Calculus Made Easy: Chapter no 6SUMS, DIFFERENCES, PRODUCTS AND QUOTIENTS.
We have learned how to differentiate simple algebraical functions such as x2 + c or ax4, and we have now to consider how to tackle the sum of two or more functions.
For instance, let
dy
what will its
dx
y = (x2 + c) + (ax4 + b);
be? How are we to go to work on this new job?
The answer to this question is quite simple: just differentiate them,
one after the other, thus:
dy = 2x + 4ax3. (Ans.)
dx
If you have any doubt whether this is right, try a more general case, working it by first principles. And this is the way.
Let y = u+v, where u is any function of x, and v any other function of x. Then, letting x increase to x + dx, y will increase to y + dy; and u will increase to u + du; and v to v + dv.
And we shall have:
y + dy = u + du + v + dv.
Subtracting the original y = u + v, we get
dy = du + dv,
and dividing through by dx, we get:
dy du dv
= + .
dx dx dx
This justifies the procedure. You differentiate each function sep- arately and add the results. So if now we take the example of the preceding paragraph, and put in the values of the two functions, we shall have, using the notation shown (p. 16),
dy d(x2 + c)
=
dx dx
d(ax4 + b)
+
dx
= 2x + 4ax3,
exactly as before.
If there were three functions of x, which we may call u, v and w, so that
y = u + v + w;
then
dy du
= +
dx dx
dv dw
+ .
dx dx
As for subtraction, it follows at once; for if the function v had itself had a negative sign, its differential coefficient would also be negative; so that by differentiating
y = u − v,
dy du dv
we should get
dx = dx − dx.
But when we come to do with Products, the thing is not quite so simple.
Suppose we were asked to differentiate the expression
y = (x2 + c) × (ax4 + b),
what are we to do? The result will certainly not be 2x × 4ax3; for it is easy to see that neither c × ax4, nor x2 × b, would have been taken into that product.
Now there are two ways in which we may go to work.
First way. Do the multiplying first, and, having worked it out, then differentiate.
Accordingly, we multiply together x2 + c and ax4 + b. This gives ax6 + acx4 + bx2 + bc.
Now differentiate, and we get:
dy = 6ax5 + 4acx3 + 2bx. dx
Second way. Go back to first principles, and consider the equation
y = u × v;
where u is one function of x, and v is any other function of x. Then, if x grows to be x + dx; and y to y + dy; and u becomes u + du, and v becomes v + dv, we shall have:
y + dy = (u + du) × (v + dv)
= u · v + u · dv + v · du + du · dv.
Now du · dv is a small quantity of the second order of smallness, and therefore in the limit may be discarded, leaving
y + dy = u · v + u · dv + v · du.
Then, subtracting the original y = u · v, we have left
dy = u · dv + v · du; and, dividing through by dx, we get the result:
dy dv
= u
dx dx
du
+ v . dx
This shows that our instructions will be as follows: To differentiate the product of two functions, multiply each function by the differential coefficient of the other, and add together the two products so obtained.
You should note that this process amounts to the following: Treat u
as constant while you differentiate v; then treat v as constant while you
dy
differentiate u; and the whole differential coefficient
dx
of these two treatments.
will be the sum
Now, having found this rule, apply it to the concrete example which was considered above.
We want to differentiate the product
(x2 + c) × (ax4 + b).
Call (x2 + c) = u; and (ax4 + b) = v.
Then, by the general rule just established, we may write:
dy
= (x2
dx
+ c)
d(ax4 + b) dx
+ (ax4
+ b)
d(x2 + c) dx
= (x2 + c) 4ax3 + (ax4 + b) 2x
= 4ax5 + 4acx3 + 2ax5 + 2bx,
dy = 6ax5 + 4acx3 + 2bx, dx
exactly as before.
Lastly, we have to differentiate quotients.
bx5 + c
Think of this example, y = . In such a case it is no use to
x2 + a
try to work out the division beforehand, because x2 + a will not divide
into bx5 + c, neither have they any common factor. So there is nothing for it but to go back to first principles, and find a rule.
u
So we will put y =
;
v
where u and v are two different functions of the independent variable x. Then, when x becomes x + dx, y will become y + dy; and u will become u + du; and v will become v + dv. So then
u + du
y + dy =
.
v + dv
QUOTIENTS 39
Now perform the algebraic division, thus:
v + dv
u
+ −
v
du
v
u · dv
v2
u + du
u + u · dv
v
—
du u · dv v
·
du dv
du +
v
v
v
u · dv − du · dv
u · dv − u · dv · dv
v v2
v
v2
— du · dv + u · dv · dv .
As both these remainders are small quantities of the second order, they may be neglected, and the division may stop here, since any further remainders would be of still smaller magnitudes.
So we have got:
v
v
v2
y + dy = u + du − u · dv ;
which may be written
v · du − u · dv
= + .
v2
Now subtract the original y =
u
, and we have left:
v
dy = v · du − u · dv ;
v2
du dv
whence
dy v dx − u dx dx = v2 .
This gives us our instructions as to how to differentiate a quotient of two functions. Multiply the divisor function by the differential coef- ficient of the dividend function; then multiply the dividend function by the differential coefficient of the divisor function; and subtract. Lastly
divide by the square of the divisor function.
bx5 + c
Going back to our example y = ,
x2 + a
write bx5 + c = u;
and x2 + a = v.
Then
dy (x2 + a)
=
d(bx5 + c) dx
5
(bx + c)
d(x2 + a) dx
dx (x2 + a)2
—
(x2 + a)(5bx4) (bx5 + c)(2x)
= (x2 + a)2 ,
dy 3bx6 + 5abx4 − 2cx
dx = (x2 + a)2 . (Answer.)
The working out of quotients is often tedious, but there is nothing difficult about it.
Some further examples fully worked out are given hereafter.
Differentiate y =
a2
a a2 a2
—
x3
b2 b x + b2 .
Being a constant, vanishes, and we have
b2
dy a
3−1 a2
1−1
dx = b2 × 3 × x − b × 1 × x .
But x1−1 = x0 = 1; so we get:
dy 3a 2 a2 dx = b2 x − b .
Differentiate y = 2a
√
bx3 −
3b√3 a
x − 2
√
ab.
Putting x in the index form, we get
2
y = 2a√bx 3 − 3b√3 ax−1 − 2√ab.
Now
= 2a√b × 3 × x −1 − 3b 3 a × (−1) × x−1−1;
2
dy 3 √
√
dx
or,
2
dy
= 3a
dx
bx +
3b√3 a x2 .
(3) Differentiate z = 1.8
θ2 − √5 θ − 27 .
r3 1 4.4 ◦
— 2 − 1 ◦
This may be written: z = 1.8 θ
The 27◦ vanishes, and we have
3 − 4.4 θ 5 − 27 .
dθ
3
3
5
5
dz = 1.8 × −2 × θ− 2 −1 − 4.4 × −1 θ− 1 −1;
dz
— 5
— 6
or,
dθ = −1.2 θ 3 + 0.88 θ 5 ;
dz 0.88 1.2
or,
dθ = √5 θ6 − √3 θ5 .
Differentiate v = (3t2 − 1.2t + 1)3.
A direct way of doing this will be explained later (see p. 66); but we can nevertheless manage it now without any difficulty.
Developing the cube, we get
v = 27t6 − 32.4t5 + 39.96t4 − 23.328t3 + 13.32t2 − 3.6t + 1;
hence
dt
dv = 162t5 − 162t4 + 159.84t3 − 69.984t2 + 26.64t − 3.6.
Differentiate y = (2x − 3)(x + 1)2.
dy = (2x − 3) d (x + 1)(x + 1) + (x + 1)2 d(2x − 3)
—
dx dx
= (2x 3) (x + 1) d(x + 1)
dx
dx
+ (x + 1) d(x + 1)
dx
+ (x + 1)2 d(2x − 3)
dx
= 2(x + 1) (2x − 3) + (x + 1) = 2(x + 1)(3x − 2);
or, more simply, multiply out and then differentiate.
Differentiate y = 0.5×3(x − 3).
dy
= 0.5
dx
x3 d(x − 3) + (x 3)
—
dx
d(x3) dx
= 0.5 x3 + (x − 3) × 3×2 = 2×3 − 4.5×2.
θ
Same remarks as for preceding example.
Differentiate w =
θ + 1 √
1
θ + √θ .
This may be written
w = (θ + θ−1
1
)(θ 2 + θ
1
2 ).
dw
= (θ + θ−1)
dθ
1
—
d(θ 2 + θ
dθ
1
2 ) 1
—
+ (θ 2 + θ
1 d(θ + θ−1)
2 )
—
dθ
−1 1 − 1 1 − 3
1 − 1 −2
= (θ + θ )( 2 θ 2 − 2 θ 2 ) + (θ 2 + θ 2 )(1 − θ )
1 1 − 3
— 1 − 5
1 − 1
— 3 − 5
= 2 (θ 2 + θ 2 − θ 2 − θ 2 ) + (θ 2 + θ 2 − θ 2 − θ 2 )
= 2
θ − √θ5
+ 2
√θ − √θ3
.
3 √
1 1 1 1
This, again, could be obtained more simply by multiplying the two factors first, and differentiating afterwards. This is not, however, always possible; see, for instance, p. 170, example 8, in which the rule for
differentiating a product must be used.
a
Differentiate y = 1 + a√x + a2x.
1 2
1 2 d(1 + ax 2 + a x)
dy (1 + ax 2 + a x) × 0 − a dx
dx = (1 + a√x + a2x)2
1 − 1 2
(1 + ax 1 + a2x)2
= − a( 2 ax
2 + a )
.
2
x2
Differentiate y = x2 + 1.
dy (x2 + 1) 2x − x2 × 2x 2x
dx = (x2 + 1)2 = (x2 + 1)2 .
a + √x
Differentiate y = a − √x.
In the indexed form, y =
1
a + x2
1 .
a − x2
1 1 − 1 1 1 − 1 1 1
dy = (a − x2 )( 2 x 2 ) − (a + x2 )(−2 x 2 ) = a − x2 + a + x2 ;
1 1 1
dx (a − x )2 2(a − x )2 x
2 2 2
dy a
hence
dx = (a − √x)2 √x.
1 − a√3 t2
Differentiate θ = 1 + a√2 t3 .
2
Now θ = 1 − at 3 .
3
1 + at 2
3 2 − 1 2 3 1
dθ = (1 + at 2 )(−3 at 3 ) − (1 − at 3 ) × 2 at 2
2
dt (1 + at 3 )2
7
√2
5a2 √6 t − √43a − 9a t
t
= 6(1 + a√2 t3)2 .
A reservoir of square cross-section has sides sloping at an angle of 45◦ with the vertical. The side of the bottom is 200 feet. Find an expression for the quantity pouring in or out when the depth of water varies by 1 foot; hence find, in gallons, the quantity withdrawn hourly when the depth is reduced from 14 to 10 feet in 24 hours.
The volume of a frustum of pyramid of height H, and of bases A
and a, is V = H (A + a + √Aa). It is easily seen that, the slope being 3
45◦, if the depth be h, the length of the side of the square surface of
the water is 200 + 2h feet, so that the volume of water is
h
[2002
3
+ (200 + 2h)2
+ 200(200 + 2h)] = 40, 000h + 400h2 +
4h3
.
3
dV = 40, 000 + 800h + 4h2 = cubic feet per foot of depth variation.
dh
dV
The mean level from 14 to 10 feet is 12 feet, when h = 12, =
dh
50, 176 cubic feet.
Gallons per hour corresponding to a change of depth of 4 ft. in 24 hours = 4 × 50, 176 × 6.25 = 52, 267 gallons.
24
The absolute pressure, in atmospheres, P , of saturated steam
at the temperature t◦ C. is given by Dulong as being P =
40 + t 5
140
as long as t is above 80◦. Find the rate of variation of the pressure with the temperature at 100◦ C.
Expand the numerator by the binomial theorem (see p. 137).
× × × ×
P = 1 (405 + 5 404t + 10 403t2 + 10 402t3 + 5 40t4 + t5); 1405
dP 1
hence
dt =537, 824 × 105
(5 × 404 + 20 × 403t + 30 × 402t2 + 20 × 40t3 + 5t4),
when t = 100 this becomes 0.036 atmosphere per degree Centigrade change of temperature.
Exercises III. (See the Answers on p. 253.)
Differentiate
x2 x3
× × ×
u = 1 + x + 1 2 + 1 2 3 + · · · .
y = ax2 + bx + c. (c) y = (x + a)2. (d ) y = (x + a)3.
2
dt
If w = at − 1 bt2, find dw .
Find the differential coefficient of
y = (x + √−1) × (x − √−1).
Differentiate
y = (197x − 34×2) × (7 + 22x − 83×3).
dx
If x = (y + 3) × (y + 5), find dy .
Differentiate y = 1.3709x × (112.6 + 45.202×2).
Find the differential coefficients of
2x + 3
y = . 3x + 2
ax + b
y =
1 + x + 2×2 + 3×3
1 + x + 2×2 .
xn + a
y = cx + d . (10) y = x−n + b .
The temperature t of the filament of an incandescent electric lamp is connected to the current passing through the lamp by the re- lation
C = a + bt + ct2.
Find an expression giving the variation of the current corresponding to a variation of temperature.
The following formulae have been proposed to express the re- lation between the electric resistance R of a wire at the temperature
t◦ C., and the resistance R0 of that same wire at 0◦ Centigrade, a, b, c
being constants.
R = R0(1 + at + bt2).
0
R = R (1 + at + b√t). R = R0(1 + at + bt2)−1.
Find the rate of variation of the resistance with regard to tempera- ture as given by each of these formulae.
The electromotive-force E of a certain type of standard cell has been found to vary with the temperature t according to the relation
E = 1.4340 1 − 0.000814(t − 15) + 0.000007(t − 15)2 volts.
Find the change of electromotive-force per degree, at 15◦, 20◦ and 25◦.
The electromotive-force necessary to maintain an electric arc of length l with a current of intensity i has been found by Mrs. Ayrton to be
E = a + bl + where a, b, c, k are constants.
c + kl
,
i
Find an expression for the variation of the electromotive-force
with regard to the length of the arc; (b) with regard to the strength of the current.
