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Lösung 1.2:3a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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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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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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{{Displayed math||<math>\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr) = {}\rlap{\frac{1}{\bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)'\,\textrm{.}}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}
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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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We can carry out the differentiation of <math>\sqrt{x}+\sqrt{x+1}</math> on the right-hand side term by term to obtain
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{{Displayed math||<math>\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{} = {}\rlap{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \bigl[ (\sqrt{x})' + (\sqrt{x+1})'\bigr]}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}
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We can carry out the differentiation of
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and it remains then only to differentiate <math>\sqrt{x}</math>, which we do directly, and <math>\sqrt{x+1}</math> (which has a simple inner derivative),
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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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{{Displayed math||<math>\begin{align}
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\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}
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&= \frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]\\[5pt]
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&= \frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot 1\Bigr]\,\textrm{.}
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\end{align}</math>}}
If we rewrite the expression inside the square brackets using a common denominator, we get
If we rewrite the expression inside the square brackets using a common denominator, we get
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{{Displayed math||<math>\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}
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= {}\rlap{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \Bigr]\,,}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}
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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 <math>\sqrt{x+1}+\sqrt{x}</math> from the numerator and denominator to get
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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>
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{{Displayed math||<math>\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}
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= {}\rlap{\frac{1}{2\sqrt{x}\sqrt{x+1}}\,\textrm{.}}\phantom{\frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot (x+1)'\Bigr]}</math>}}

Version vom 11:41, 15. Okt. 2008

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

ddxlnx+x+1=1x+x+1x+x+1.

We can carry out the differentiation of x+x+1  on the right-hand side term by term to obtain

=1x+x+1(x)+(x+1) 

and it remains then only to differentiate x , which we do directly, and x+1  (which has a simple inner derivative),

=1x+x+112x+12x+1(x+1)=1x+x+112x+12x+11.

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

=1x+x+12xx+1x+1+x 

and we can then eliminate the factor x+1+x  from the numerator and denominator to get

=12xx+1.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3a“