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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>.
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<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,

sin

and then an inner part, =x2.

When we differentiate compound expressions, we first differentiate the outer part, sin, as if were the variable that we differentiate with respect to, and then we multiply with the derivative of the inner part  , so that

ddxsinx2=cosx2x2=cosx22x. 
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2a“