Answer
$$\frac{d}{{dx}}\left[ {\cosh x} \right] = \sinh x$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\cosh x} \right] = \sinh x \cr
& {\text{By the definition of }}\cosh x \cr
& \frac{d}{{dx}}\left[ {\cosh x} \right] = \frac{d}{{dx}}\left[ {\frac{{{e^x} + {e^{ - x}}}}{2}} \right] \cr
& {\text{Differentiating}} \cr
& = \frac{d}{{dx}}\left[ {\frac{{{e^x}}}{2}} \right] + \frac{d}{{dx}}\left[ {\frac{{{e^{ - x}}}}{2}} \right] \cr
& = \frac{{{e^x}}}{2} - \frac{{{e^{ - x}}}}{2} \cr
& = \frac{{{e^x} - {e^{ - x}}}}{2} \cr
& {\text{Where }}\frac{{{e^x} - {e^{ - x}}}}{2} = \sinh x.{\text{The differentiation formula has}} \cr
& {\text{been proved}}{\text{.}} \cr
& = \sinh x \cr} $$
