Derivative of sin x, Algebraic Proof A speciﬁc derivative formula tells us how to take the derivative of a speciﬁc function: if f (x) = xn then f � (x) = nxn−1 . We’ll now compute a speciﬁc formula for the derivative of the function sin x. As before, we begin with the deﬁnition of the derivative: d sin(x + Δx) − sin(x) sin x = lim dx Δx→0 Δx You may remember the following angle sum formula from high school: sin(a + b) = sin(a) cos(b) + sin(b) cos(a) This lets us untangle the x from the Δx as follows: d sin x cos Δx + cos x sin Δx − sin(x) sin x = lim . dx Δx→0 Δx We can simplify this expression using some basic algebraic facts: � � d sin x cos Δx − sin x cos x sin Δx sin x = lim + Δx dx Δx→0 Δx � � sin x(cos Δx − 1) cos x sin Δx + = lim Δx→0 Δx Δx � � � � �� cos Δx − 1 sin Δx = lim sin x + cos x Δx→0 Δx Δx � � � � d cos Δx − 1 sin Δx sin x = lim sin x + lim cos x dx Δx→0 Δx Δx→0 Δx We now have two familiar functions – sin x and cos x – and two ugly looking fractions to deal with. The fractions may be familiar from our discussion of removable discontinuities. cos Δx − 1 Δx→0 Δx sin Δx lim Δx→0 Δx lim = 0 = 1. Using these (as yet unproven) facts, � � � � cos Δx − 1 sin Δx lim sin x + lim cos x Δx→0 Δx Δx→0 Δx simpliﬁes to sin x · 0 + cos x · 1 = cos x. We conclude: d sin x = cos x dx