Lösung 1.2:3b

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The outer function in the expression is "the square root of something",
The outer function in the expression is "the square root of something",
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{{Displayed math||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}
+
{{Abgesetzte Formel||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } }</math>}}
and differentiating with the chain rule gives
and differentiating with the chain rule gives
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{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}</math>}}
We establish the inner derivative by using the quotient rule,
We establish the inner derivative by using the quotient rule,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}
\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}}
&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt]
&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt]
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where we have used the simplification
where we have used the simplification
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{{Displayed math||<math>\frac{\sqrt{x-1}}{(x-1)^2}
+
{{Abgesetzte Formel||<math>\frac{\sqrt{x-1}}{(x-1)^2}
= \frac{(x-1)^{1/2}}{(x-1)^2}
= \frac{(x-1)^{1/2}}{(x-1)^2}
= (x-1)^{1/2-2}
= (x-1)^{1/2-2}
= (x-1)^{-3/2}
= (x-1)^{-3/2}
= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}
= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}

Version vom 12:54, 10. Mär. 2009

The outer function in the expression is "the square root of something",

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

and differentiating with the chain rule gives

\displaystyle \frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\frac{x+1}{x-1} } } = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\dfrac{x+1}{x-1} } } }\cdot \Bigl( \frac{x+1}{x-1}\Bigr)'\,\textrm{.}

We establish the inner derivative by using the quotient rule,

\displaystyle \begin{align}

\frac{d}{dx}\,\sqrt{\frac{x+1}{x-1}} &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot\frac{(x+1)'\cdot (x-1) - (x+1)\cdot (x-1)'}{(x-1)^2}\\[5pt] &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{1\cdot (x-1) - (x+1)\cdot 1}{(x-1)^2}\\[5pt] &= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{-2}{(x-1)^2}\\[5pt] &= -\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{(x-1)^2}\\[5pt] &= -\frac{1}{(x-1)^{3/2}\sqrt{x+1}}\,, \end{align}

where we have used the simplification

\displaystyle \frac{\sqrt{x-1}}{(x-1)^2}

= \frac{(x-1)^{1/2}}{(x-1)^2} = (x-1)^{1/2-2} = (x-1)^{-3/2} = \frac{1}{(x-1)^{3/2}}\,\textrm{.}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3b“