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Lösung 1.2:2b

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.2:2b moved to Solution 1.2:2b: Robot: moved page)
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The whole expression consists of two parts: the outer part, "
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<center> [[Image:1_2_2b.gif]] </center>
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<math>e</math>
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raised to something",
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<math>e^{\left\{ \left. {} \right\} \right.}</math>
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where "something" is the inner part
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<math>\left\{ \left. {} \right\} \right.=x^{2}+x</math>. The derivative is calculated according to the chain rule by differentiating
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<math>e^{\left\{ \left. {} \right\} \right.}</math>
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with respect to
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<math>\left\{ \left. {} \right\} \right.</math>
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and then multiplying by the inner derivative
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<math>\left( \left\{ \left. {} \right\} \right. \right)^{\prime }</math>, i.e.
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<math>\frac{d}{dx}e^{\left\{ \left. x^{2}+x \right\} \right.}=e^{\left\{ \left. x^{2}+x \right\} \right.}\centerdot \left( \left\{ \left. x^{2}+x \right\} \right. \right)^{\prime }</math>
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The inner part is an ordinary polynomial which we differentiate directly:
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<math>\frac{d}{dx}e^{\left\{ \left. x^{2}+x \right\} \right.}=e^{\left\{ \left. x^{2}+x \right\} \right.}\centerdot \left( 2x+1 \right)</math>

Version vom 12:56, 11. Okt. 2008

The whole expression consists of two parts: the outer part, " e raised to something",


e 


where "something" is the inner part =x2+x. The derivative is calculated according to the chain rule by differentiating e  with respect to and then multiplying by the inner derivative  , i.e.


ddxex2+x=ex2+xx2+x 


The inner part is an ordinary polynomial which we differentiate directly:


ddxex2+x=ex2+x2x+1 

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2b“