Lösung 1.2:2b
Aus Online Mathematik Brückenkurs 2
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Version vom 10:03, 11. 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{.} |
