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17. If we had the powers of logarithms, or even logarithms of logarithms, it would be easy to find their fluxions. For example, ylt"=m-1

lx we should obtain y=mambo (lt)" + noml; (2x) ** = 2m10 (13) (n + mix), &c. Again let y=llx; make lx=2, and we shall have

xlx

18. The equation flux (ix) =-, gives - = Aux (lx). Consequently the flusion of any quantity is equal to the product of that quantity by the the fluxion of its logarithm. This rule may be applied to facilitate finding the fluxions of quantities of any sort, even

mrr if algebraic. For example flux (**) = x flux lx"=

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Flux (xy)=xy** + y)=yš+xj. Flux ($) = _%) = y*+ xy

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The same principle may be applied with У ya success in finding the fluxions of exponential quantities ; by which name is to be understood, such quantities as have variable exponents,

Such are a*, x, &c, which are of the first order; ry, which is of the second, &c.

The fluxion of a", will be according to this rule a® flux la'a ffux (xla) = a*ila. Therefore if e be the number 2.7182818, of which the logarithm is 1, we shall have flux (e) = er. Similarly flux (xy = x flux ( ylx) = xy (ylx + 4%), &c. 19. We might also have found these Auxions in the following We have seen that n = 1 + In +

(In)? _ (lnys

2 2.3 Suppose now that n=a", and substitute this value in place of n, and we shall find a*=l+la" +

(la)? (la")

十. + &c. Now la*= xla, and 2 2.3

xila, da (la") = (xla)' = 2?l?a; therefore a'=1+xla + +

2 2.3

**x 13a and consequently flux (a*) = xla + xx l'a +

+ &c. =

xia

2 x?la x?la 1 + xla +

2.3

manner.

+ &c.

+ &c.

With respect to such exponentials as to their fluxion is easily found: for we have

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+

=yt, y fluxion of e?.

In a similar manner we may find the second, third, &c. fluxions of logarithmic and exponential quantities, but it is needless to dwell longer on this head.

OF THE FLUXIONS OF SINES, COSINES, &c. AND OTHER CIRCULAR FUNCTIONS.

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20. Let sin x=y, we shall have y+y=sin (x+x) = sin x cos i + sin coss. Now : representing an infinitely small arc, we shall have 1st, cos c=1; 2dly, sin =s. Therefore y+y=sin x +ë cos 2, or y=flux sin x=x cos x. Consequently the fluxion of the sine of any arc is equal to the fluxion of that are multiplied by its cosine.

21. Since flux sin x=x cos x, if we make x = 90°—y we shall have a =- - y, and flux cos y - y sin y, a formula which we might have obtained in either of the two following manners.

ist. Sin? x + cos x =1. Therefore sin x (flux sin x) + cos e (fus cos x) =0, and fluz cos x = - (flux sin x)=-o sin x .... or 2dly, flux cos x =cos (+2) - cos x = cos x cos is — sin

cos X=-I sin x. Hence the fusion of the cosine of any arc is equal to the negative fluxion of that arc multiplied by its sine. 22. Let lang x = %

we shall have ;= cos a (flux sin x) - flux cos i x sin x * cos 2x + če sin 2x

sin

COS I

ż sin I

sin

COS I

COS X

COST

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= flux tang 1.

Hence the fluxion of the tangent of any arc is equal to the fluxion of that arc divided by the square of its cosine.

If instead of supposing the radius = 1, we had supposed it = a, we should have had fiux tang x =

aar

cos ar

23. Let x=90°—y, we shall have flux cot y =

; similarly

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COS #

sin x

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Flux (sin x cos x)== cos sin = cos 2x. Since v + cos s) =cos i e, we have flurnt

1 + cos

= flux cos 2

2 =-**sins.

Similarly we shall find that flux (cos log x)= -flur log i sin log * =- sin log *; and that flux (z sin =)=xsin x + 2x cos z.

flux cos 24. If x be any arc, its Auxion : =. Pur sin s

flux tan flux tan a cos ** flux tan *=

flux cot x sin 's sec ** 1+tang flux cot a -flur cots

&c. &c. cosec x 1+cot APPLICATION OF FLUXIONS TO THE

THEORY OF CURVES. 25. Of all the problems that can be proposed respecting a curve, the most simple is that which requires us to draw a tangent to any point of the curve.

Suppose the curve to be AM, its axis, AP; its coa ordinates AP and PM; it is evident that to draw a

MT

В. tangent to the point M, we have only to determine the subtangent PT.

Let us imagine the arc Mm to be infinitely small, draw the ordinate mp, infinitely near to AP, and sup- T

T

A PP pose Mr parallel to Pp. Let, as usual, AP=, PM=y, and we shall have Pp, or Mrz, mrry; and by similar triangles Mrm, TPM we have mr :M::MP: PT ory::::y: PT=YL. Consequently we

y have only to throw the equation of the curve into Auxions, in order to obtain the value of.

and then to substitute this value in the formula for the subtangent just found, and PT will be determined. 26. The expression for the tangent MT is v (+ y ) = y

y? y (*+); that of the subnormal PN is

y yy, the normal

PT' MN

and if through the point A we draw the line

or

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8 =7( + );

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AQ parallel to MP, we shall have PT : AT::MP: AQ, that is yx : yt

::y: AQ=y my These valucs of AQ and y y AT enable us to find the asymptotes of the curve AM if it have any ; for if, after having substituted in these two values that of ý obtained from the equation of the curve, we suppose x infinite, there will be as many asymptotes as there are different values of the lines AQ and AT. The position of the asymptotes will always be determined by the points T and Q. We shall now apply these formula to a few examples.

The equation of the circle is yʻ=a*--*"; therefore yy=-21, and yx or the subtangent ==y'-(a'—-*)

The sign indicates y that the subtangent must be taken in the same direction as the abscissa, because in the construction of the formula it was taken in a contrary direction. If we had taken the vertex of the curve for the origin of the abscissæ, the equation would have been y2 = 2ax-xx, and this would have given a positive result as in the formula.

T'he equation y*=a*–** gives yy or the subnormal =-*; and

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the normal = v(y2 + y* y) = v (+*+90=a= the radius, as it evidently should be.

In the parabola yo=px; therefore yy = $ p, the subnormal ; and

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ar

and yx

2ax +33

a + x y We have also AT (see preceding figure) = an expression

atx which is reduced the quantity a, if we suppose z infinite. On the same supposition we find that AQ=y

=y

(a +x)

a’y a'ya_b*z (a +x) box

->, becomes simply=b. Thest 2a +

ty

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afy

ay

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