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Lösung 1.1:2d

Aus Online Mathematik Brückenkurs 2

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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
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
-
{{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>}}
+
{{Abgesetzte Formel||<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>}}
The answer can also be written as
The answer can also be written as
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{{Displayed math||<math>f^{\,\prime}(x) = \frac{1}{2\sqrt{x}}</math>}}
+
{{Abgesetzte Formel||<math>f^{\,\prime}(x) = \frac{1}{2\sqrt{x}}</math>}}
since <math>x^{-1/2} = \bigl(x^{1/2}\bigr)^{-1} = \bigl(\sqrt{x}\,\bigr)^{-1} = \frac{1}{\sqrt{x}}\,</math>.
since <math>x^{-1/2} = \bigl(x^{1/2}\bigr)^{-1} = \bigl(\sqrt{x}\,\bigr)^{-1} = \frac{1}{\sqrt{x}}\,</math>.

Version vom 12:50, 10. Mär. 2009

If we write x  in power form x12 , we see that the square root is a function having the appearance of xn and its derivative is therefore equal to

f(x)=ddxx=ddxx12=21x12−1=21x−12. 

The answer can also be written as

f(x)=12x

since x−12=x12−1=x−1=1x .

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:2d“