Answer
$f'\left( x \right) = {e^x}\cos x + {e^x}\sin x - \sin x$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {e^x}\sin x + \cos x \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\sin x} \right] + \frac{d}{{dx}}\left[ {\cos x} \right] \cr
& {\text{Use the product rule for derivatives }}\left( {gh} \right)' = gh' + hg' \cr
& {\text{Let }}g = {e^x}{\text{ and }}h = \sin x,{\text{ then}} \cr
& f'\left( x \right) = {e^x}\frac{d}{{dx}}\left[ {\sin x} \right] + \sin x\frac{d}{{dx}}\left[ {{e^x}} \right] + \frac{d}{{dx}}\left[ {\cos x} \right] \cr
& {\text{Use the Derivatives of Trigonometric Functions }} \cr
& {\text{and }}\frac{d}{{dx}}\left[ {{e^x}} \right] = {e^x},{\text{ then}} \cr
& f'\left( x \right) = {e^x}\left( {\cos x} \right) + \sin x\left( {{e^x}} \right) + \left( { - \sin x} \right) \cr
& {\text{Simplifying}} \cr
& f'\left( x \right) = {e^x}\cos x + {e^x}\sin x - \sin x \cr} $$
