Calculus and Probability for Actuarial Students

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CHAPTER XI
DIFFERENTIAL CALCULUS. STANDARD FORMS.
PARTIAL DIFFERENTIATION

1. The differential coefficient of any particular function can, of course, be obtained by direct calculation, but the process can usually be simplified by the application of the following general rules. The general similarity to the propositions already demonfor Finite Differences will be apparent.

I.The differential coefficient of any constant term is zero. This is evident since a constant is a quantity which does not change in value in any mathematical operation.

The differential coefficient of the product of a constant and a function of x is equal to the product of the constant and of the differential coefficient of the function.

Thus

[c . f (x)]

fio

h — c

(x)

= Lt

(x + h)

—

(x)

nyoh

^df(^x)

c(1).

—

III.
The differential coefficient of the algebraic sum of a number of functions of x is the sum of the differential coefficients of the several functions.

Let

+ ..., where

u, v, w, ...

are functions of

then

+ ...

and

Du+Ov+Aw+...,

and

Ox Ox Lx

which, by proceeding to the limit, becomes

dy _ du

dx+dx+

_ dx +

(2).

IV.
The differential coef^ficient of the product of two functions is the sum of the products of each function and the differential coefficient of the other,

H. T. B. L

66    DIFFERENTIAL CALCULUS
Let    y = uv,
where u and v are both functions of x.
Then    Dy=(u+Au)(v+w)—uv =u0v+vLu+&(Ov =u0v+(v+Ov)Au
and    x=u^A+(v+4v) ^Au Ax'
whence, taking the limit, when v + Ov v, dy _ dv du
dx —udx +v_dx     (3),
which may be written

1dy Ida ldv

ydx udx + vdx     (4). This result may be extended to include the product of any number of functions.
For if y =uvw; let vw = z, then y = uz.
Whence    1 dy 1 du + 1 dz . ydx udx zdx
But    ldz ldv₊ldw_. z dx vdx wdx

1dy_—1du ldv 1 dw

Therefore    ydx = u dx + vdx + wdx """"""...(5). Multiplying by uvw, we obtain
dy =vwdu+wuv+uv dw
    (6)
dx    dx    ^dx--dx
Similarly for the product of any number of functions.

V. The differential coefficient of the quotient of two functions is (Di Coeff. of Numr.) (Denr.) — (Di,'f. Coef . of Denr.) (Numr.) Square of Denominator

Let    u
y v^'
u -i-AuuvLu — uLv
Then    D y =    _ v -r Av v v (v + Av )
Du Av A ^vLx^—uAx
and         y     Ax v (v + Av)

STANDARD FORMS    67
whence, taking the limit,
du dv

dy dx dx

dx v
which may be written

ldy ldu ldv

y dx u, dx v dx (8).

VI. The differential coefficient of y with respect to x, where y is a function of u and u is a function of x, is the product of the difcoefficients of y with respect to u and u with respect to x.

For Ay Ay Du

Ax — Du Ax'

whence, taking the limit,

dy dy du

dx du' dx     (9).
Similarly    ^d?/ _ dy du dv
x du'dv.dx     (10), d
and so for any number of functions.

2. Various standard forms can now be developed, mainly from first principles. It is instructive to note the points of analogy with the corresponding forms for Finite Differences.

(ⁱ)    y = xⁿ,
dy_=Lt(x+)r)ⁿ—xⁿ
    dx _h-~0h
xⁿ(1+^le)ⁿ -1
    = Lt     x
    hyo    h
Expanding by the Binomial we have
^x—ⁿ    s+...1
dx    _r^x

=Ltnxⁿ-' L1 ₊n — lh_+...l

h-1.0
x
= n_aaa—1
5—2

68 DIFFERENTIAL CALCULUS

(ii)y = a^x

dy=Lt ax+h —

dx h~o

-1

'Lt

h-.0 it

=ahLt

h[1+hloge

2(loge a)2+...—1

log

a + (log

a)2 + ...

h-.0

= a'' log

y=ex,

_dx=exloge

e=e

^fD

(iii)y = log,,x.

Then

But

(aY)

—

d (a') dy

dxdy dx

Hence, using the result established in (ii) and remembering that

we have

dx —

a'log

_dx

or1 =xloge

^a.

Whence

d_y =

1 —

dx x

log

^log

- -

dx x

log

(iv)y = [f (^x)]^"; ^y= e¹²' ; ^y= log, f(^x)•

(

)]

dy —

[

(x)]"

df (x)

dxdf(x)dx

ⁿ

[f (x)]"-1.

f '

(x).

Similarly

ify=e

^f()

dx=ef(x).f'(x),

and if

log

f (x),

^dy

(

)

_{f (x)}

STANDARD FORMS 69

(v)y = sin x,

dry

_—Lt

sin

—

sin

_h>o

2sin-.cos(x+₂)
= Lt
hyo    h
h
sin 2    /    h)
= Lt    . cos I x + 2)
hy0
2
= cos x.

(vi)y = sin^—1 x.

Thensin

whence

= cos y = 11 — sin

y =

VT: .

Therefore

dy _

dx dx

1 — x

3. The values of differential coefficients for the other trigonometrical functions can be found by methods similar to those employed in (v) and (vi). The results are given in the table below and should be verified as an exercise by the student.

Function	DifferentialCoefficient	Function	DifferentialCoefficient
	nx^n—1	sin^-1x	1
a^x	a^x log, a		1 - x²
e^x	e^x	cos-¹	1
log_ax	¹		1-x^s
	xlog, a	tan-^lx	¹
log,x	1		1+x9
	x	cot-¹ x	1
sin x	cos x		1 +x2
cos x	- sin x	sec-¹ x	1
tan x	eec² x		xx^a - 1
cot x	- cosec² x	cosec^—1 x	-
sec xcosec x	sec x . tan x - cosec x . cot x		x,./x² -1

70 DIFFERENTIAL CALCULUS

Logarithmic Differentiation. This method is of special value in two cases. Thus if y = uvw ..., where u, v, w, ... are functions of x, then

log y = loglog

logw+...,

dy l

and- -

+...

ydx udx vdx wdx

or    ^dy_=uvw1 du + 1 dv + 1 dw +
dx    udx vdx wdx
a result which agrees with that already obtained in § 1. Secondly if y = u", u and v both being functions of x, log y = v log u,
and    ydx⁼udx^+login
or    dx=^vu"-'--^+uvlogudx^.

We will now give some miscellaneous examples of differen

- b+x

By the ordinary rule for a quotient

dy (b + x) d

(dx

x2)

(a +

x2)

d (b

(b+x)'

(b+x)2x-(a+x2)

+2bx-a

(b + x)

- (b +

7x-1

= 1 - 5x + 6x^2'

This can best be treated by resolving the expression into partial fractions. Then
4 – 5
= 1 – 3x 1 – 2x^'
dy_-4d(1-3x)-' d(1-3x)_–5d(1-2x)-1 d(1-2x) dx d (1 - 3x) dx d (1 - 2x) dx

"= 12 _ 10
(1 - 3x)² (1 - 2x)²

STANDARD FORMS 71

(iii)y = ZVa + x.

dyd(a+x)"

d(a+x)1

dx d(a+x)dxn

^(a+x)

(iv)y = log (log x).

d log (log x) d_(log x)

dx – d

(log

x)dxx log cc

(v)y = tan

x'–1

_dy

(tan-

Vx'–1

d(x'–1)

d(x'–1)

'Vx'–1

.–

(a'—1)-g.2ro

¹⁺

x'–1

x'—1

x-2

(vi)^y– (x 3) .

logy=

logx–2–

logx–3,

ldy1d(log

x–2)

d(x–2)

+logx–2.

1 d(logx–3)

d(x–3)

x d

x–3

' dx

ogx–

1– 1 to

x–2

x(x–2) x(x–3) x'

x–3'

Whencedy - - x – 2

+ 1

_to

–

dx(x–3) [x

(x–2)(x–3)

x–3J

72    DIFFERENTIAL CALCULUS
(vii)    y = a^x . b^cx.
log y = x log a + c^xlog b,
^{1 d}=loga+c^xlogclogb, y ^dxdx=a^z.V(log a+c^xlogclogb).
(viii)    y = x^x + x^x.
In this case logarithmic differentiation must be used, but for this purpose the two terms must be taken separately.
Let    x^x = u and x^x = . dy_–du dv
Then    + d_x dx dxx
log u = x log x,
u d u
=xx+logx,
x ^du- =x^x(1+logx).
dx
Also    log v = ¹log x, x
v _dx=x.x~-logx    ,
dv    1
dx- =xx. — (1 — log x).
Therefore    dx = x^x (1 + log x) + x^x (1 — log x).
(ix) Differentiate log_e x with regard to x'.
Let    y = log, x and z = x².

Then dy — dy dx d_y 1 dz dx^' dz dx'dz^' dx

Therefore dy = 1 1 — 1 dz x' 2x 2x²

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. Let

= f (x, y).

Then

u+Su=

f(x+h,

y+k)

and

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.

Proceeding

to the limit when

and

each —> 0,

f(x+It,y+k)—

f(x,

y+k)_

f(x,y+k)

(x, y),

since

k -

Therefore

+ ^u.

Sy,

and

ax ay

au or, when Sx — 0,

ay .

dx (11).

(x, y)

0, dx

and

au au

°-ax

⁺

ay'

Whence

_dx

⁼

–d

(12).

Example.

then

ax=2x+y,

a=x+2y,

and

dy2x+

dxx+2y