Page 174 - Foundations Of Differential Calculus

8. On the Higher Differentiation of Differential Formulas 157
Then we have
2
2
3

4 ps dx 4 10p − 3 qr dx 4 15p − 13 pq dx 4
d x = − 2 + 2 − 3 .
2
2
1+ p (1 + p ) (1 + p )
Since we are assuming that dy = pdx, when this is differentiated, we have
2
2 2 2 2 p qdx 2 qdx 2
d y = qdx + pd x = qdx − = ,
1+ p 2 1+ p 2
2
2
3 2pq dx 3 2qdx d x
rdx
3
d y = − + ,
2
1+ p 2 (1 + p ) 2 1+ p 2
so that
2
3 rdx 3 4pq dx 3
d y = − .
2
1+ p 2 (1 + p ) 2
When we differentiate again we have
3
2

4 sdx 4 13pqr dx 4 4 6p − 1 q dx 4
d y = − 2 + 3 .
2
2
1+ p 2 (1 + p ) (1 + p )
Hence all higher differentials of both x and y are expressed through fi-
nite quantities and powers of dx. After these substitutions, the resulting
expression is completely free of second differentials.
270. Now that we have given the method for stripping second and higher
differentials from expressions, it is fitting that we illustrate this material
with some few examples.
2
2
I. Let the given expression be xd y/dx , in which dx is set constant.
2
2
Hence we let dy = pdx and dp = qdx, so that d y = qdx and the
given expression becomes this finite quantity xq.
2 2 2
II. Let the given expression be dx + dy /d x, in which dy is set con-
2 2
stant. We let dx = pdy, dp = qdy. Since d x = qdy , we obtain
2
1+ p /q. However, if we should wish, as before, to let dy = pdx,
2
dp = qdx, since dy is constant, we have 0 = pd x + dp dx and
2 2 2
d x = −qdx /p. Hence the given expression becomes −p 1+ p /q.
III. Let the given expression be
2
2
yd x − xd y
,
dx dy
in which ydx is set constant. We let dy = pdx and dp = qdx; from
2
2
paragraph 268 we have d x = −pdx /y and
2
2 2 p dx 2
d y = qdx − .
y