Lösung 1.2:2c

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When we see this expression, we should think "root of something",
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<center> [[Image:1_2_2c.gif]] </center>
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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 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
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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
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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>

Version vom 13:06, 11. Okt. 2008

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


\displaystyle \sqrt{\left\{ \left. {} \right\} \right.}


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

\displaystyle \left\{ \left. {} \right\} \right.=\cos x,


\displaystyle \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 },

where we have used the differentiation rule


\displaystyle \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}}.

Thus, we obtain


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

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