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Lösung 1.2:2a

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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The expression is composed of two parts: first, an outer part,
The expression is composed of two parts: first, an outer part,
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{{Displayed math||<math>\sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,}</math>}}
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{{Abgesetzte Formel||<math>\sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,}</math>}}
and then an inner part, <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,} = x^{2}\,</math>.
and then an inner part, <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,} = x^{2}\,</math>.
Zeile 9: Zeile 9:
<math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}</math> were the variable that we differentiate with respect to, and then we multiply with the derivative of the inner part <math>\bigl(\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}\bigr)'</math>, so that
<math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}</math> were the variable that we differentiate with respect to, and then we multiply with the derivative of the inner part <math>\bigl(\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}\bigr)'</math>, so that
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{{Displayed math||<math>\frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\,x^2\,} = \cos \bbox[#FFEEAA;,1.5pt]{\,x^2\,}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\,x^2\,}\bigr)' = \cos x^2\cdot 2x\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\,x^2\,} = \cos \bbox[#FFEEAA;,1.5pt]{\,x^2\,}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\,x^2\,}\bigr)' = \cos x^2\cdot 2x\,\textrm{.}</math>}}

Version vom 12:52, 10. Mär. 2009

The expression is composed of two parts: first, an outer part,

\displaystyle \sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,}

and then an inner part, \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,} = x^{2}\,.

When we differentiate compound expressions, we first differentiate the outer part, \displaystyle \sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}, as if \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,} were the variable that we differentiate with respect to, and then we multiply with the derivative of the inner part \displaystyle \bigl(\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}\bigr)', so that

\displaystyle \frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\,x^2\,} = \cos \bbox[#FFEEAA;,1.5pt]{\,x^2\,}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\,x^2\,}\bigr)' = \cos x^2\cdot 2x\,\textrm{.}
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2a“