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.
