





PROFESSOR KELLAND ON GENERAL DIFFERENTIATION 
ix. 2. y—m x ce a 
ad x? 
ae He EN eel 
This gives meas (Lim /-1 art) xX 
yt /—D 
a asi (lamV=1 i a a 
gee |r 
Cor. If =n, this expression must be reduced, as in Ex. 2, Class 2, to 
A b- loge 1 eee +#) -1 1 
a” mrt 2x i eee (1tmv=1 cage oi) x 
. dn In 
Cor. 2. As a ae case, the solution of 
dz Ser a 
a a Vag is 
d as 
Ag al 86 i 
Pe GEN, EP F 

These equations might have been included in the preceding Class, to which 
to that Class 
Ex. 3. 
both in their form and in the mode of their solution, they are very analogous 
They are, however, particular cases of Example 5, below, which does not belong 
ytaVvx a 
— +b al 
The equation in 0 is (by C 





ga 
— /-D+3 —D+1 
yravat 2s yo 770 
—/—-D+3 ,/-D+1 
or {1sav=1 =F — 28 \ _y=0 
Suppose y=%a,e "’; then 
za, f1+av—7 "tt 
_,imtd 
|m |n 
\ —n Peat 0 by ( A) 
Hence any value of » which will satisfy the equation 
live «aun ed 2g 
In |” 
will give a term in the solution. 
Lote PEs is Fy 
Cor. le If 5 Tee ee aaa 
