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Lösung 1.2:3b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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
-
{{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,
-
{{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]
Zeile 20: Zeile 20:
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",

x−1x+1 

and differentiating with the chain rule gives

ddxx−1x+1=12x−1x+1x−1x+1.

We establish the inner derivative by using the quotient rule,

ddxx−1x+1=12x−1x+1(x−1)2(x+1)(x−1)−(x+1)(x−1)=12x−1x+1(x−1)21(x−1)−(x+1)1=12x−1x+1−2(x−1)2=−x+1x−11(x−1)2=−1(x−1)32x+1

where we have used the simplification

x−1(x−1)2=(x−1)2(x−1)12=(x−1)12−2=(x−1)−32=1(x−1)32.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3b“