101]
DERIVATIVES AND INTEGRALS
203
In this formula D x °yi, P> x y 2 must be interpreted as meaning simply y x
and y 2 . To prove the theorem we observe that
Dx iyiVi)=I>*y 1 -¿/2+.?/!• D x y 2,
D x 2 (.2/11/2)=D 2 y 1 ■l/2 J rZDxy l . D x y-i+y x D 2 y 2 ;
and so on. It is obvious that by repeating this process we arrive at a
formula of the type
D x n (lh3/2) = Dx n yi-V2 + a n , 1 D x n ~ l y x .D x y 2 + a n , %D x n ~ 2 y l .D 2 y 2 +...+y l .D x n y 2 .
Let us assume that a n>r = for r— I, 2, ... n— 1, and show that if this
is so then a n+1>r = f° r r== l> 2, ... ». It will then follow by the
principle of mathematical induction that a n>r — (^j for all values of n and r
in question.
When we form J/ X n + I (y x y 2 ) by differentiating D x n (y x y 2 ) it is clear that the
coefficient of B x n + 1 ~ r y 1 .D x r y 2 is
&n, r "b «n. r — 1
This establishes the theorem.
7. Find </>(”)(x) when </> (x) is any one of the functions x/(l +x), x 2 /{ 1 — x),
.rsinj?, x 1 cos x. Verify, in the case of the first two functions, that the
results agree with those obtained by the method of partial fractions.
8. The nth derivative of x m f(x) is
m !
(m — n) !
x m ~ n f{x) + n
, x
(m — 7i +1) !
n(n- 1)
+-
m !
X/m-n + 2
1.2 (m~n + 2)!'
the series being continued for n +1 terms or until zero terms occur.
9. Differentiate 1/(1 -# 2 ) = {1/(1+#)}{1/(1— #)} n times by means of
Leibniz’ theorem. [The result is n! % {( —l) r /(l+.n) r + 1 (l-.r) n_r + 1 }. It is
r=0
easy to verify that this result agrees with that of Ex. 5.]
10. Prove that D x n cosx=cos{x+%mr), D x "sin.r=sin(tf+^njr).
11. If y=A cos mx + B sin mx, then D x 2 y + m 2 y=0. And if
y = A cos mx + B sin mx + P n (x),
where P n (x) is a polynomial of degree n, then D x n + 3 y + m 2 T) x n + 1 y = 0.
12. If x 2 D x z y+xDy+y — 0, then
x 2 D x n + 2 y + (211 -f 1) .r D n + D/ + (n 2 +1) D x n y—0.
[Differentiate n times by Leibniz’ Theorem.]
13. If U n denotes the nth derivative of (Lx + M)/(x 2 - 2Px + C), then
u ^ + ^r (*<"*• T "P- 1900 -)