where fis homogeneous and linear in the system of quaternions dq, dr, ds,...... and distributive with respect to each of them. Thus, in differentiating any power, product, &c. of one or more quaternions, each factor is to be differentiated as if it alone were variable; and the terms corresponding to these are to be added for the complete differential. This differs from the ordinary process of scalar differentiation solely in the fact that, on account of the non-commutative property of quaternion multiplication, each factor must be differentiated in situ. Thus

d(qr) = dq.r+qdr, but not generally = rdq+qdr.

133. As Examples we take chiefly those which lead to results of constant use to us in succeeding Chapters. Some of the work will be given at full length as an exercise in quaternion transformations.

(1) (Tp)=-p2.

The differential of the left-hand side is simply, since Tp is a scalar, 2 Tp dTp.

2TqdTq = d(qKq) = £„~[(1 + d2) K (q + 1⁄47) − qKq ],

Hence

since

dTq=S.UKqdq= S.Uq'dq

Tq=TKq, and UKq=Uq,

If q=p, a vector, Kq=Kp=p, and the formula becomes

qdr+drKq = (q2 + Tq2 +2 Sq.q) dq;

whence dq, i. e. dr3, is at once found in terms of dr. This cess is given by Hamilton, Lectures, p. 628.

pro

Comparing, we have

dSq = Sdq, dVq = Vdq.

Since KqSq-Vq, we find by a similar process

dKq= Kdq.

134. Successive differentiation of course presents no new difficulty.

Thus, we have seen that

d(q2) = dqq+qdq.

Differentiating again, we have

d2 (q2) = d2q.q+2(dq)2+qd3q,

and so on for higher orders.

If q be a vector, as p, we have, § 133 (1),

d(p2) = 2 Spdp.

Hence d'(p) 2 (dp)+2 Spd p, and so on.

=

Up

=

Τρο

· ((Vpdp)2 — p2 Vpd2p+2 VpdpSpdp).

[This may be farther simplified; but it may be well to caution the student that we cannot, for such a purpose, write the above expression as

135. If the first differential of q be considered as a constant quaternion, we have, of course,

d2q

d'q = 0,

d3q = 0, &c.,

and the preceding formulæ become considerably simplified.

Hamilton has shown that in this case Taylor's Theorem admits of an easy extension to quaternions. That is, we may write

x2

ƒ (q+xdq) = ƒ (q)+xdƒ (q) + 2.2 d2 ƒ (q) + ......

if d'q=0; subject, of course, to particular exceptions and limita-
tions as in the ordinary applications to functions of scalar
variables. Thus, let

ƒ(q) = q3, and we have

df (q) = q2 dq+qdqq+dqq2,

d2f (q) = 2 dq qdq+2q(dq)2+2(dq)'q,
d3f (q) = 6 (dq)3,

and it is easy to verify by multiplication that we have rigorously (y + xdq)3 = q3 +x (q2dq+qdqq+dqq2)+x2 (dqqdq+q(dq)2+(dq)3q)+x3 (dq)3 ; which is the value given by the application of the above form of Taylor's Theorem.

As we shall not have occasion to employ this theorem, and as the demonstrations which have been found are all too laborious for an elementary treatise, we refer the reader to Hamilton's works, where he will find several of them.

136. To differentiate a function of a function of a quaternion we proceed as with scalar variables, attending to the peculiarities already pointed out.

137. A case of considerable importance in geometrical applications of quaternions is the differentiation of a scalar function. of p, the vector of any point in space.

where F is a scalar function and C an arbitrary constant, be
the equation of a series of surfaces. Its differential,

f(p, dp) = 0,