Prove the derivative of sinx = cosx

Prove the derivative of sinx = cosx

Prove using both methods:

(1) definition of the limit

(2) rules for derivatives

Help Please!!!!!

By the definition of the derivative:
f'(x) = lim (h-->0) [f(x + h) - f(x)]/h.

So, the derivative of sin(x) is given by the limit:
d/dx sin(x) = lim (h-->0) [sin(x + h) - sin(x)]/h.

Using the sine addition formula, this becomes:
lim (h-->0) [sin(x)cos(h) + cos(x)sin(h) - sin(x)]/h
= lim (h-->0) [sin(x)cos(h) - sin(x) + cos(x)sin(h)]/h, by re-arranging
= lim (h-->0) {sin(x)[cos(h) - 1] + cos(x)sin(h)]/h, by factoring out sin(x)
= lim (h-->0) {sin(x)[cos(h) - 1]/h + cos(x)[sin(h)/h]}.

Using the special limits:
lim (h-->0) [cos(h) - 1]/h = 0 and lim (h-->0) sin(h)/h = 1,

this limit equals sin(x)(0) + cos(x)(1) = cos(x). Therefore:
d/dx sin(x) = cos(x), as required.

As for your second question, what do you mean by "rules for derivatives?" The only proofs of the derivative of sin(x) that are not fallacious use the definition of the derivative.

limit (sin(x+t)-sinx)/t = limit 2cos((x+t+x)/2)sin((x+t-x)/2)/t
= cos (x+t/2) * sin(t/2) / (t/2)
when t -> 0 sint/t -> 1
so the limit = cosx