POLYNOMIAL FUNCTIONS.

163

The ratio of the increment of any continuous function to that of its variable, differs from the derivative of the same variable, by a quantity which diminishes continuously with the increment of the variable, so as to vanish at the same instant.

PROPOSITION II.

To find the derivative of a polynomial, the terms of which are continuous functions of the same variable.

Let yu+v+...+ constant, be the polynomial, where y, u, are functions of x, viz.,

y = fx, u = f1x, v = ƒ1⁄2 x, ... ;

and let the corresponding increments be k, i, j, ..., h; we have y+k=(u+i)+(v + j) + ... + constant ;

and

or (292)

Y'y=sz + z = (U'u=s1 = +%1) + (0′o=128 +22) + .....,

becoming 0 when h = 0,

and, Z, Z1, Z2)

=

ƒ'x = (ƒ1x+ƒ1⁄2 x +ƒ3 x + ... + constant)'

=fix+fxx+fs'x+...; i. e.,

(293)

The derivative of a polynomial, consisting of continuous functions of the same variable, may be found by forming the algebraical sum of the derivatives of its several terms.

Illustration. The student has already had particular examples of this proposition, as in (247), (248), where xa, x3, xo,

tions of x.

... are func

Cor. If the functions f1, ƒ ƒ3, be all the same and a in number, (293) reduces to

The derivative of a multiple or submultiple function, is equal to the same multiple or submultiple of its derivative.

Illustration. The derivative of x is mam-1, of Axm isA • mxTM-1.

PROPOSITION III.

To find a derivative by the aid of intermediate functions. When y is a function of u, and u a function of x, the corresponding increments being k, i and h, (292) gives

for

for

k k

=

i

=

h i h (Y'y=s1u) • (U'u=Sq7) + (Y'v=s1u) • %q

When several variables are successively functions of each other, the continued product of their derivatives, taken in the same order of succession, will be the derivative of the first, regarded as a function of the last.

Illustration. See Proposition II., Section First, y = f1 z = AzTM,

1

z=f2x=x".

Cor. The derivatives of converse functions are reciprocals of each other.*

(296)

Illustration. In (285) let x and y change places, and compare (286) with (268).

The chord and arc of a circle are conversely functions of each other.

To find the derivative of a product, the factors of which are continuous functions of the same variable.

Given y = fr; y = uv, u = ƒ, x, v=f, x; to find y'y-fz •

We have

but

and

and

or

2

y+k=(u+i) (v +j) = y + vi+uj +ij,
k = vi+uj + ij;

z z,
=U'us1z + z1 or i = (u,' + z1)h,

j

1⁄2 = v ́‚—‚ ̧ + z„ or j = (v.' + za)h ;

h

k = v(u,' + z1)h + u(v,' +z2)h + (u,' + z1) (0% + z2)h2 ;

k

= v(u,' + z1) + u(v,' + z2) + (uz' +21) (v2' +z2)h,

=

h

y.' = ['1⁄2/ ] = (uv)' = vu;' + uvés

(uv)'

uv

39

omitting the variable x, and employing only the symbols of operation, fi, fa fa ..., which may be done, since they are equally applicable to any quantity which may be made the independent variable; thus, instead of xxx, we may write // = √√, as a general rule. We may enunciate (297) :

The derivative of a product of continuous functions, divided by the product itself, is equal to the sum of the derivatives of the functions divided by these functions severally.

PROPOSITION V.

To find the derivative of any real power of a continuous function.

If in (297) we make the functions s, t, u,..., n in number, we find

all the same and

From a comparison of (a), (b), (c), it appears that the same rule holds for the derivative of a power of a function, whatever real quantity the exponent may be; and we may write

or

[( ƒx)"]' = n(fx)”—' (fx)',
[(ƒ)"]' = n(ƒ)11 (ƒ)' ; i. e.,

(298)

The rule found in the first section for the derivative of any power of a variable, holds for the power of a function.

Indeed (298) embraces (245); for if fx = x, then (fx) = x' = 1, and (298) reduces to (x")' = nx”—'.

To find the derivative of a fraction, the terms of which are continuous functions of the same variable.

The derivative of a fraction whose terms are continuous functions of the same variable, is equal to the denominator multiplied into the derivative of the numerator minus, the numerator multiplied into the derivative of the denominator, divided by the square of the denominator.

In finding the derivative of any continuous function (300) y = fx, we may replace the increments k, h, one or both, by such quantities, k1, h1, as are separately functions of k and h, and such that the final or vanishing ratios k, k, h, h, become that of unity.

For, by hypothesis, we have

k1

k

=1+1, z1 being such as to reduce to zero when k and k become =

0;