From Förberedande kurs i matematik 2

Revision as of 14:20, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 1.1:2d moved to Solution 1.1:2d: Robot: moved page) ← Previous diff		Current revision (11:58, 14 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we write <math>\sqrt{x}</math> in power form <math>x^{1/2}</math>, we see that the square root is a function having the appearance of <math>x^n</math> and its derivative is therefore equal to
-	<center> [[Image:1_1_2d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>f^{\,\prime}(x) = \frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2}\,\textrm{.}</math>}}
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		+	The answer can also be written as
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		+	{{Displayed math||<math>f^{\,\prime}(x) = \frac{1}{2\sqrt{x}}</math>}}
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		+	since <math>x^{-1/2} = \bigl(x^{1/2}\bigr)^{-1} = \bigl(\sqrt{x}\,\bigr)^{-1} = \frac{1}{\sqrt{x}}\,</math>.

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Current revision

If we write \displaystyle \sqrt{x} in power form \displaystyle x^{1/2}, we see that the square root is a function having the appearance of \displaystyle x^n and its derivative is therefore equal to

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

The answer can also be written as

\displaystyle f^{\,\prime}(x) = \frac{1}{2\sqrt{x}}

since \displaystyle x^{-1/2} = \bigl(x^{1/2}\bigr)^{-1} = \bigl(\sqrt{x}\,\bigr)^{-1} = \frac{1}{\sqrt{x}}\,.