From Förberedande kurs i matematik 2

Revision as of 08:39, 4 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_2f-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:1_2_2f-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (10:49, 15 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	The entire expression is made up of several levels,
-	<center> [[Bild:1_2_2f-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:1_2_2f-2(2).gif]] </center>	+	and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}
		+
		+	and differentiate this using the chain rule,
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}</math>}}
		+
		+	In the next differentiation, we have "the square root of something",
		+
		+	{{Displayed math||<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}
		+
		+	where we have used the differentiation rule,
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
		+
		+	for the outer derivative.
		+
		+	The whole differentiation in one go becomes
		+
		+	{{Displayed math||<math>\begin{align}
		+	\frac{d}{dx}\cos\sqrt{1-x}
		+	&= -\sin\sqrt{1-x}\cdot\frac{d}{dx}\,\sqrt{1-x}\\[5pt]
		+	&= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot\frac{d}{dx}\,(1-x)\\[5pt]
		+	&= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot (-1)\\[5pt]
		+	&= \frac{\sin\sqrt{1-x}}{2\sqrt{1-x}}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

The entire expression is made up of several levels,

\displaystyle \cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }

and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",

\displaystyle \cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,

and differentiate this using the chain rule,

\displaystyle \frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}

In the next differentiation, we have "the square root of something",

\displaystyle \bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,

where we have used the differentiation rule,

\displaystyle \frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}

for the outer derivative.

The whole differentiation in one go becomes

\displaystyle \begin{align} \frac{d}{dx}\cos\sqrt{1-x} &= -\sin\sqrt{1-x}\cdot\frac{d}{dx}\,\sqrt{1-x}\\[5pt] &= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot\frac{d}{dx}\,(1-x)\\[5pt] &= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot (-1)\\[5pt] &= \frac{\sin\sqrt{1-x}}{2\sqrt{1-x}}\,\textrm{.} \end{align}