Lösung 1.2:3a

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There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:
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<math>\frac{d}{dx}\ln \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)=\frac{1}{\left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right.}\centerdot \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)^{\prime }</math>
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<center> [[Image:1_2_3a-2(2).gif]] </center>
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We can carry out the differentiation of
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<math>\sqrt{x}+\sqrt{x+1}</math>
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on the right-hand side term by term to obtain
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<math>\quad =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \left( \sqrt{x} \right)^{\prime }+\left( \sqrt{x+1} \right)^{\prime } \right]</math>
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and it remains then only to differentiate
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<math>\sqrt{x}</math>,which we do directly, and
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<math>\sqrt{x+1}</math>
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which a simple inner derivative)
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<math></math>
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<math>\begin{align}
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& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot \left( x+1 \right)^{\prime } \right] \\
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& =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot 1 \right] \\
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\end{align}</math>
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If we rewrite the expression inside the square brackets using a common denominator, we get
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<math>=\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \right]</math>,
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and we can then eliminate the factor
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<math>\sqrt{x+1}+\sqrt{x}</math>
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from the numerator and denominator to get
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<math>=\frac{1}{2\sqrt{x}\sqrt{x+1}}</math>

Version vom 14:17, 11. Okt. 2008

There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:


\displaystyle \frac{d}{dx}\ln \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)=\frac{1}{\left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right.}\centerdot \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)^{\prime }


We can carry out the differentiation of \displaystyle \sqrt{x}+\sqrt{x+1} on the right-hand side term by term to obtain


\displaystyle \quad =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \left( \sqrt{x} \right)^{\prime }+\left( \sqrt{x+1} \right)^{\prime } \right]


and it remains then only to differentiate \displaystyle \sqrt{x},which we do directly, and \displaystyle \sqrt{x+1} which a simple inner derivative)


\displaystyle

\displaystyle \begin{align} & =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot \left( x+1 \right)^{\prime } \right] \\ & =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot 1 \right] \\ \end{align}


If we rewrite the expression inside the square brackets using a common denominator, we get


\displaystyle =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \right],

and we can then eliminate the factor \displaystyle \sqrt{x+1}+\sqrt{x} from the numerator and denominator to get


\displaystyle =\frac{1}{2\sqrt{x}\sqrt{x+1}}

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