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CHAPTER II.

CONSTRUCTION OF RULES FOR THE DIFFERENTIATION
OF FUNCTIONS.

SECTION 1.-The differentiation of explicit functions of one variable.

25.] THE nature of infinitesimals, their mode of discovery, the means of expressing them, and the kind of problems to which they are applicable, have been explained in the preceding Chapter. I propose therefore to investigate in the first place certain properties which are generally true of all functions, and in the second place to consider certain special forms; because we shall thereby construct rules which will be useful subordinately, and we shall be spared the labour of continually having recourse to first principles. The convenience of this method is manifest.

If a constant is connected with a function of a variable by the symbol of addition or subtraction, it disappears in the process of differentiation.

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This result is manifest from the very nature of constants, which do not admit of increase, and therefore have the same value in both states in which the function is considered, and therefore disappear in the subtraction; thus we may say, that the differential of a constant is zero.

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26.] A constant connected with a function of a variable by the processes of multiplication or division is not changed by differentiation.

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27.] The differentiation of an algebraical sum of functions.

Let y = f(x)+F(x)+(x)+......

y+ay = f(x+▲x) + F(x+▲x) + P(x+▲x) + ......

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▲y = f(x+▲x) − f(x) + {F(x+▲x) − F(X)}

Ay

+ {4(x+▲x)−P(x)} +......

.. dy = f(x+dx) − f(x) ± {r(x+dx) − F(x)}

+ {(x+dx)—$(x)} ± ......

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Hence the differential of the algebraical sum of functions is equal to the sum of the differentials of the functions.

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From the last two Articles it is plain, that if the function to be differentiated is of the form

y = f(x) + √−1¢(x),

one of the functions being what is commonly called imaginary or impossible,

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Whence it appears that, in the differentiation of impossible quantities, we may treat the symbol of impossibility in the same way as we treat an ordinary constant or symbol of affection.

28.] Differentiation of a product of two functions of a variable. y = f(x)x(x),

Let

.. y+▲y = f(x+▲x) × 4(x+▲X),

... Ay = ▲y = f(x+▲x) × 4(x+▲x) − f(x) × $(x); whence, adding and subtracting the required quantities, {ƒ(x+▲x)−ƒ(x)} $(x)+{$(x+▲x)—$(x)}f(x)

Ay =

.. dy =

+ {ƒ(x+▲x)− f(x)} {$(x+▲x) — $(x)}, d.f(x) × (x) +d.p(x) × f(x) + d.f(x) × d.p(x); and omitting the last term, which is an infinitesimal of the second order, we have

dy=d.f(x) x p(x) + d.p(x) ׃(x),

= f'(x) $(x)dx + '(x) f(x) dx.

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Hence it follows that, the differential of the product of two functions is the sum of the products of each function and the differential of the other.

Ex. 1.

y = (b+cx) (e—ax),

.. dy cdx (e-ax)-a dx (b+cx),

=

= (ec-ab-2a cx) dx.

29.] Differentiation of the quotient of two functions.

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4(x+▲x) φ (α)

f(x+▲x) $(x)− f(x) $(x+▲x)

4(x) 4(x+▲x)

{f(x+▲x)−ƒ(x)} $(x) − {4(x+▲x)−4(x)} f(x)

(x)+(x+▲x)

d.f(x) × 4(x) — d.¢(x) × f(x)

{(x)}2

;

omitting the infinitesimal da in the denominator, because it is added to the finite quantity ; and therefore

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Let ax and z become infinitesimal; then by Cor. I, Lemma I, Art. 21,

dx

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therefore when ▲æ is infinitesimal, (1+4)-1 is to be re

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Therefore to differentiate x" we have the following rule: Multiply by the exponent, diminish the exponent by unity, and multiply by dx.

Suppose the exponent to be negative, then

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dy

Also, since = nxn-1, it is manifest that if f(x) = x",

dx

f'(x)=nx-1; that is, the derived function of x" is nx"-1.

The above rule for differentiating " might also have been determined as follows:

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-1

dy = nbx11 dx (c— exTM) — mexm−1 dx (a + bx”).

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