Answer
$e^{2x}$
Work Step by Step
Given: $f(x)=e^x \cosh x$
Now, $f'(x)=\dfrac{d}{dx}(e^x \cosh x)$
Apply the product rule:
$f'(x)=\cosh x(\frac{d}{dx}e^x)+(e^x)( \frac{d}{dx}\cosh x)$
or, $=e^x \cosh x+e^x\sinh x$
or, $=e^x(\cosh x+\sinh x)$
Since, we have $\cosh(x) = \dfrac{e^{x} + e^{-x}}{2}$ and
$\sinh(x) = \frac{e^{x} - e^{-x}}{2}$
Thus, $e^x(\cosh x+\sinh x)=e^{x}(\dfrac{e^{x} + e^{-x}}{2}+ \frac{e^{x} - e^{-x}}{2})$
or, $=e^{2x}$
