Lösung 1.2:3c

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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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<math>\frac{1}{x\sqrt{1-x^{2}}}=\left( x\sqrt{1-x^{2}} \right)^{-1}</math>,
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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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and then we see that we have "something raised to
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{{Displayed math||<math>\begin{align}
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<math>-\text{1}</math>", which can be differentiated one step by using the chain rule:
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\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}
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&= {}\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]
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&= -\frac{1}{\bigl(x\sqrt{1-x^2}\bigr)^2}\cdot \bigl(x\sqrt{1-x^2}\bigr)'\\[5pt]
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&= -\frac{1}{x^2(1-x^2)}\cdot\bigl(x\sqrt{1-x^2}\bigr)'\,\textrm{.}
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\end{align}</math>}}
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The next step is to differentiate the product <math>x\cdot\sqrt{1-x^2}</math> using the product rule,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \frac{d}{dx}\left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)^{-1}=-1\centerdot \left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)\centerdot \left( \left\{ \left. x\sqrt{1-x^{2}} \right\} \right. \right)^{\prime } \\
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\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
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& =-\frac{1}{\left( x\sqrt{1-x^{2}} \right)^{2}}\centerdot \left( x\sqrt{1-x^{2}} \right)^{\prime } \\
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&= {}\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]
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( x\sqrt{1-x^{2}} \right)^{\prime } \\
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&= -\frac{1}{x^2(1-x^2)}\cdot \Bigl(1\cdot\sqrt{1-x^2} + x\cdot (\sqrt{1-x^2})'\Bigr)\,\textrm{.}
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\end{align}</math>
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\end{align}</math>}}
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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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The next step is to differentiate the product
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{{Displayed math||<math>\begin{align}
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<math>x\centerdot \sqrt{1-x^{2}}</math>
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\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
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using the product rule:
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&= -\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]
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&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} + x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot ( -2x)\Bigr)\\[5pt]
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&= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\Bigr)\,\textrm{.}
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\end{align}</math>}}
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We write the expression on the right over a common denominator,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \left( x \right)^{\prime }\sqrt{1-x^{2}}+x\left( \sqrt{1-x^{2}} \right)^{\prime } \right) \\
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\phantom{\frac{d}{dx}\,\bigl( \bbox[#FFEEAA;,1.5pt]{x\sqrt{1-x^2}}\,\bigr)^{-1}}{}
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( 1\sqrt{1-x^{2}}+x\left( \sqrt{1-x^{2}} \right)^{\prime } \right) \\
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&= {}\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]
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\end{align}</math>
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&= -\frac{1}{x^2(1-x^2)}\cdot \frac{1-x^2-x^2}{\sqrt{1-x^2}}\\[5pt]
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&= -\frac{1-2x^2}{x^2(1-x^2)^{3/2}}\,\textrm{.}
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\end{align}</math>}}
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The expression
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Note: When we make simplifications of the form <math>\bigl(\sqrt{1-x^2} \bigr)^2 = 1-x^2</math>, we assume that both sides are well defined (i.e. in this case that <math>x</math> lies between -1 and 1).
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<math>\sqrt{1-x^{2}}</math>
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is of the type "root of something", so we use the chain rule to differentiate,
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<math>\begin{align}
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\left( 1-x^{2} \right)^{\prime } \right) \\
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}+x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( -2x \right) \right) \\
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\left( \sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}} \right) \\
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\end{align}</math>
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We write the expression on the right over a common denominator:
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<math>\begin{align}
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{\left( \sqrt{1-x^{2}} \right)^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\
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& =-\frac{1}{x^{^{2}}\left( 1-x^{2} \right)}\centerdot \left( \frac{1-x^{2}-x^{2}}{\sqrt{1-x^{2}}} \right) \\
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& =-\frac{1-2x^{2}}{x^{2}\left( 1-x^{2} \right)^{{3}/{2}\;}} \\
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\end{align}</math>
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NOTE: When we make simplifications of the form
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<math>\left( \sqrt{1-x^{2}} \right)^{2}=1-x^{2}</math>, we assume that both sides are well defined (i.e. in this case that
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<math>x</math>
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lies between
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<math>-\text{1}</math>
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and
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<math>\text{1}</math>
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).
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Version vom 12:55, 15. Okt. 2008

We can write the expression as

\displaystyle \frac{1}{x\sqrt{1-x^{2}}} = \bigl(x\sqrt{1-x^2}\,\bigr)^{-1}\,\textrm{,}

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

\displaystyle \begin{align}

\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] &= -\frac{1}{\bigl(x\sqrt{1-x^2}\bigr)^2}\cdot \bigl(x\sqrt{1-x^2}\bigr)'\\[5pt] &= -\frac{1}{x^2(1-x^2)}\cdot\bigl(x\sqrt{1-x^2}\bigr)'\,\textrm{.} \end{align}

The next step is to differentiate the product \displaystyle x\cdot\sqrt{1-x^2} using the product rule,

\displaystyle \begin{align}

\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] &= -\frac{1}{x^2(1-x^2)}\cdot \Bigl(1\cdot\sqrt{1-x^2} + x\cdot (\sqrt{1-x^2})'\Bigr)\,\textrm{.} \end{align}

The expression \displaystyle \sqrt{1-x^2} is of the type "square root of something", so we use the chain rule to differentiate,

\displaystyle \begin{align}

\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}}\cdot ( -2x)\Bigr)\\[5pt] &= -\frac{1}{x^2(1-x^2)}\Bigl(\sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}\Bigr)\,\textrm{.} \end{align}

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

\displaystyle \begin{align}

\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] &= -\frac{1}{x^2(1-x^2)}\cdot \frac{1-x^2-x^2}{\sqrt{1-x^2}}\\[5pt] &= -\frac{1-2x^2}{x^2(1-x^2)^{3/2}}\,\textrm{.} \end{align}


Note: When we make simplifications of the form \displaystyle \bigl(\sqrt{1-x^2} \bigr)^2 = 1-x^2, we assume that both sides are well defined (i.e. in this case that \displaystyle x lies between -1 and 1).

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