The way to differentiate this function would be to use the chain rule. There are different versions, but the form which is most appropriate to this situation is:
#([f(x)]^n)^'=n[f(x)]^(n-1)f^'(x)#
#(sin^3x)^'=3sin^2x(sinx)^'=3sin^2xcosx#
We can also derive #(sinx)^'# using the definition of a derivative:
#(sinx)^'=lim_(hrarr0) (sin(x+h)-sinx)/h=#
#lim_(hrarr0)(sinxcos h+cosxsin h-sinx)/h=#
#lim_(hrarr0) (sinx(cos h-1))/h+lim_(hrarr0)(cosxsin h)/h=cosx#
Now, the way I would explain the above derivation is this: As #h rarr0#, #cos h rarr 1#. This means that we are essentially subtracting #1# from #1#,
so #lim_(hrarr0)(sinx(cos h-1))/h=(sinx(1-1))/h=0#
Also, as #h rarr 0#, #sin h rarr 0#. So we are essentially dividing a very small number by another very small number, which is approx. equal to one. (The result can be made rigorous by using the sandwich theorem.)
Note: #lim_(hrarr0)sin h/h=1# only when #h# is in radians. So #(sinx)'# is only possible when #x# is also in radians.
So #lim_(hrarr0)=(cosxsin h)/h=cosx xx sin h/h=cosx xx1 = cosx#
#lim_(hrarr0) (sinx(cos h-1))/h+lim_(hrarr0)(cosxsin h)/h=0+cosx=cosx#
