FLUXIONS. 
Mood. by rule (A), du=—t2—'de=—yertds = 
—~ S8dz : : ; 
ine ; 
3. uma +b/r— crm Rule (E) applies to 
br hoon ral and as the term a is eines ager 
dusbd(s ab now, by rule (A), d(a® )= 
zx 
=I }am 3 — — —* 
aya? amd d (2 j= — 2c de=- ze therefore, 
d _bdx “edz 
“=O a Pes 
Ex. 4. Letu= =, In this case, we applyrule(H), 
and asd (#9)= 22d andd (1=2)=—d z, we have 
pes Q(i—az)edapardx (Q—x)edx, _ 
Ex. La pee a a then 
5. a zm). ta z=, 
y=u". This ct NE Dina i 
as there directed, we have, by rule (A), dy=n w"—"d u, 
and from the equation u=a+5 x”, by rules (E) and (A), 
du=mb2x"-1dz. Therefore, 
> dy=mn bla+b x"! 21 dz. 
Ex. 6. pe Y= ete "Paty gem’ and 
we have y=r u. youd«+ 
_ edu, Meaadiaiet a 6 Hes Iain dine 
vma?+2%, and then u==/s, therefore by rules (I), (B), 
and (A), dumeeazdux 0” 
you = eeey 
Hence, ; 
d 2 
PP aN ey Wan orev nts), re? ot 
ghee: peti aie) e! Now 
‘Seger hets—e ay « 
rules and (A), therefore, after substituting for 
u, d ot (Und ‘ ca alues, and reducing, we find - 
d _ (a 4ate*—42')dz 
I=" Ve) 
We bars re sowe’ letters for complex func- 
tions, with a view er operation to a particu- 
lar rule ; but, in actual calculation, this will not always 
be necessary. Thus, in the 5th example, in which 
oe ae by retaining the symbol a-++-b 2”, instead 
u for it, we have sgt ap baa 30 
Sexier), and as d(a+62")=b d(2")—mb x"~1d x, 
"px 8 Lety (a tqb)-42,0 (aa) (2d 
ees Pooh +6)*+24/( ) x (#—#). 
d {(ax+0)} =2%Xa2)48) xd(a245) = 
6ax*(ax+b)dz. 
Again, afoy(a—a) x (2—s)} 
aida) x d(a—b) + (2—b) Kd } 24/(a*—2") 
But d(2—})=d2, and d fey(a—e) ~ aol 
therefore, upon the whole, “i a 
dy=6a2*(a234b)d x + 20—tU+2ba)d x 
ae—z* 
33, Wed next. give exomplesof thm fune- 
one putting 1.(z) for the Napierean log. of any func~ 
z 
405 
. Ex. 1. Letu=l. (Fay). Put Tay =* 
then by rule (B),du=“* But by the rule for the 
fluction of a fraction, 
dnd (FS 
Therefore, warns S40 
zea) 
Ex 2. Letwnt, $V (t+#) + (1—e) 
1 i 
Put Y=1-+42, 2= Seoayh meee ore 
Then u = i(¢=) =1.(y+2) —L. (y—z); 
of LN aoe. 
y 
Now eedeeuty, oni S08 —dv. 
Therefore dy= and dex ="; 
Hence we have 
dyt+dz_  y—z dx 
ait3 ype Vyz 
_ te) de 
PN nate a ate 
ay yAP—*) 
But y? + vs y Aa 2s and yexy/(i—e): 
Therefore, du= z/(1—a") 
In the next two examples we shall merely give the 
results, 
@edx 
=¢= 
fy": 
and du= 
dy—dz_y4+2 dx _ 
Ye yz Qyz” 
— 
Ex. 3, wxifervigeyhs ;du=—__—__—_ 
Vey) 
ee ER Vee teet 4 dx 
(148) Vapey 
As an example of an exponential quantity, let 
; supposing y and z to be any functions of a va- 
rial ae then log. u=zl.(y);_ There- 
fore, (Article 28.) d-{1.(u) } =a feuy) bs now 
aj 1 (1 wh 4 and af 21, (9) } saei) +74 
by rules (B) and (F); Therefore, substituting y* for «, 
we have 
du=y fast) +222}. 
34. Examples in circular Wiactione .. 
Ex. 1. Letu=tan. 2. Because tan. preren 
fore, dud (= 2) =e dE (sin) sin. 2d eos.) 
But d (sin. x) = dx cos. x, and d (cos. 2) = — de sin. 2, 
role (D) ; theesborn dat ene) in 
cos, 2.x cos.7a" 
In the three following examples, we shall, for the 
sake of brevity, merely state the results. 
Ex. 2.. ux sec. x, du = dx tan. « sec, a. 
Ex. 3. u= cot.x,du=— Se . 
sin. * x 
Ex, 4 ux cosec, x, du =—d a cot. xcosec. «. 
35. In the above examples, the tangen 
are considered as functions. of the are : ‘but*we may re- 
verse the h and consider the are as a function 
of the tangent, or secant, &c. 
Ex. 1. Let it be required to find the fluxion of an arc 
Direct’ . 
Method. 
—o— 
