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Lösung 1.2:4a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Lösning 1.2:4a moved to Solution 1.2:4a: Robot: moved page)
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We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives
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<center> [[Image:1_2_4a-1(2).gif]] </center>
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<math>\begin{align}
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<center> [[Image:1_2_4a-2(2).gif]] </center>
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& \frac{d}{dx}\frac{x}{\sqrt{1-x^{2}}}=\frac{\left( x \right)^{\prime }\sqrt{1-x^{2}}-x\left( \sqrt{1-x^{2}} \right)^{\prime }}{\left( \sqrt{1-x^{2}} \right)^{2}} \\
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& \\
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& =\frac{1\centerdot \sqrt{1-x^{2}}-x\left( \sqrt{1-x^{2}} \right)^{\prime }}{1-x^{2}} \\
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\end{align}</math>
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We determine the derivative
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<math>\left( \sqrt{1-x^{2}} \right)^{\prime }</math>
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by using the chain rule
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<math>\begin{align}
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& =\frac{\sqrt{1-x^{2}}-x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( 1-x^{2} \right)^{\prime }}{1-x^{2}} \\
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& \\
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& =\frac{\sqrt{1-x^{2}}-x\centerdot \frac{1}{2\sqrt{1-x^{2}}}\centerdot \left( -2x \right)}{1-x^{2}} \\
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\end{align}</math>
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We simplify the result as far as possible, so as to make the second differentiation easier:
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<math>\begin{align}
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& =\frac{\sqrt{1-x^{2}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\
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& \\
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& =\frac{\frac{\left( \sqrt{1-x^{2}} \right)^{2}}{\sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\
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& \\
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& =\frac{\frac{1-x^{2}+x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}} \\
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& \\
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& =\frac{1}{\left( 1-x^{2} \right)^{{3}/{2}\;}} \\
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\end{align}</math>
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The second derivative is:
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<math>\begin{align}
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& \frac{d^{^{2}}}{dx^{^{2}}}\frac{x}{\sqrt{1-x^{2}}}=\frac{d}{dx}\frac{1}{\left( 1-x^{2} \right)^{{3}/{2}\;}} \\
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& \\
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& =\frac{d}{dx}\left( 1-x^{2} \right)^{-{3}/{2}\;}=-\frac{3}{2}\left( 1-x^{2} \right)^{-\frac{3}{2}-1}\centerdot \left( 1-x^{2} \right)^{\prime } \\
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& \\
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& =-\frac{3}{2}\left( 1-x^{2} \right)^{{-5}/{2}\;}\centerdot \left( -2x \right)=3x\left( 1-x^{2} \right)^{{-5}/{2}\;} \\
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& \\
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& =\frac{3x}{\left( 1-x^{2} \right)^{{5}/{2}\;}} \\
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\end{align}</math>

Version vom 14:04, 12. Okt. 2008

We are to differentiate the expression two times, so we start by differentiating once. The quotient rule gives


ddxx1−x2=1−x22x1−x2−x1−x2=1−x211−x2−x1−x2


We determine the derivative 1−x2  by using the chain rule


=1−x21−x2−x121−x21−x2=1−x21−x2−x121−x2−2x


We simplify the result as far as possible, so as to make the second differentiation easier:


=1−x21−x2+x21−x2=1−x21−x21−x22+x21−x2=1−x21−x21−x2+x2=11−x232


The second derivative is:


d2dx2x1−x2=ddx11−x232=ddx1−x2−32=−231−x2−23−11−x2=−231−x2−52−2x=3x1−x2−52=3x1−x252

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:4a“