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CHAPTER X
DIFFERENTIAL CALCULUS
DEFINITIONS; STANDARD FORMS;
SUCCESSIVE DIFFERENTIATION
1. We have seen in the previous chapter that if y = f (x) be a
continuous function of x, the average rate of change of y with x is
f (x    + h) - f (x)
h    ,
or Ay/Ox, where Ox = h.
The limit of this function when the interval tends to zero (which we may call the rate of change) is called the differential coefficient of y with respect to x. If we denote this result by Dy, we have
    
Dv = Lt Ay/Ox = Lt f (—x + h)    — f (x)
.1x- 0    h--> 0    h
and we are said to have " differentiated y with respect to x."
The usual notation for the differential coefficient is dx, where dx is an operator analogous to A, E, E, ... in finite differences. For convenience in working alternative methods of denoting dx are used. For example,
y', D y ; f ' (x), Df (x) 4 dx f (x) .
represent the same result.
It should be noted that, although Ay/Ox is the result of the division of a definite quantity Dy by another definite quantity Ox,
dx represents an operation performed on the function y, the operator being dx. At this stage neither dy nor dx should be
considered to have a separate meaning.
The differential coefficient of y with respect to x is sometimes called the "first derivative" or the "first derived function" of y with respect to x.

GEOMETRICAL INTERPRETATION OF DERIVATIVE 151
2. Before proceeding to examine the values of the differential coefficients of various functions, it will be of advantage to consider the geometrical interpretation of the operation of differentiation.
Let B1AB represent the continuous curve y = f (x) and let the coordinates of a point A on the curve be (x, y) or {x, f (x)}. Let B be a near point on the curve whose coordinates are {x + h, f (x + h)}.
Fig. 21.
Then it is evident from the figure that, if B be the angle BAK, tang=BK/AK=BN—NKBN—MA MN ON — OM

f (x + h) — f (x)
h
Now as the point B moves along the curve so as ultimately to coincide with the point A, the secant BA takes up the position of the tangent AL to the curve (see Ex. 1 of Chapter Ix). The angle B then becomes the angle which the tangent AL makes with the axis of x.
But the limit of f -(x +    hh —    f (x) when B coincides with A is
the limit of this expression as h -> o.
Also    Lt f (x + h) — f (x) d (x).
n->o    h    dx

152    DIFFERENTIAL CALCULUS

Therefore    dx f (x) = tan :G
= the tangent of the angle that the tangent to the curve y = f (x) at the point (x, y) makes with the x-axis.
The tangent to a curve at any point measures the slope or gradient of the curve at that point. The differential coefficient of y with respect to x is often referred to as the gradient of the curve y = f (x) at the point (x, y).
It may happen that near the point A of the curve the curve is continuous as x increases, but that there is a discontinuity in the other direction—as in Fig. 22. If we were to consider the effect
x
Fig. 22.
of allowing the'point B1 to approach A,--the x coordinate of which is the same as that of the point A—the value of the differential coefficient might be different from that found by assuming B to coincide with A.
For this reason it is probably better to define the differential coefficient thus :
If f (x) be a continuous function of x and if Lt f (x + h) — f (x)
h->0    h
is equal to Lt f (x)    (x — h) , then either of these limits is called h-*0
the differential coefficient of y with respect to x.
Another method of obtaining dx (when it exists) is to consider
two points B1 {
(x—h), f(x—h)}
and B {
(x + h), f (x + h)}


Y

GENERAL PROPOSITIONS    X53
which approach A simultaneously. By reference to Fig. 21 it will be seen that
tan i = dx = h_>.() f (x + h)     zlzf (x — h) .
This form is of advantage when the evaluation of f (x + h) — f (x — h)
is simpler than the evaluation off (x + h) — f (x).
3. The following propositions are of general application :
 The limit of this expression as Ax -)- o is
dy =      d f (x) + d (x) + d (x) + ....
dx dx    dx    dx
(iv) If y = uv, where u, v are both functions of x, then
dx— udx+vdx'

DIFFERENTIAL CALCULUS

°y = ° (uv)

= (u+.,u) (v+°v)—uv = u°v + v°u + &°v.
°y=u°v+v°u+°u °v °x    °x    °x    AX

AV AU
= (u + °u) °x +
Therefore since in the limit °x -> o, u + °u will become u,
dxudxdv du +vdx A slightly different proof follows from the relation established in para. 4 of Chapter VIII:
°y = °uxvx = vx°ux + ux+,x°vx (where the interval of differ
encing is °x).
so that when °x -+ o we have
dy    du dv
_
dx—vdx+udx'
which is the same formula as the above.
As a corollary we have, by successive applications of (iv),
d    d dxuvw... =uv.. dw+vw... dx+uw...dx f ....
( du    dv
(v) If y = u, then dv
_
v    vdx 2
u dx dx    v
Now    ° ° ul u + ~u _ u
y    \v/ v + °v v (u+°u)v—(v+°v)u
v (v + °v)
v°u—u°v
v(v+°v)'
154
°y _ °u    °vx °x = vx Ax + u.+:1. °x'

GENERAL PROPOSITIONS
Du Ov v— u
Ay Lx Lx Ax v (v + Ov)
du    dv
v — u
dy dx dx dx    v2
 (vi) If y is a function of x and z is a function of y, then dz dz dy
dx dy dx
Oz Az Ay Since    ox = Ay . Ox ,
dz dz dy dx - dy dx when :fix --> o.
It follows that
dz dz dy dx du dr ds
dt dy ' dx' du' dv ' ' ds' dt' where s is a function of t, r a function of s, etc.
 dy
dx dx dy
. If Ox be not equal to zero,
Ay

156    DIFFERENTIAL CALCULUS
a given value of y, then in taking the changes in value Ox, Ay, Az


we must keep them consistent in assuming that 0'x =    . Oz Ay , or oy D,a
that
Ay
Ax=0xAy
4. Standard forms: Algebraic.
We will now proceed to obtain the differential coefficients of some standard functions.
(i) Y = xn.
yq=xP=z, say.
dz
We have    dx = px
dz    from (a) above.
and    dy = TYq
But    dy dy dz i dz dx dz. dx dz ' dx dy
pxP-1 p xP y qyq-1 4'Y° x
_p y_p x~
_q x_q
= nxn-1.

STANDARD FORMS    157
(c) n negative.
Let n = — m where m is positive (integral or fractional).
y_xn_x-m_
xm'
xmy = 1,
and    x (xmy) – mxm-1y + xm x ,
since m is positive.
But xmy = I = constant.
(xmy) = o. mxm-1y + xm dx = o.
dy    my
dx    xm    x
= nxn-1.
de'
I.e. d~ = nxn-1 for all values of n positive or negative, integral
or fractional.
For example,    dx x5 = 5x4, d
dx x-5 – 5x-8,
d    1 _4
dx x°    5x °, d
dx
(ii) y = e~.
ex--h    — x    I
ex    ex    eh -
-- es = Lt
dx    h*O    h    —eh to h
eh — = es, since Lt    h
h—0
x
Corollary:    dx = as loge a.
I
= I (p. 14.6).

158    DIFFERENTIAL CALCULUS
For ax = ex log a and dax = dex log a _ dex log a d (x log a)
dx    dx    d (x log a).    dx
= ex log a . log a = ax log a.
Now put - = k so that if x + o, hLto is the same ash Lt . Then
h


Lt log ( +      x) = Lt x log I + k = x Lt II? log (1 + k
h h~o    x >~    (    x~~    )~
=
X Ltwflog(1+k)k =x logs Lt (1+k)k~
I    I
=-loge=-.
x    x
d
dx logo x = loge a. x
5. Standard forms: Trigonometrical.
(i) y ----- sin x.
For the differentiation of sin x we adopt the alternative form
dx f (x) = Lt f (x + h) - f (x - h) .
(x + h) - sin (x - h) dx sin x = Lt sin    -    - -
h-)-0    2h
Lt 2COSsinh h–)-0    zh
= cos x Lt sin h
h~0 h


= cos x, since Lt sin h =
I h-*0 la
(iii) y = loge x.
d dx loge x = Lt log (x + h) - log x
h-/0    h
log x + log (1 + x) - log x
= Lt
    h ->O    h
log (1 +      x)
    = Lt     h

h~0
Corollary:
Then
(P• 4)
(P• 139)-

EXAMPLES OF DIFFERENTIATION
 160    DIFFERENTIAL CALCULUS
d    1    _ d (a2 — x2)4 d (a2 — x2)
 (x2 — 3x + 2)2 – 2x2 – 1ox + 19
(x2– 3x+2)2
or, alternatively, we can split    2 2x + $     into x—3x+2 differentiate each fraction separately;
dy d 2x+5 _ d    2x+5
dx dxx2—3x+2 dx(x—1)(x–2)
_ d
dx ( x — 9    7 2 x — 1)    (x      9     z)2 + (x 7     1)2
– 9x2 + 18X – 9 + 7x2 – z8x + 28

Example 3.
Find where y
The differentiation can be performed at once by treating y as the quotient of two functions of x, thus :
dy (x2 – 3x + 2) dx (2x + 5) – (2x + 5) dx (x2 – 3x + 2) dx    (x2 — 3x + 2)2
2(x2—3x+2)—(2x+5)@x—3)
partial fractions and
(x – 2)2 (x – 1)2

- - 2X2 – lox + 19 as before. (x2 — 3X + 2)2
y = b°l . Find dx.

    EXAMPLES OF DIFFERENTIATION    161


For this type of function it is useful to employ the process known as logarithmic differentiation. Here we take logarithms of both sides of the equation before differentiating and write


log y = cx log b.


    Let    z = log y.


    Then    dz dz dy d (logy) dy    dy
d x dy' d x    d y dx y dx
Also    dx (cx log b) = cx log c log b.
y dx = cx log c log b.
dx = ycx log c log b = bexcx log c log b.
Example 4.
Differentiate tan-1 -        with respect to x.
1/x2 —
dy d tan-1(x2 — I)-I d (x2 — I)-I d (x2 — I)
dx    d (x2 _ 1)-I    d (x2 — I)    dx
I
(x2 - I) .2x
I +
x2 - I


I Example 5.
Find dx where y = xx + xx.
It is important to note that dx x" = nxi-1 only where n is a constant, and it is therefore incorrect to state that d xx = x.xx 1. To obtain dx xx we must employ the method of logarithmic differentiation. More


over, y is the sum of two functions of x, and if we are to employ this method we must differentiate each of the functions separately. It would be incorrect to take logarithms of each side of the equality as it stands, for if y = u + v then logy + log u + log v.
    
P    II

162    DIFFERENTIAL CALCULUS
1
Let    y=xx+xx=u+v.
Then    log u = x log x and log v = I log x. x
u—=x.-+logx=I+logx. dx    x
du —

dx I's (1 + log x) ; "
Idv I +logx.— 2= x2
v    x-    (1-logx).
dx x'    x
1 i
dx=xx•x2(I-logx).
dx=xx(I+log x)+xx(I     _ x2°g x). Example 6.
Differentiate x sin x with respect to tan x. d (x sin x) d (x sin x)    dx
d tan x    dx    • d tan x
d (x sin x)
dx    xcosx+sinx
= x cos' x + sin x cos2x. d tan x    sec2x
dx
7. Successive differentiation.
If we differentiate dy/dx with respect to x we obtain a new function which is called the second differential coefficient of y with respect to x. By analogy with the symbolic notation adopted in finite differences, we write
d dy as d2y dx dx    dx2'
where, it should be remembered, the independent variable is still x and not x2. Similarly, the third differential coefficient of y with respect to xis dx, and if y is differentiated n times with respect to
x, the nth differential coefficient is dx . In the alternative notation we have
D2y, Day, ... Dny,
f" (x), f,,, (x), ... f cn~ (x).
A notation frequently employed for the nth derivative is yn.
similarly
SUCCESSIVE DIFFERENTIATION 163 8. Successive differential coefficients of many simple functions can be found by an inductive process.
Example 7.
dn Find dxn log x. y = log x; Yl = 1/x; Y2 = (— 1). I/x2,
Y3 = (— I) (— 2). I/x3; Y4 = (— 1) (— 2) (— 3).1/x4;
and so on.
Therefore by induction yn = (— 1)n-1(n - 1)
x
Example 8.
y = (2x + 5)/(X2 — 3X + 2). Find yn .
It is imperative where higher differential coefficients than the first are required to use the second method given in Example 2 of para. 6, and to express y in partial fractions before differentiating.
9         7
Y        —
x—2 x — I'
    9 I    I

Y1 = .(x— 2)2 + (x — I)2'
9.I.2    7.I.2
Y2 (x — 2)3 (x — 1)3'
_ — 9.I.2.3    7.I.2.3
Y3    (x — 2)4 + (x — 1)4'
...........................
Yn=(— I)nn![(x—2)n+1 (x — I)n+11 Example 9.
Show that if y be a rational integral function of the nth degree in x, then the nth differential coefficient of y with respect to x is constant. If    y=a+bx+cx2+...+kxn,
Y1=b+2cx+...+knxn__1,
y2=zc+...+kn(n— I)xn-2,
and each time that we differentiate we lower the degree of the function by unity. Hence after n differentiations we shall lower the degree of the function by n.
.•, yn=kn(n—1)(n—2)...(n—n—I)xn—n = kn ! which is constant.
II-2

164    DIFFERENTIAL CALCULUS
If we denote dxn by Dny we may write D'nxm = m !, which is analogous to Amx(m) = m
9. Leibnitz's Theorem.
In Chapter VIII, para. 4, we discussed the compound function y = uxvx and the expansion resulting from differencing this function n times. The method of obtaining Onuxvx was to introduce the symbols 01 and z 2 (representing the operation of differux and vx separately) and then to expand the resulting function by the method of separation of symbols.
dn In an exactly similar way we may find dxn (uv), where u and v are functions of x, by using the alternative notation D for dx.


Let D1 and D2 represent the operations of differentiating u and v respectively, where D1 operates only on u and its successive differential coefficients and D2 operates only on v and its successive differential coefficients.


Then    Duv = uDv + vDu
= D1 (uv) + D2 (uv).


D nD1+D2, so that


Dn - (Di + D2)n
Din + n1D1n-1 D2 + n2 D,n-2 D22 + ... + nrD1n_r Der + ... + D.
do-r    dr But D1n-r Der (uv) = dxn_r u . dxr v.
dny = Dn (uv) = dnu v n dn-1u dv + n do-2u d2v + ... dxn    dxn    i dxn-1'dx    2 dxn-2' dx2


do-ru dry    d"v + nrdxn_r'dxr+... +udxn.
This is Leibnitz's Theorem for the successive differentiation of the product of two functions of x.

LEIBNITZ'S THEOREM
10. Application of Leibnitz's Theorem. Example 10.
Ify=x2exfind dxy'
In the general expansion above let x2 = v and ex = u. Then    v = x2,    u = ex,
dv    du


dx = 2x,    _ dx — ex'
d2v =--    d2u
dx22'    dx2=ex.
3
d 33 and higher derivatives are zero.
n
n(n — I)
    dxn (x2ex) = x2ex + n.2x.ex +    2    .2ex = ex (x2 + 2nx + n2 — n). Example 11.
If    y = log {(x — I)I + (x + I)I}
165
prove that
(x2 — I) dy+xdy=o
x2    dx
and that (x2— I)dxn    ddxn i +n2dx' y = log {(x — 1)I + (x + 1)1}.
dx (x - I)i + (x + 1)i.{k (x - 1)-i + 2 (x + 1) #} = (x2-10.
ax = dx (x2 - 1)} _ - Zx (x2 - I)-1. d2    d

(x2—I)dx2+xd =°;
Differentiate each of the products (x2 — 1) d 2 and x d n times by Leibnitz's Theorem.
Then    (x2 - I) y n+2 n. 2X    + n (n — 1) 2
n+2    yn+I    2    yn
+ x.yn+i + nyn = °;
i.e.    (x2 — I) yn+2 + (2n + 1) xyn+l + n2yn = 0.

166    DIFFERENTIAL CALCULUS
11. We will conclude this chapter with some miscellaneous examples of differentiation.
Example 12.
If y, z are both functions of x and if y2 +•z2 = k2 prove that d y1    2
' dx (+ WC (z) z dx '
•d(yl d(z2l_    d    y    d    :2
ydx \kl+dx k)    ' dx ,/y2+z2+dx1/y2+z2
dy dz'    2 dy dzl
dy 1/y2 + z2      f ydx+zdxi    2z dz V2 + z2 z t dx+zdxJ
—    -    —
dx    -62 + z2    dx y    •\/y2 +

ll    (y2 + z2)
(YZ 1x2)1 [y d (y2 + z2) y2 (y dx + z dxl 2z dx (y2 + z2)
    d    dz
— z2 (y d + z which simplifies to
1    `z(y2+z2)Y=2' dx
(y2 + x2)1 I dx    JJ1    k' dx
Example 13.
Prove that if n be a positive integer, and a and b have any values, then (a + b) (a + b — 1) ... (a + b — n + I)
l
    pE [    n! a(a—1)...(a—p+1).b(b—I)...(b—q+I)]
o p!q!
    where p + q = n.    (Vandermonde's Theorem.)
Now Da (uv) = E [ n! DD(u)Dl(v)J by Leibnitz's Theorem. pq!
Let u = xa and v = xb so that uv = xa+b. Then
(a+b)(a+b— 1)... (a+b—n+ 1)xa+b-n
=E [a(a_I)...(a_p+r)P.b(b_I)...(b_q+I)xQ]
    
n    
E [    q1 a(a — I)...(a—p+I).b(b— I)...(b—q+ I)xa+b-p-q~
Since p + q = n we may divide through by xa+b-n and the proposition is proved.

EXAMPLES    167

Example 14.
3
If y = f (x) obtain dye and yx in terms of d, d 2 and d2


d2y d3y
Letyi, y2, .1'3 stand for dx' dx2' -d-xi respectively. dx t
_
I    I    Y2
_ — ,VlY2'yl — y13.
d3x d (d2xl d ( y2 l dx
 168    DIFFERENTIAL CALCULUS
EXAMPLES    169
\/x2 — a2    (bxn)#    .
 I70    DIFFERENTIAL CALCULUS
32. Differentiate log ex + \/e2x — a2 with respect to x. ex — 1/e's — a2
Find dx where y (y2 + x2) = x + y.
Differentiate (i) log sin x; a tan-1 4x (1 a x2)1

O    -6x+xJ}
Determine the coefficients a0, a1, a2, .., an, so that
d    xn + a1 xn