Chapter no 4

Calculus Made Easy

SIMPLEST CASES.

Now let us see how, on first principles, we can differentiate some simple algebraical expression.

Case 1.

Let us begin with the simple expression y = x2. Now remember that the fundamental notion about the calculus is the idea of growing. Mathematicians call it varying. Now as y and x2 are equal to one another, it is clear that if x grows, x2 will also grow. And if x2 grows, then y will also grow. What we have got to find out is the proportion between the growing of y and the growing of x. In other words our task

is to find out the ratio between dy and dx, or, in brief, to find the value

of .

Let x, then, grow a little bit bigger and become x + dx; similarly,

y will grow a bit bigger and will become y+dy. Then, clearly, it will still be true that the enlarged y will be equal to the square of the enlarged x. Writing this down, we have:

y + dy = (x + dx)2.

Doing the squaring we get:

y + dy = x2 + 2x · dx + (dx)2.

‌What does (dx)2 mean? Remember that dx meant a bit—a little bit—of x. Then (dx)2 will mean a little bit of a little bit of x; that is, as explained above (p. 4), it is a small quantity of the second order of smallness. It may therefore be discarded as quite inconsiderable in comparison with the other terms. Leaving it out, we then have:

y + dy = x2 + 2x · dx.

Now y = x2; so let us subtract this from the equation and we have

left

dy = 2x · dx.

Dividing across by dx, we find

= 2x.

Now this* is what we set out to find. The ratio of the growing of y

to the growing of x is, in the case before us, found to be 2x.

* N.B.—This ratio

is the result of differentiating y with respect to x. Dif-

ferentiating means finding the differential coefficient. Suppose we had some other

function of x, as, for example, u = 7x2 + 3. Then if we were told to differentiate this

with respect to x, we should have to find

, or, what is the same thing,

d(7x2 + 3)

On the other hand, we may have a case in which time was the independent variable

(see p. 14), such as this: y = b + 1 at2. Then, if we were told to differentiate it, that means we must find its differential coefficient with respect to t. So that then our

business would be to try to find

dy d(b + 1 at2)

, that is, to find 2 .

dt dt

Numerical example.

Suppose x = 100 and ∴ y = 10, 000. Then let x grow till it becomes 101 (that is, let dx = 1). Then the enlarged y will be 101 × 101 = 10, 201. But if we agree that we may ignore small quantities of the second order, 1 may be rejected as compared with 10, 000; so we may round off the enlarged y to 10, 200. y has grown from 10, 000 to 10, 200; the bit added on is dy, which is therefore 200.

dy 200

dx 1

= 200. According to the algebra-working of the previous

paragraph, we find

= 2x. And so it is; for x = 100 and 2x = 200.

But, you will say, we neglected a whole unit.

Well, try again, making dx a still smaller bit. Try dx = 1 . Then x + dx = 100.1, and

(x + dx)2 = 100.1 × 100.1 = 10, 020.01.

Now the last figure 1 is only one-millionth part of the 10, 000, and is utterly negligible; so we may take 10, 020 without the little decimal

at the end. And this makes dy = 20; and still the same as 2x.

Case 2.

dy 20

dx 0.1

= 200, which is

Try differentiating y = x3 in the same way.

We let y grow to y + dy, while x grows to x + dx. Then we have

y + dy = (x + dx)3.

Doing the cubing we obtain

y + dy = x3 + 3x2 · dx + 3x(dx)2 + (dx)3.

Now we know that we may neglect small quantities of the second and third orders; since, when dy and dx are both made indefinitely small, (dx)2 and (dx)3 will become indefinitely smaller by comparison. So, regarding them as negligible, we have left:

y + dy = x3 + 3x2 · dx.

But y = x3; and, subtracting this, we have:

dy = 3x2 · dx,

and

= 3x2.

Case 3.

Try differentiating y = x4. Starting as before by letting both y and x

grow a bit, we have:

y + dy = (x + dx)4.

Working out the raising to the fourth power, we get

y + dy = x4 + 4x3 dx + 6x2(dx)2 + 4x(dx)3 + (dx)4.

Then striking out the terms containing all the higher powers of dx, as being negligible by comparison, we have

y + dy = x4 + 4x3 dx.

Subtracting the original y = x4, we have left

dy = 4x3 dx,

and

= 4x3.

Now all these cases are quite easy. Let us collect the results to see if

we can infer any general rule. Put them in two columns, the values of y

in one and the corresponding values found for

in the other: thus

x2 x3

2x 3x2 4x3

‌Just look at these results: the operation of differentiating appears to have had the effect of diminishing the power of x by 1 (for example in the last case reducing x4 to x3), and at the same time multiplying by a number (the same number in fact which originally appeared as the power). Now, when you have once seen this, you might easily conjecture how the others will run. You would expect that differentiating x5 would give 5x4, or differentiating x6 would give 6x5. If you hesitate, try one of these, and see whether the conjecture comes right.

Try y = x5.

Then y + dy = (x + dx)5

= x5 + 5x4 dx + 10x3(dx)2 + 10x2(dx)3

+ 5x(dx)4 + (dx)5.

Neglecting all the terms containing small quantities of the higher orders, we have left

y + dy = x5 + 5x4 dx,

and subtracting y = x5 leaves us

dy = 5x4 dx,

whence

= 5x4, exactly as we supposed.

Following out logically our observation, we should conclude that if we want to deal with any higher power,—call it n—we could tackle it in the same way.

Let y = xn,

then, we should expect to find that

dy = nx(n−1). dx

For example, let n = 8, then y = x8; and differentiating it would

give

= 8x7.

And, indeed, the rule that differentiating xn gives as the result nxn−1

is true for all cases where n is a whole number and positive. [Expanding (x + dx)n by the binomial theorem will at once show this.] But the question whether it is true for cases where n has negative or fractional values requires further consideration.

Case of a negative power.

Let y = x−2. Then proceed as before:

y + dy = (x + dx)−2

= x−2

dx −2

1 + .

Expanding this by the binomial theorem (see p. 137), we get

= x−2

“1 −

2 dx

2(2 + 1) dx 2

— etc.#

x 1 × 2 x

= x−2 − 2x−3 · dx + 3x−4(dx)2 − 4x−5(dx)3 + etc.

So, neglecting the small quantities of higher orders of smallness, we have:

y + dy = x−2 − 2x−3 · dx.

Subtracting the original y = x−2, we find

dy = −2x−3dx,

—

= 2x−3.

And this is still in accordance with the rule inferred above.

Case of a fractional power.

Let y = x2 . Then, as before,

y + dy = (x + dx) 2 = x2

1 +

= x + √ − √ +

√

1 dx 1 (dx)2

2 x 8 x x

terms with higher powers of dx.

Subtracting the original y = x2 , and neglecting higher powers we

have left:

1 dx 1 − 1

dy = 2 √x = 2 x 2 · dx,

and

dy 1

dx 2

—

2 . Agreeing with the general rule.

‌Summary. Let us see how far we have got. We have arrived at the following rule: To differentiate xn, multiply by the power and reduce the power by one, so giving us nxn−1 as the result.

—

‌Exercises I. (See p. 252 for Answers.) Differentiate the following:‌