4. Let y = ƒ (2) where z = 0 (x), it is required to find the first differential co-efficient of y considered as a function of x

When a becomes x + h let z become z + k y' = f (x + k)

Substitute this value of k in equation (1)

y=f(z)+A (Ah + B'h2 + ...) + B (Ah + B'h2 + ...)' + · =f()+ AAh + Ph® + Qh® +.

y'

dy

=AA'

dx

...

But A is the first differential co-efficient of y considered as a function of z, and A' is the first differential of z considered as a function of x, hence we have the first differential co-efficient of y considered as a function of x, or

This theorem will be found of great use in differentiating many complicated functions, thus,

The differential coefficients of many complicated functions may be found by first taking the Napierian logarithms of the functions.

f(x) is of such a complicated form that, in order to find its first differential co-efficient, it is necessary to simplify it by the substitution of two new variables, u, z, each of which is, of course, a function of x. Hence arises the following problem :

required the first differential co-efficient of y, considered as a function of x.

In order to discover this, it is manifest that we must substitute a + h for x in each of the two functions u, and z, and find the co-efficient of the simple power of h in the developement of the compound function F (u, z).

When a becomes x + h let u become u + k, and let z become z + ↳

It now remains for us to substitute u+k, z + 1, for u and z in F (u, z), but it is manifest that if we make these two substitutions in succession, we shall obtain the same result as if we make them both at once, since u and z are con sidered altogether independent of each other in these substitutions.

Let us then, in the first instance, suppose that u becomes u + k, and that z remains constant in the equation y = F (u, z).

y' or F (u + k, z) = F (u, z) + A ̧ k + B, k2 + (4) Let us now suppose that z becomes z + 1, and that u remains constant, then F (u, z) becomes F (u, z + 1) = F (u, z) + A‚1 + B, 1o + (5) Since A, k involves z, it now becomes a function of zl, and being expanded as a function of (A, k + })

Substitute then these values of F (u, z), A ̧ k, B, k2, in (4), and we have y' or F (u+k, z + 1) = F (u, z) + A, k + A, l + terms in kl, k2, l2, . . . Substitute for k and 7 their values from (2) and (3)

.. y' = F (u, z) + A, (A'h + B'h2 + ...) + A2 (A”h + B'2 + . . .) + .. Arranging according to powers of h

= F (u, z) + (A, A' + A, A”) h + Ph2 + Qh3 +

Hence by definition

y = ƒ (t, u, z) where t = F (x), u = ¢ (x), z = ¥ (x) dy dy dt dy du dy dz dx dt dx du dx dz dx and so for any number of functions.

=

+

+

Hence we deduce the following general conclusion :

The first differential co-efficient of a function composed of different particu lar functions, will be sum of the first differential co-efficients of each of these functions considered separately and independent of each other, according to the rule established in the last article.

This principle, combined with the preceding one, will enable us to determine the first differential co-efficients of all functions of one variable, however complicated in form.

Ex. 1. Let y = (ax3 + bx2 + cx + d )TM (ex2 + nx3 + rx2)".

= mum-1 z11 (3ax2 + 2bx + c) + nzñ--1 um (4ex2 + 5nx1 + 7rx®) Substituting for u and z their values

= m (ax3 + bx2 + cx + d)TM-1 (ex1 +nx3 +rx2)" (3ax2+2bx+c) +n(ex1+nx2+rx2)"−1(ax3+bx2+cx+d)TM(4ex3+5nx1+7rx®)

By dint of practice, however, the student will be able to obtain the first differential co-efficients of all functions without actually performing the process of substitution.

On finding the differential co-efficients of equations.

We have hitherto supposed the function ≈ to be given under the form

y = f(x)

but it frequently happens that y is given only by an equation between x and y

of the form

F(x, y) = ?

The resolution indeed of this equation for y would give us y under the form

y = f(x)

but this solution is seldom possible, and wholly unnecessary for our present purpose.

If the equation, then be of the form

F(x, y) = 0

and if we suppose y = f(x) to be the value of y, which would be obtained

from the solution of the equation, it is manifest that if we substitute this value for y in the proposed equation, we shall arrive at an identical equation

u or F {x, ƒ (x)} = 0

whatever may be the value of x, and hence, if we substitute x + h for x, the equation will still be identically =0, whatever may be the value of h.

whatever be the value of h, hence necessarily each individual term must be u = 0, Ah = 0, Bh2 = 0..

= 0, and

But since

..