Lösung 1.2:2b

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The whole expression consists of two parts: the outer part, "
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The whole expression consists of two parts: the outer part, "''e'' raised to something",
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<math>e</math>
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raised to something",
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{{Displayed math||<math>e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,</math>}}
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<math>e^{\left\{ \left. {} \right\} \right.}</math>
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where "something" is the inner part <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.
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The derivative is calculated according to the chain rule by differentiating <math>e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}</math> with respect to <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}</math> and then multiplying by the inner derivative <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)'</math>, i.e.
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where "something" is the inner part
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{{Displayed math||<math>\frac{d}{dx}\,e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}} = e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,x^2+x\,} \bigr)'\,\textrm{.}</math>}}
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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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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( \left\{ \left. x^{2}+x \right\} \right. \right)^{\prime }</math>
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{{Displayed math||<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</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>
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Version vom 08:02, 15. Okt. 2008

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

\displaystyle e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,

where "something" is the inner part \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x.

The derivative is calculated according to the chain rule by differentiating \displaystyle e^{\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}} with respect to \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} and then multiplying by the inner derivative \displaystyle \bigl( \bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} \bigr)', i.e.

\displaystyle \frac{d}{dx}\,e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}} = e^{\,\bbox[#FFEEAA;,1.5pt]{\,x^2+x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,x^2+x\,} \bigr)'\,\textrm{.}

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

\displaystyle \frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2b“