From Förberedande kurs i matematik 2

Revision as of 08:34, 4 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_1f-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_1f-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (14:22, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by <math>\sin x</math>". As a first step, we therefore use the quotient rule,
-	<center> [[Bild:1_2_1f-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:1_2_1f-2(2).gif]] </center>	+	We can, in turn, differentiate the expression <math>x\ln x</math> by using the product rule,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\begin{align}
		+	(x\ln x)'
		+	&= (x)'\ln x + x\,(\ln x)'\\[5pt]
		+	&= 1\cdot\ln x + x\cdot\frac{1}{x}\\[5pt]
		+	&= \ln x+1\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	All in all, we thus obtain
		+
		+	{{Displayed math||<math>\begin{align}
		+	\Bigl(\frac{x\ln x}{\sin x}\Bigr)'
		+	&= \frac{(\ln x+1)\cdot\sin x - x\ln x\cdot \cos x}{(\sin x)^2}\\[5pt]
		+	&= \frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin^2\!x}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

In this case, we have a slightly more complicated expression, but if we focus on the expression's outer form, we have essentially "something divided by \displaystyle \sin x". As a first step, we therefore use the quotient rule,

\displaystyle \Bigl(\frac{x\ln x}{\sin x}\Bigr)' = \frac{(x\ln x)'\cdot \sin x - x\ln x\cdot (\sin x)'}{(\sin x)^2}\,\textrm{.}

We can, in turn, differentiate the expression \displaystyle x\ln x by using the product rule,

\displaystyle \begin{align} (x\ln x)' &= (x)'\ln x + x\,(\ln x)'\\[5pt] &= 1\cdot\ln x + x\cdot\frac{1}{x}\\[5pt] &= \ln x+1\,\textrm{.} \end{align}

All in all, we thus obtain

\displaystyle \begin{align} \Bigl(\frac{x\ln x}{\sin x}\Bigr)' &= \frac{(\ln x+1)\cdot\sin x - x\ln x\cdot \cos x}{(\sin x)^2}\\[5pt] &= \frac{\ln x+1}{\sin x}-\frac{x\ln x\cos x}{\sin^2\!x}\,\textrm{.} \end{align}