Lösung 1.2:3e
Aus Online Mathematik Brückenkurs 2
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Version vom 10:06, 11. 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} |
