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
