Answer
$$y'={e^{e^x+x}}$$$$y''=(e^x+1){e^{e^x+x}}$$
Work Step by Step
$y'=\frac{d{e^{e}}^x}{dx}$
Using the chain rule:
$y'=\frac{d{e^{e}}^x}{de^x}
\times \frac{d{e^x}}{dx}$
$={e^{e}}^x
\times e^x$
$={e^{e^x+x}}$
$y''=\frac{d{e^{e^x+x}}}{dx}$
Using the chain rule:
$y''=\frac{d{e^{e^x+x}}}{d e^x+x}
\times \frac{de^x+x}{dx}$
$={e^{e^x+x}}
\times (e^x+1)$
$=(e^x+1){e^{e^x+x}}$
