#sin^5x# is a shorthand for #(sinx)^5#. There are two functions involved here: #g(x)=sinx# and #f(x)=x^5#, composed such that #y=f(g(x))#, which can be differentiated using the chain rule.

The chain rule states that #y'=f'(g(x))*g'(x)#. #f'(x)=5x^4# and #g'(x)=cosx#. Therefore, #y'=5(sinx)^4*cosx=5sin^4xcosx#.

Alternatively, if you want you can do this with no intermediate steps by using the chain rule in the form that #(dy)/(dx)=(dy)/(du) * (du)/(dx)# for any function #u(x)#. You can say that #(d((sinx)^5))/(dx)=(d((sinx)^5))/(d(sinx))*(d(sinx))/(dx)=5(sinx)^4*cosx=5sin^4xcosx#