Lösung 1.2:3e

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At first sight, the expression looks like “''e'' raised to something” and therefore we differentiate using the chain rule,
At first sight, the expression looks like “''e'' raised to something” and therefore we differentiate using the chain rule,
-
{{Displayed math||<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =
+
{{Abgesetzte Formel||<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =
e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}
e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}</math>}}
Then, we differentiate “sine of something”,
Then, we differentiate “sine of something”,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)'
e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)'
&= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt]
&= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt]
&= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{.}
&= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 12:54, 10. Mär. 2009

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

\displaystyle \frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}} =

e^{\bbox[#FFEEAA;,1.5pt]{\sin x^2}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\sin x^2}\bigr)'\,\textrm{.}

Then, we differentiate “sine of something”,

\displaystyle \begin{align}

e^{\sin x^2}\cdot \bigl( \sin \bbox[#FFEEAA;,1.5pt]{x^2} \bigr)' &= e^{\sin x^2}\cdot \cos\bbox[#FFEEAA;,1.5pt]{x^2} \cdot \bigl( \bbox[#FFEEAA;,1.5pt]{x^2}\bigr)'\\[5pt] &= e^{\sin x^2}\cdot \cos x^2\cdot 2x\,\textrm{.} \end{align}

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