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

Illustration. The derivative of xTM is mxTM-1, of AxTM isA • mx”~!.

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

h

=

k

y=f1u,

=

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

+(U'u-sqx) • 21+%1%2;

•••_Y'y=sz= (Y'y=s1u) • (U'u=ƒ1⁄2ï),

making h, and, consequently, k, Z1, Z2, = 0.

So from

we find

and generally

y=f1u, u=f2v, v = ƒ3x,

Y'y=ƒz = (Y'y—su) • (U'u-sqv) • (D'o—sgx) ;

y=fx

y = y; • si

...

i. e.,

(295)

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=f, z = AzTM, z=f2x=x".

Cor. The derivatives of converse functions are recipro- (296) cals of each other.*

For, let y and x be functions of each other, or

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=fx; y = uv, u = f1x, v=f2; to find y'y-fz.

k = v(u,' + z1)h + u(v,' + z2)h + (U2' + z1) (0% + z2)h2 ;

k

h

and

or

=

= v(u;' + z1) + u(v,' + z2) + (U1⁄2' + z1) (0%' +z2)h,

y.' = ['/] = (uv)' = vu,' + u0%,

|

(uv)'

In the same way, if v be

have

uv

=

omitting the variable x, and employing only the symbols of operation, fi, f2, fз, ..., which may be done, since they are equally applicable to any quantity which may be made the independent variable; thus, instead of 2/x • 3√√x = x, we may write 2/3/5/, as a general rule. We may enunciate (297) :

3

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, ..., all the same and n in number, we find

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

[(fx)"]' = n(fx)”—' (ƒx)',
[(ƒ)"]' = n(ƒ)”~1 (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”-1.

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 k1 : k, h1 : h, become that of unity.

For, by hypothesis, we have

k1

k

=1+z1, z, being such as to reduce to zero when k and k1

PROPOSITION VIII.

To find the expansion of a continuous function, such that its successive derivatives all become finite when the independent variable reduces to zero.

Let y = fx be the function. It follows, from a process of reasoning precisely like that employed in the demonstration of the Binomial Theorem, that no other than integral additive powers of x can enter into the expansion; and it only remains (and is sufficient) to see if the assumption

y = A。 + A1x + А2x2 + А‚Ã3 + ...,

is possible, or, what amounts to the same thing, if the coëfficients Ao, A1, A2, A3, ..., are determinable, and, therefore, real. Taking the successively derived functions, we find

y' = A1 • 1 + A2 • 2x + A3 • 3x2 + ...,

y" = A, 2.1+A, 3. 2x + A1 · 4 · 3x2 +....

y'"' = A, • 3 • 2 • 1 + 4 • 4 • 3 • 2x + A5 • 5 • 4 • 3x2 + ....., yiv = A • 4 • 3.2. 1 + A5 • 5 • 4. 3. 2x + ....

Now, if in the above equations we make x = 0, and indicate the corresponding finite values of

there results yo = A, yʊ=A ̧ • 1, yő' = A, • 2 • 1, yő" =A, • 3 • 2 • 1, ... ;

This is essentially Maclaurin's Theorem, and is very serviceable in expansions, being more general than the Binomial, which it becomes simply by putting y = (a + x)".