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What is the derivative of sin x?

Updated: 12/22/2022
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βˆ™ 14y ago

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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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