STANDARD FORMS 73
- In dealing with cases where a function of two variables is involved it is convenient to adopt methods similar to those used in Finite Differences (see Chapter IX, 4). Thus we define Partial Differentiation as the process of differentiating a function of several variables with reference to any one of them, treating the other variables as constants.
This process is denoted by the symbols ax , sy , etc. We will also use the symbol Sx to denote a small change in the value of x.Letu = f (x, y).Thenu+Su= f(x+h, y+ k) andSu = f (x + h, y + k) — f (x, y)—f(x+h,y+k)—f(x,y+k).h+f(x,y+k)—.f(x,y).k. h Proceeding to the limit when h and k each —> 0,f(x+h,y+k)— f(x, y+k)_haxf(x,y+k)ax f (x, y), since k - 0. ThereforeSu =aua. Sx + u . Sy, yandSu au + au Sy Sx ax ay axdu au or, when Sx — 0,dx ax + ay . dx (11).7. If f (x, y) = 0, dx = 0 andau au dy°—ax+ ay' yau dyaxWhencedx=–du (12).ay Example. Ifx' + xy + y' = 0,thenax=2x+y, a=x+2y,anddy2x+,y andxx+2y'
CHAPTER XII
DIFFERENTIAL CALCULUS. SUCCESSIVE
DIFFERENTIATION
- When dx is differentiated with respect to x, we obtain a further function of x which is known as the second differential coefficient or second derived function of y. This operation represented by x (mil is generally written dx or f" (x). If the function is differentia//ted n times with respect to x, the result is called the nth differential coefficient or nth derived function and is written ny
dxn. Other symbols for the nth differential coefficient aref(n) (x), Dny, yn.
- In many cases the value of the nth differential coefficient can be found readily by inductive reasoning.
Example 1.y=log(x+a).11(—1)(—2).yl=x+a; y2=(x+a)2' y$(x+a)3 and by analogy(n—1 !yn=(—1)n-1(x+a)''Example 2.y=a+bx+ cx2+dx'+... +kx71,yl =b + 2cx + 3dx2 + ... + n . kxn—1,y2=2c+2.3dx +...+n(n—1)kxn—a,yn=n1k. Example 3.7x—1_y 1—5x+6xf
SUCCESSIVE DIFFERENTIATION 75
Making use of the method of partial fractions,
4 _ 5
= 1 — 3x 1—2x'
4.3 5.2
y1 _ (1 — 3x)2 (1 — 2x)2'
4.3n.n! 5.2".n!
(1 — 3x)"+' — 1— 2x)n+i
3. Leibnitz's Theorem.
n d This theorem gives an expression for dxY' where y is the product of two functions of x, say u and v.
It should be noted first that, as shown in Chapter XI, § 1, the operator dx or D obeys the distributive and index laws and is commutative in regard to constants. In these respects it is similar to 0 (see Chapter II, § 6).
Now d (uv) v du + u dv
dx dx dx '
and we may therefore write
D (vv) = D, (uv) + D2 (uv)
=(D,+D,)uv,
where D, operates only on u and differential coefficients of u and D, operates only on v and differential coefficients of v.
Therefore
D" (uv) = (D, + D2)" uv
= [D," + n D,"-1D, + (2) D,"-2 D22 + ... + D2"] uv. Now
u
D,"(uv)=v dx" ,
dv\/1 _ dv do-'u
D,"—1 D, (uv) = D,"—' (u Cdx/ dx • 1 de-1'
etc. Hence
d" (uv) d"u dv d"-lu nl d2v d"yu d"v dx" =v d"+ndx• dx"-1+( dx2 "-a+...+u dx"
(1).
76 DIFFERENTIAL CALCULUS
This formula can also be established by induction, but the proof by this method is left as an exercise to the student.
4. Examples of the application of the formula are given below. Example 1. y = x'eax.
do eax
Sincen can be written at once as aneax, it is convenient to take x' as the factor v, and eax as the factor u.
Then dx = 3x', eta
Therefore
yn = x'aneax + n. 3x'. an-'eax + (2) . 6x. an-'eax + (3) .6 . an-3e,. Example 2. If y = xn (log x)',
prove that
x'yn+s + xyn+, = 2 x n!.
We have y1= xn 21og x + (log x)' n . x"-' x
or xy, = 2xn log x + nxn (log x)'
=2x" log x + ny (i) Differentiating both sides of the equation
2xn
xy2+y,= x +2 log x.nxn-'+ny,
or x'y2 + xy, = 2xn + 2nxn log x + nxy,. Substituting for xn log x from equation (i) we obtain x'y2+xy,= 2xn + nxy, - n'y + nxy,
or x'y2-(2n-1)xy,+n'y=2xn (ii).
By differentiating each term of this equation n times, making use of Leibnitz's Theorem, we arrive at the result
x2yn+2 + nyn+, . 2x + n (n -1) yn -(2n-1)yn+,.x-n(2n-1)yn
+ n'yn
O Yn+s + xyn+, = 2 x n!.