102 3Ir. Adams's Demonstration [Feb. 



Example 10. — To find the increment of — . 



Put to = d X = A X, a constant quantity, then will 

 d (~) = d (X-") = - m «; a;-"" + " = - m . -^, , 



\X"J ^ ' 0,'"+' 



d(— mw X '"'+") -7- 2 = X , 



d (w (m + 1) tc^ .r-"' + ^'j H- 6 = -^ ^ '- x ^i;;^, 



&c 



Therefore a(-L)= -m.~^ + ^«±nj^ ,Jf- _ &c. = 



, , m w X"-' + H^^Lziljf! ^.".- 4. &c. 



(* + »)"• x"> 4:"'(,i- + tc)"' 



Example 11. — To find the increment of log. x. 



Put w = d X = A X, a constant quantity, and y = log. x, 

 then 



erefcre 

 + &c. 



x' 2 2 j'' 6 3 j3' 24 4 j.^ 



therefore d y + ~ + -^ + — + Sec. = — H — 



y ' 2 ^ 6 ^ i;4 ^ *• 2j^ ^ 3J.3 4 j« 



Ax A x'^ . A a-i A x< 



«„ A /I \ ^ -^ A X' , A X ' A x' . o 



or A (log;, .r) = h • h &c. 



^ * '^ X 2 X* ^ 3 X' Ax* ^ 



Example 12. — To find the increment of .r log. x. 

 Put w = d X = A X, a constant quantity, andj/ = x log. x, then 



rfj/ = i« log. .r + io=w (log. a; + 1), -i' = __, _J1 = - 

 ^^^ IT = 12^' IFo = ~ W^' ^^' Therefore, (iy + ^- + 

 2 — 3 "^ 2~~3 — 4 "*■ • ~ "^ ( S* *^' + 1) + '^" ; or A (a- log. x) 

 = A .r (loff. .r + 1) H . 



V to ^ ^ ^ 2 . X 2 . 3 . x^ 3 . 4 . x' 4.3.x* 



+ &C. 



Example 13. — To find the increment of (log. x)". 

 Put y = log. J- = / x, then will A (log. x)" — A ( ?/") ; but by 



example 4, A (y") = my""' A j/ + ""'" ~ '\j "''^'''' + &c. by 



substituting for ?/, we have A (log. x)" = /« (/ .r)""' A I x + 



»» (m - n ('-r)'"--(A/x? ™ , e \ 7 /A 7 \. 



+ Sec. Ihe values of A ( x, (A Ixp, 



(A / xy, &c. may be expanded into a series, if necessary, by 

 example 11. 



Example 14. — To find the increment of a'. 

 Suppose c/ .r = A x, a constant quantity, and y = a'. 



