Aus Online Mathematik Brückenkurs 2

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}