From Förberedande kurs i matematik 2

Revision as of 08:37, 4 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_2d.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:21, 15 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We can see the expression as "ln of something",
-	<center> [[Bild:1_2_2d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\ln \bbox[#FFEEAA;,1.5pt]{\phantom{\ln x}}\,,</math>}}
		+
		+	where "something" is <math>\ln x</math>.
		+
		+	Because we have a compound expression, we use the chain rule and obtain, roughly speaking, the outer derivative multiplied by the inner derivative,
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,\ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} = \frac{1}{\bbox[#FFEEAA;,1.5pt]{\,\ln x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)'\,,</math>}}
		+
		+	where the first factor on the right-hand side <math>1/\bbox[#FFEEAA;,1.5pt]{\,\ln x\,}</math> is the outer derivative of <math>\ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,}</math> and the other factor <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)'</math> is the inner derivative. Thus, we get
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,\ln\ln x = \frac{1}{\ln x}\cdot \frac{1}{x} = \frac{1}{x\ln x}\,\textrm{.}</math>}}

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Current revision

We can see the expression as "ln of something",

\displaystyle \ln \bbox[#FFEEAA;,1.5pt]{\phantom{\ln x}}\,,

where "something" is \displaystyle \ln x.

Because we have a compound expression, we use the chain rule and obtain, roughly speaking, the outer derivative multiplied by the inner derivative,

\displaystyle \frac{d}{dx}\,\ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} = \frac{1}{\bbox[#FFEEAA;,1.5pt]{\,\ln x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)'\,,

where the first factor on the right-hand side \displaystyle 1/\bbox[#FFEEAA;,1.5pt]{\,\ln x\,} is the outer derivative of \displaystyle \ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} and the other factor \displaystyle \bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)' is the inner derivative. Thus, we get

\displaystyle \frac{d}{dx}\,\ln\ln x = \frac{1}{\ln x}\cdot \frac{1}{x} = \frac{1}{x\ln x}\,\textrm{.}