Lösung 1.2:3a
Aus Online Mathematik Brückenkurs 2
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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 | + | There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm, |
| + | {{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>}} | ||
| - | <math> | + | We can carry out the differentiation of <math>\sqrt{x}+\sqrt{x+1}</math> on the right-hand side term by term to obtain |
| + | {{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>}} | ||
| - | + | 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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| - | and it remains then only to differentiate | + | |
| - | <math>\sqrt{x}</math>,which we do directly, and | + | |
| - | <math>\sqrt{x+1}</math> | + | |
| - | which a simple inner derivative) | + | |
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| + | {{Displayed math||<math>\begin{align} | ||
| + | \phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{} | ||
| + | &= \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] | ||
| + | &= \frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot 1\Bigr]\,\textrm{.} | ||
| + | \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 | ||
| + | {{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[\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>}} | ||
| - | + | 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 | + | |
| - | <math>\sqrt{x+1}+\sqrt{x}</math> | + | |
| - | from the numerator and denominator to get | + | |
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| - | <math>=\frac{1}{2\sqrt{x}\sqrt{x+1}}</math> | + | {{Displayed math||<math>\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{} |
| + | = {}\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,
| \displaystyle \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]} |
We can carry out the differentiation of \displaystyle \sqrt{x}+\sqrt{x+1} on the right-hand side term by term to obtain
| \displaystyle \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]} |
and it remains then only to differentiate \displaystyle \sqrt{x}, which we do directly, and \displaystyle \sqrt{x+1} (which has a simple inner derivative),
| \displaystyle \begin{align}
\phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{} &= \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] &= \frac{1}{\sqrt{x}+\sqrt{x+1}}\cdot \Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\cdot 1\Bigr]\,\textrm{.} \end{align} |
If we rewrite the expression inside the square brackets using a common denominator, we get
| \displaystyle \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[\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]} |
and we can then eliminate the factor \displaystyle \sqrt{x+1}+\sqrt{x} from the numerator and denominator to get
| \displaystyle \phantom{\frac{d}{dx}\,\ln\bigl( \bbox[#FFEEAA;,1.5pt]{\sqrt{x}+\sqrt{x+1}} \bigr)}{}
= {}\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]} |
