Answer
$F'(x)=e^{rx}r$
or
$F'(x)=re^{rx}$
Work Step by Step
Since, $e^{x+iy}=e^{x}e^{iy}=e^{x}(cosy+isiny)$
Given: $F(x)=e^{rx}=e^{(a+ib)x}=e^{(ax)}e^{(ibx)}$
On differentiating , we get
$F'(x)=e^{ax}(be^{ibx)}+e^{ibx}(ae^{ax})$
$F'(x)=e^{ax}e^{ibx}(a+ib)$
Because $e^{ax}e^{ibx}=e^{rx}$ and $(a+ib)=r$
Then
$F'(x)=e^{rx}r$
or
$F'(x)=re^{rx}$
