Page 133 - Foundations Of Differential Calculus

116 6. On the Differentiation of Transcendental Functions
Hence we conclude as above (paragraph 177) that the general formula will
be
d n+1 y = 1 · 2 · 3 ··· n
dx n+1 (1 − x ) n+1/2
2

1 n (n − 1) n−2
n
× x + · x
2 1 · 2
1 · 3 n (n − 1) (n − 2) (n − 3) n−4
+ · x
2 · 4 1 · 2 · 3 · 4
1 · 3 · 5 n (n − 1) (n − 2) (n − 3) (n − 4) (n − 5) n−6
+ · x
2 · 4 · 6 1 · 2 · 3 · 4 · 5 · 6

+ ··· .

201. There remain some quantities that arise as inverses of these func-
tions, namely the sines and tangents of given arcs, and we ought to show
how these are differentiated. Let x be a circular arc and let sin x denote its
sine, whose differential we are to investigate. We let y = sin x and replace
x by x + dx so that y becomes y + dy. Then y + dy = sin (x + dx) and
dy = sin (x + dx) − sin x.

But
sin (x + dx) = sin x · cos dx + cos x · sin dx,

and since, as we have shown in Introduction,
3 5
z z z
sin z = − + − ··· ,
1 1 · 2 · 3 1 · 2 · 3 · 4 · 5
z 2 z 4
cos z =1 − + − ··· ,
1 · 2 1 · 2 · 3 · 4
when we exclude the vanishing terms, we have cos dx = 1 and sin dx = dx,
so that

sin (x + dx) = sin x + dx cos x.

Hence, when we let y = sin x,wehave
dy = dx cos x.

Therefore, the differential of the sine of any arc is equal to the product of
the differential of the arc and the cosine of the arc.