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Lösung 1.2:3c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We can write the expression as
We can write the expression as
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{{Displayed math||<math>\frac{1}{x\sqrt{1-x^{2}}} = \bigl(x\sqrt{1-x^2}\,\bigr)^{-1}\,\textrm{,}</math>}}
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{{Abgesetzte Formel||<math>\frac{1}{x\sqrt{1-x^{2}}} = \bigl(x\sqrt{1-x^2}\,\bigr)^{-1}\,\textrm{,}</math>}}
and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,
and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}
\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}
&= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
&= {}\rlap{-1\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-2}\bigl(\bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)'}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
Zeile 14: Zeile 14:
The next step is to differentiate the product <math>x\cdot\sqrt{1-x^2}</math> using the product rule,
The next step is to differentiate the product <math>x\cdot\sqrt{1-x^2}</math> using the product rule,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \Bigl( (x)'\sqrt{1-x^2} + x(\sqrt{1-x^2})'\Bigr)}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
Zeile 22: Zeile 22:
The expression <math>\sqrt{1-x^2}</math> is of the type "square root of something", so we use the chain rule to differentiate,
The expression <math>\sqrt{1-x^2}</math> is of the type "square root of something", so we use the chain rule to differentiate,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt]
&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)\\[5pt]
Zeile 31: Zeile 31:
We write the expression on the right over a common denominator,
We write the expression on the right over a common denominator,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \frac{\bigl(\sqrt{1-x^2}\bigr)^2-x^2}{\sqrt{1-x^2}}}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]
&= {}\rlap{-\frac{1}{x^2(1-x^2)}\cdot \frac{\bigl(\sqrt{1-x^2}\bigr)^2-x^2}{\sqrt{1-x^2}}}\phantom{-\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}(1-x^2)'\Bigr)}\\[5pt]

Version vom 12:54, 10. Mär. 2009

We can write the expression as

1x1−x2=x1−x2−1,

and then we see that we have "something raised to -1", which can be differentiated one step by using the chain rule,

ddxx1−x2−1=−1x1−x2−2x1−x2=−1x1−x22x1−x2=−1x2(1−x2)x1−x2.

The next step is to differentiate the product x1−x2  using the product rule,

=−1x2(1−x2)(x)1−x2+x(1−x2)=−1x2(1−x2)11−x2+x(1−x2).

The expression 1−x2  is of the type "square root of something", so we use the chain rule to differentiate,

=−1x2(1−x2)1−x2+x121−x2(1−x2)=−1x2(1−x2)1−x2+x121−x2(−2x)=−1x2(1−x2)1−x2−x21−x2.

We write the expression on the right over a common denominator,

=−1x2(1−x2)1−x21−x22−x2=−1x2(1−x2)1−x21−x2−x2=−1−2x2x2(1−x2)32.


Note: When we make simplifications of the form 1−x22=1−x2 , we assume that both sides are well defined (i.e. in this case that x lies between -1 and 1).

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