\(\displaystyle{f}'{\left({x}\right)}=\lim_{{{h}\rightarrow{0}}}{\frac{{{\sin{{\left({x}+{h}\right)}}}-{\sin{{\left({x}\right)}}}}}{{{h}}}}\)

Let's use representation of a difference of sin functions as a product of sin and cos, and we get:

\(\displaystyle{f}'{\left({x}\right)}=\lim_{{{h}\rightarrow{0}}}{\frac{{{2}\times{\sin{{\left({\frac{{{h}}}{{{2}}}}\right)}}}{\cos{{\left({x}+{\frac{{{h}}}{{{2}}}}\right)}}}}}{{{h}}}}\)

\(\displaystyle{f}'{\left({x}\right)}=\lim_{{{h}\rightarrow{0}}}{\frac{{{\sin{{\left({\frac{{{h}}}{{{2}}}}\right)}}}}}{{{\frac{{{h}}}{{{2}}}}}}}\times\lim_{{{h}\rightarrow{0}}}{\cos{{\left({x}+{\frac{{{h}}}{{{2}}}}\right)}}}\)

\(\displaystyle{f}'{\left({x}\right)}={1}\times{\cos{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}\)

Thus, the derivative is \(\displaystyle{f}'{\left({x}\right)}={\cos{{\left({x}\right)}}}\)

Let's use representation of a difference of sin functions as a product of sin and cos, and we get:

\(\displaystyle{f}'{\left({x}\right)}=\lim_{{{h}\rightarrow{0}}}{\frac{{{2}\times{\sin{{\left({\frac{{{h}}}{{{2}}}}\right)}}}{\cos{{\left({x}+{\frac{{{h}}}{{{2}}}}\right)}}}}}{{{h}}}}\)

\(\displaystyle{f}'{\left({x}\right)}=\lim_{{{h}\rightarrow{0}}}{\frac{{{\sin{{\left({\frac{{{h}}}{{{2}}}}\right)}}}}}{{{\frac{{{h}}}{{{2}}}}}}}\times\lim_{{{h}\rightarrow{0}}}{\cos{{\left({x}+{\frac{{{h}}}{{{2}}}}\right)}}}\)

\(\displaystyle{f}'{\left({x}\right)}={1}\times{\cos{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}\)

Thus, the derivative is \(\displaystyle{f}'{\left({x}\right)}={\cos{{\left({x}\right)}}}\)