Lösung 1.2:3b

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
Zeile 1: Zeile 1:
-
The outer function in the expression is "the root of something",
+
The outer function in the expression is "the square root of something",
-
 
+
-
 
+
-
<math>\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}</math>
+
 +
{{Displayed math||<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>}}
-
<math>\frac{d}{dx}\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}=\frac{1}{2\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}}\centerdot \left( \frac{x+1}{x-1} \right)^{\prime }</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}
-
<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{\frac{x+1}{x-1}}}\centerdot \frac{\left( x+1 \right)^{\prime }\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot \left( x-1 \right)^{\prime }}{\left( x-1 \right)^{2}} \\
+
&= \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{\frac{x+1}{x-1}}}\centerdot \frac{1\centerdot \left( x-1 \right)-\left( x+1 \right)\centerdot 1}{\left( x-1 \right)^{2}} \\
+
&= \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{\frac{x+1}{x-1}}}\centerdot \frac{-2}{\left( x-1 \right)^{2}} \\
+
&= \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}\cdot \frac{-2}{(x-1)^2}\\[5pt]
-
& =-\sqrt{\frac{x-1}{x+1}\centerdot }\frac{1}{\left( x-1 \right)^{2}} \\
+
&= -\sqrt{\frac{x-1}{x+1}}\cdot\frac{1}{(x-1)^2}\\[5pt]
-
& =-\frac{1}{\left( x-1 \right)^{{3}/{2}\;}\sqrt{x+1}} \\
+
&= -\frac{1}{(x-1)^{3/2}\sqrt{x+1}}\,,
-
\end{align}</math>
+
\end{align}</math>}}
-
 
+
where we have used the simplification
where we have used the simplification
-
 
+
{{Displayed math||<math>\frac{\sqrt{x-1}}{(x-1)^2}
-
<math>\frac{\sqrt{x-1}}{\left( x-1 \right)^{2}}=\frac{\left( x-1 \right)^{{1}/{2}\;}}{\left( x-1 \right)^{2}}=\left( x-1 \right)^{\frac{1}{2}-2}=\left( x-1 \right)^{-\frac{3}{2}}=\frac{1}{\left( x-1 \right)^{\frac{3}{2}}}</math>
+
= \frac{(x-1)^{1/2}}{(x-1)^2}
 +
= (x-1)^{1/2-2}
 +
= (x-1)^{-3/2}
 +
= \frac{1}{(x-1)^{3/2}}\,\textrm{.}</math>}}

Version vom 11:53, 15. Okt. 2008

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“