An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

64 Successive differentiation of explicit functions. Bk. I. ch. VII.
We have: =1 + - 2 (cos y)2 cos y. sin (y -+ zi), therefore:
y y' — si y sin y s (y + z) + - cosycos(y -+ ~ z)} = y'cos (2 y + 3-t)
= (cos y)2 sin 2(y + 1r),
y"'"= y' {- 2 cosy sinysin2(y + ~r) + 2(cosy)2cos2 (y + -C) }
-- 2y'cosycos(3y + 2-) = 2(cosy)3sin3(y -+ ~z).
By reasoning from n to n + 1 it is proved, that in general:
y(n) -= I- (cosy) sinn(y -+ 2 I), (| =1).