We are given
[tex]f(x)=e^{x} cosh(x)[/tex]
Since, [tex]e^{x}[/tex] and [tex]cosh(x)[/tex] are multiplied
so, we will use product rule of derivative
[tex]\mathrm{Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'[/tex]
[tex]f'(x)=\frac{d}{dx}\left(e^x\right)\cosh \left(x\right)+\frac{d}{dx}\left(\cosh \left(x\right)\right)e^x[/tex]
we know that
[tex]\frac{d}{dx}\left(e^x\right)=e^x[/tex]
and
[tex]\frac{d}{dx}\left(\cosh \left(x\right)\right)=\sinh \left(x\right)[/tex]
so, we can plug these values
and we get
[tex]f'(x)=e^x\cosh \left(x\right)+\sinh \left(x\right)e^x[/tex]...............Answer
