Lösung 1.2:3e

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At first sight, the expression looks like “
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<center> [[Image:1_2_3e.gif]] </center>
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<math>e</math>
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raised to something” and therefore we differentiate using the chain rule:
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<math>\frac{d}{dx}e^{\left\{ \left. \sin x^{2} \right\} \right.}=e^{\left\{ \left. \sin x^{2} \right\} \right.}\centerdot \left( \left\{ \left. \sin x^{2} \right\} \right. \right)^{\prime }</math>
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Then, we differentiate “sine of something”:
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<math>\begin{align}
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& e^{\sin x^{2}}\centerdot \left( \left\{ \left. \sin x^{2} \right\} \right. \right)^{\prime }=e^{\sin x^{2}}\centerdot \cos \left\{ \left. x^{2} \right\} \right.\centerdot \left( \left\{ \left. x^{2} \right\} \right. \right)^{\prime } \\
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& =e^{\sin x^{2}}\centerdot \cos x^{2}\centerdot 2x \\
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\end{align}</math>

Version vom 13:27, 12. Okt. 2008

At first sight, the expression looks like “ \displaystyle e raised to something” and therefore we differentiate using the chain rule:


\displaystyle \frac{d}{dx}e^{\left\{ \left. \sin x^{2} \right\} \right.}=e^{\left\{ \left. \sin x^{2} \right\} \right.}\centerdot \left( \left\{ \left. \sin x^{2} \right\} \right. \right)^{\prime }


Then, we differentiate “sine of something”:


\displaystyle \begin{align} & e^{\sin x^{2}}\centerdot \left( \left\{ \left. \sin x^{2} \right\} \right. \right)^{\prime }=e^{\sin x^{2}}\centerdot \cos \left\{ \left. x^{2} \right\} \right.\centerdot \left( \left\{ \left. x^{2} \right\} \right. \right)^{\prime } \\ & =e^{\sin x^{2}}\centerdot \cos x^{2}\centerdot 2x \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3e“