trig identities go as such,

remember that f' = [f(x+h) - f(x)]/h as h ->0

sin(x+h) = sin x cos h + cos x sin h

therefore:

f' = lim (sin x cos h + cos x sin h - sin x )/h as h ->0

using the properties of limits:

f' = lim (sin x cos h - sin x)/h + lim cos x sin h/h as h->0

factor sin x:

lim sin x(cos h - 1)/h + lim cos x sin h/h as h->0

based on the fact that sin x and cos x have nothing to do with h(i.e. they are not involved in the limit):

sin x lim (cos h-1)/h + cos x lim sin h/h as h->0

computing the limit of (cos h - 1)/h and sin h/h as h -> 0 we have

sin x * 0 + cos x * 1 = cos x

EDIT: I know you can find out for yourself, just wanted to save you some trouble :P

and about chain rule, you are right, it is pretty intuitive:

if y = f(x)

and z = g(x)

and h(x) = f(g(x))

dy/dx = dy/dz * dz/dt

It is just that doing it this way tends to make people think of leibnix notation as a fraction, which in some ways it is, but most of the time it is better to think of it as just a notation.