An even periodic extension of $e^x$ will be represented by the following Fourier Cosine Series:
\begin{align*} e^x &\sim A_0 + \sum\limits_{n=1}^\infty A_n \cos \frac{n \pi x}{L} \\ \end{align*}
The easy way to solve for the coefficients is to use simple trigonometric orthogonality properties:
\begin{align*} A_0 &= \frac{1}{L} \int_0^L e^x \, dx = \frac{e^L-1}{L} \\ A_n &= \frac{2}{L} \int_0^L e^x \cos \frac{n \pi x}{L} \, dx \\ \end{align*}
However, for this textbook problem I am explicitly supposed to use term-by-term differentiation approach to solve this.
Term by Term Differentiation yields:
\begin{align*} e^x &\sim - \sum\limits_{n=1}^\infty \frac{n \pi}{L} A_n \sin \frac{n \pi x}{L} \\ \end{align*}
My work:
Over the interval of $[0,L]$, these two representations of $e^x$ will be identical:
\begin{align*} A_0 + \sum\limits_{n=1}^\infty A_n \cos \frac{n \pi x}{L} &= - \sum\limits_{n=1}^\infty \frac{n \pi}{L} A_n \sin \frac{n \pi x}{L} \\ \end{align*}
\begin{align*} A_0 + \sum\limits_{n=1}^\infty A_n \left( \frac{n \pi}{L} \sin \frac{n \pi x}{L} + \cos \frac{n \pi x}{L} \right) &= 0 \\ \end{align*}
I'm really stumped on what to do from here.