Lösung 1.2:2f

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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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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{{Abgesetzte Formel||<math>\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }</math>}}
and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cosine of something",
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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{{Abgesetzte Formel||<math>\cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,</math>}}
and differentiate this using the chain rule,
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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{{Abgesetzte Formel||<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>}}
In the next differentiation, we have "the square root of something",
In the next differentiation, we have "the square root of something",
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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>}}
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{{Abgesetzte Formel||<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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{{Abgesetzte Formel||<math>\frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
for the outer derivative.
for the outer derivative.
Zeile 23: Zeile 23:
The whole differentiation in one go becomes
The whole differentiation in one go becomes
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\cos\sqrt{1-x}
\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{d}{dx}\,\sqrt{1-x}\\[5pt]

Version vom 12:53, 10. Mär. 2009

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“