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f'[x] = lim(h->0) (f[x+h]-f[x])/h
lim(h->0) (sin[x+h]-sin[x])/h
By angle-addition formula, we have:
lim(h->0) (sin[x]cos[h]+sin[h]cos[x]-sin[x])/h
lim(h->0) (sin[x]cos[h]-sin[x])/h + lim(h->0) (sin[h]cos[x])/h
sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h
In a calculus class, it is shown that:
lim(h->0) (cos[h]-1)/h = 0 and that lim(h->0) sin[h]/h is 1. So,
sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h becomes
sin[x]*0 + cos[x]*1
cos[x]
So, if f[x] = sin[x], f'[x] = cos[x]
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