Lösung 1.2:2c

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When we see this expression, we should think "root of something",
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When we see this expression, we should think "square root of something",
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{{Displayed math||<math>\sqrt{\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}}}\,\textrm{,}</math>}}
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<math>\sqrt{\left\{ \left. {} \right\} \right.}</math>
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and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression <math>\bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x</math>,
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{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,</math>}}
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and in order to differentiate it, we should first differentiate the outer function , "the root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression
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<math>\left\{ \left. {} \right\} \right.=\cos x</math>,
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<math>\frac{d}{dx}\sqrt{\left\{ \left. \cos x \right\} \right.}=\frac{1}{2\sqrt{\left\{ \left. \cos x \right\} \right.}}\centerdot \left( \left\{ \left. \cos x \right\} \right. \right)^{\prime }</math>,
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where we have used the differentiation rule
where we have used the differentiation rule
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{{Displayed math||<math>\frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}</math>}}
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<math>\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2}\;}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}</math>.
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Thus, we obtain
Thus, we obtain
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{{Displayed math||<math>\frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}</math>}}
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<math>\frac{d}{dx}\sqrt{\cos x}=\frac{1}{2\sqrt{\cos x}}\centerdot \left( -\sin x \right)=-\frac{\sin x}{2\sqrt{\cos x}}</math>
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Version vom 08:15, 15. Okt. 2008

When we see this expression, we should think "square root of something",

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

and in order to differentiate it, we should first differentiate the outer function , "the square root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression \displaystyle \bbox[#FFEEAA;,1.5pt]{\phantom{\cos x}} = \cos x,

\displaystyle \frac{d}{dx}\,\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}} = \frac{1}{2\sqrt{\bbox[#FFEEAA;,1.5pt]{\cos x}}}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\cos x}\bigr)'\,,

where we have used the differentiation rule

\displaystyle \frac{d}{dx}\,\sqrt{x} = \frac{d}{dx}\,x^{1/2} = \tfrac{1}{2}x^{1/2-1} = \tfrac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\,\textrm{.}

Thus, we obtain

\displaystyle \frac{d}{dx}\,\sqrt{\cos x} = \frac{1}{2\sqrt{\cos x}}\cdot (-\sin x) = -\frac{\sin x}{2\sqrt{\cos x}}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2c“