From Förberedande kurs i matematik 2

Revision as of 14:24, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 1.2:2a moved to Solution 1.2:2a: Robot: moved page) ← Previous diff		Current revision (14:50, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The expression is composed of two parts: first, an outer part,
-	<center> [[Image:1_2_2a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,}</math>}}
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		+	and then an inner part, <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,} = x^{2}\,</math>.
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		+	When we differentiate compound expressions, we first differentiate the outer part,
		+	<math>\sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}</math>, as if
		+	<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
		+
		+	{{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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Current revision

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{.}