Lösung 1.2:2f

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The entire expression is made up of several levels,
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<center> [[Image:1_2_2f-1(2).gif]] </center>
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<math>\cos \left\{ \left. \sqrt{\left\{ \left. 1-x \right\} \right.} \right\} \right.</math>
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<center> [[Image:1_2_2f-2(2).gif]] </center>
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and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cos of something",
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<math>\cos \left\{ \left. {} \right\} \right.</math>
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and differentiate this using the chain rule:
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<math>\frac{d}{dx}\cos \left\{ \left. \sqrt{1-x} \right\} \right.=-\sin \left\{ \left. \sqrt{1-x} \right\} \right.\centerdot \left( \left\{ \left. \sqrt{1-x} \right\} \right. \right)^{\prime }</math>
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In the next differentiation, we have "the root of something",
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<math>\left( \sqrt{\left\{ \left. 1-x \right\} \right.} \right)^{\prime }=\frac{1}{2\sqrt{1-x}}\centerdot \left( 1-x \right)^{\prime }</math>
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where we have used the differentiation rule,
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<math>\frac{d}{dx}\left( \sqrt{x} \right)=\frac{1}{2\sqrt{x}}</math>
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for the outer derivative.
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The whole differentiation in one go becomes:
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<math>\begin{align}
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& \frac{d}{dx}\cos \sqrt{1-x}=-\sin \sqrt{1-x}\centerdot \frac{d}{dx}\sqrt{1-x} \\
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& =-\sin \sqrt{1-x}\centerdot \frac{1}{2\sqrt{1-x}}\centerdot \frac{d}{dx}\left( 1-x \right) \\
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& =-\sin \sqrt{1-x}\centerdot \frac{1}{2\sqrt{1-x}}\centerdot \left( -1 \right) \\
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& =\frac{\sin \sqrt{1-x}}{2\sqrt{1-x}} \\
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\end{align}</math>

Version vom 13:53, 11. Okt. 2008

The entire expression is made up of several levels,


\displaystyle \cos \left\{ \left. \sqrt{\left\{ \left. 1-x \right\} \right.} \right\} \right.


and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cos of something",


\displaystyle \cos \left\{ \left. {} \right\} \right.


and differentiate this using the chain rule:


\displaystyle \frac{d}{dx}\cos \left\{ \left. \sqrt{1-x} \right\} \right.=-\sin \left\{ \left. \sqrt{1-x} \right\} \right.\centerdot \left( \left\{ \left. \sqrt{1-x} \right\} \right. \right)^{\prime }


In the next differentiation, we have "the root of something",


\displaystyle \left( \sqrt{\left\{ \left. 1-x \right\} \right.} \right)^{\prime }=\frac{1}{2\sqrt{1-x}}\centerdot \left( 1-x \right)^{\prime }


where we have used the differentiation rule,


\displaystyle \frac{d}{dx}\left( \sqrt{x} \right)=\frac{1}{2\sqrt{x}}


for the outer derivative.

The whole differentiation in one go becomes:


\displaystyle \begin{align} & \frac{d}{dx}\cos \sqrt{1-x}=-\sin \sqrt{1-x}\centerdot \frac{d}{dx}\sqrt{1-x} \\ & =-\sin \sqrt{1-x}\centerdot \frac{1}{2\sqrt{1-x}}\centerdot \frac{d}{dx}\left( 1-x \right) \\ & =-\sin \sqrt{1-x}\centerdot \frac{1}{2\sqrt{1-x}}\centerdot \left( -1 \right) \\ & =\frac{\sin \sqrt{1-x}}{2\sqrt{1-x}} \\ \end{align}

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