From Förberedande kurs i matematik 2

Revision as of 08:36, 4 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_2c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:15, 15 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	When we see this expression, we should think "square root of something",
-	<center> [[Bild:1_2_2c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{,}</math>}}
		+
		+	and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,
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		+	{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}
		+
		+	where we have used the differentiation rule
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
		+
		+	Thus, we obtain
		+
		+	{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}</math>}}

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Current revision

When we see this expression, we should think "square root of something",

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

and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression \displaystyle \bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x,

\displaystyle \frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,

where we have used the differentiation rule

\displaystyle \frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}

Thus, we obtain

\displaystyle \frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}