du)=d/(du)2e^u=2e^u=2e^-x# #(du)/(dx)=d/(dx)-x=-1# #:.(dy)/(dx)=-1*2e^-x=-2e^-x#]]></description>
										<content:encoded><![CDATA[<h1 class="questionTitle">Wie finden Sie die Ableitung von # 2e ^ -x #?</h1>
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<p>#(dy)/(dx)=-2e^-x#</p>
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<p>Erinnere dich daran #d/dxe^x=e^x#</p>
<p class="gt-block">Mit <a href="https://socratic.org/calculus/basic-differentiation-rules/chain-rule">Kettenregel</a>,  #(dy)/(dx)=(dy)/(du)*(du)/(dx)#,</p>
<p>Lassen #u=-x#</p>
<p>#(dy)/(du)=d/(du)2e^u=2e^u=2e^-x#</p>
<p>#(du)/(dx)=d/(dx)-x=-1#</p>
<p>#:.(dy)/(dx)=-1*2e^-x=-2e^-x#</p>
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