Lösung 1.2:3b

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The outer function in the expression is "the root of something",
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<center> [[Image:1_2_3b-1(2).gif]] </center>
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<math>\sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}</math>
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<center> [[Image:1_2_3b-2(2).gif]] </center>
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and differentiating with the chain rule gives
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<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>
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We establish the inner derivative by using the quotient rule,
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<math>\begin{align}
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& \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}} \\
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& =\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}} \\
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& =\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\centerdot \frac{-2}{\left( x-1 \right)^{2}} \\
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& =-\sqrt{\frac{x-1}{x+1}\centerdot }\frac{1}{\left( x-1 \right)^{2}} \\
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& =-\frac{1}{\left( x-1 \right)^{{3}/{2}\;}\sqrt{x+1}} \\
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\end{align}</math>
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where we have used the simplification
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<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>

Version vom 12:45, 12. Okt. 2008

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


\displaystyle \sqrt{\left\{ \left. \frac{x+1}{x-1} \right\} \right.}


and differentiating with the chain rule gives


\displaystyle \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 }


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{\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{\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{\frac{x+1}{x-1}}}\centerdot \frac{-2}{\left( x-1 \right)^{2}} \\ & =-\sqrt{\frac{x-1}{x+1}\centerdot }\frac{1}{\left( x-1 \right)^{2}} \\ & =-\frac{1}{\left( x-1 \right)^{{3}/{2}\;}\sqrt{x+1}} \\ \end{align}


where we have used the simplification


\displaystyle \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}}}

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