Lösung 1.2:2f

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The entire expression is made up of several levels,
The entire expression is made up of several levels,
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{{Displayed math||<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}
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<math>\cos \left\{ \left. \sqrt{\left\{ \left. 1-x \right\} \right.} \right\} \right.</math>
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and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",
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{{Displayed math||<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}
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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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and differentiate this using the chain rule,
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{{Displayed math||<math>\frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}</math>}}
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<math>\cos \left\{ \left. {} \right\} \right.</math>
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In the next differentiation, we have "the square root of something",
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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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{{Displayed math||<math>\bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,</math>}}
where we have used the differentiation rule,
where we have used the differentiation rule,
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{{Displayed math||<math>\frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
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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.
for the outer derivative.
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The whole differentiation in one go becomes:
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The whole differentiation in one go becomes
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<math>\begin{align}
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{{Displayed math||<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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\frac{d}{dx}\cos\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}\cdot\frac{d}{dx}\,\sqrt{1-x}\\[5pt]
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& =-\sin \sqrt{1-x}\centerdot \frac{1}{2\sqrt{1-x}}\centerdot \left( -1 \right) \\
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&= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot\frac{d}{dx}\,(1-x)\\[5pt]
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& =\frac{\sin \sqrt{1-x}}{2\sqrt{1-x}} \\
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&= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot (-1)\\[5pt]
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\end{align}</math>
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&= \frac{\sin\sqrt{1-x}}{2\sqrt{1-x}}\,\textrm{.}
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\end{align}</math>}}

Version vom 10:49, 15. Okt. 2008

The entire expression is made up of several levels,

\displaystyle \cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }

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

\displaystyle \cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,

and differentiate this using the chain rule,

\displaystyle \frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}

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

\displaystyle \bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,

where we have used the differentiation rule,

\displaystyle \frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}

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}\cdot\frac{d}{dx}\,\sqrt{1-x}\\[5pt] &= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot\frac{d}{dx}\,(1-x)\\[5pt] &= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot (-1)\\[5pt] &= \frac{\sin\sqrt{1-x}}{2\sqrt{1-x}}\,\textrm{.} \end{align}

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