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"= mzolx. Flux (xy)= XY + y 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") + + &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 + =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. 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 = 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 COS # sin x 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) = - flux log i sin log * = sin log *; and that flux (x sin =)=xsin #+22 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 Mr=, mr=y; and by similar triangles Mrm, TPM we have mr :M::MP : PT ory::::y: PT=YE. Consequently we 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 (*+); that of the subnormal PN is yy, the normal PT' MN and if through the point A we draw the line or |