Step 1

Given

\(\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}}\)

The first five derivatives are

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

\(\displaystyle{f}{''}{\left({x}\right)}=-{\sin{{x}}}\)

\(\displaystyle{f}^{{{\left({3}\right)}}}=-{\cos{{x}}}\)

\(\displaystyle{f}^{{{\left({4}\right)}}}={\sin{{x}}}\)

\(\displaystyle{f}^{{{\left({5}\right)}}}={\cos{{x}}}\)

From this it can be concluded that every fourth one repeats .

Step 2

The eighth derivative will be

\(\displaystyle{{f}^{{{\left({8}\right)}}}{\left({x}\right)}}={\sin{{x}}}\)

so the ninth derivative will be

\(\displaystyle{{f}^{{{\left({9}\right)}}}{\left({x}\right)}}={\cos{{x}}}\)

The hundredth derivative will be

\(\displaystyle{{f}^{{{\left({100}\right)}}}{\left({x}\right)}}={\sin{{x}}}\)

Then,

\(\displaystyle{{f}^{{{\left({102}\right)}}}{\left({x}\right)}}=-{\sin{{x}}}\)

Given

\(\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}}\)

The first five derivatives are

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

\(\displaystyle{f}{''}{\left({x}\right)}=-{\sin{{x}}}\)

\(\displaystyle{f}^{{{\left({3}\right)}}}=-{\cos{{x}}}\)

\(\displaystyle{f}^{{{\left({4}\right)}}}={\sin{{x}}}\)

\(\displaystyle{f}^{{{\left({5}\right)}}}={\cos{{x}}}\)

From this it can be concluded that every fourth one repeats .

Step 2

The eighth derivative will be

\(\displaystyle{{f}^{{{\left({8}\right)}}}{\left({x}\right)}}={\sin{{x}}}\)

so the ninth derivative will be

\(\displaystyle{{f}^{{{\left({9}\right)}}}{\left({x}\right)}}={\cos{{x}}}\)

The hundredth derivative will be

\(\displaystyle{{f}^{{{\left({100}\right)}}}{\left({x}\right)}}={\sin{{x}}}\)

Then,

\(\displaystyle{{f}^{{{\left({102}\right)}}}{\left({x}\right)}}=-{\sin{{x}}}\)