Lösung 1.2:2b

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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The whole expression consists of two parts: the outer part, "''e'' raised to something",
The whole expression consists of two parts: the outer part, "''e'' raised to something",
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{{Displayed math||<math>e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,</math>}}
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{{Abgesetzte Formel||<math>e^{\,\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,}}\,,</math>}}
where "something" is the inner part <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x+x}\,} = x^2+x</math>.
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.
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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{{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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{{Abgesetzte Formel||<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>}}
The inner part is an ordinary polynomial which we differentiate directly,
The inner part is an ordinary polynomial which we differentiate directly,
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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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{{Abgesetzte Formel||<math>\frac{d}{dx}\,e^{x^2+x} = e^{x^2+x}\cdot (2x+1)\,\textrm{.}</math>}}

Version vom 12:53, 10. Mär. 2009

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“