Lösung 1.1:2d

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If we write
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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
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<math>\sqrt{x}</math>
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in power form
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<math>x^{{1}/{2}\;}</math>, we see that the square root is a function having the appearance of
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<math>x^{n}</math>
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and its derivative is therefore equal to
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{{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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<math>{f}'\left( x \right)=\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2}\;}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}</math>
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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>}}
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<math>{f}'\left( x \right)=\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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because
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<math>x^{-\frac{1}{2}}=\left( x^{\frac{1}{2}} \right)^{-1}=\left( \sqrt{x} \right)^{-1}=\frac{1}{\sqrt{x}}</math>
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Version vom 11:58, 14. Okt. 2008

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}}\,.

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