FUNCTION. 
be put for x, and the epee of the fecond term of 
the expanded expreffion is to be taken. 
To, find the entire  diferene, and the differential of y, a 
funGtion o oF Ps Q 1 $y &C. when Np, J» tS, Kc. are fimple 
_ functions 
when x becomes x + A x, p becomes as t Ap» 
Piecone qg+4q, &c. andy becomes ytAy 
yt Aye Pt Dy dap Dry (Ax) + 
ptrAp=pt+ Dp. Ax tDrp (Ax) + 
gq+Ag=gt DyAwt Mg (Ax) + Dg (Aa) 
rpArar4+Dr.Axc+ Mr (Ax)? 4+ Pir (Az)! 
Therefore in the oe ie “a ou of y pipes 
the new values of f, g, o-efficient o 
equal D y, or a differential eran of y. 
Example if.—Lety=aptg+r& 
Then fincey + Dy. Ax 4+ D'y as 
=p+Dp.Ax+ Dp (Ax) ke. 
+q9+Dq.Ax4D'q (Ax) &e. 
tr+Dr-Ar+ Dr (Ax) &e. 
Pa é x) 
Ds p (A#)3 
Dy= Dp Dgq+ Dr &e. 
Therefore 
dy=Dy.dx=Dp.de4Dq.dxer4+D.r.dak&e. 
_dy _ dp dy dr 
=7,:d*= 75 ~— 7 ae da + 7 Pa 
The entire diff Ay = evidentl 
A(ptqtr+ &e.} = (Dp +Daq+ Dr) Ax 
+ (Pp 4+ 0940 r&c) Ax’ + &e. 
Let.y be a fimple funtion of 2, then » becomes p + Ap; 
dy. AP y 
dp’ 1.2 ip 1.2.3 
If p be at the fame time fuppofed a funétion of x, ip 
# becoming x + A x, y becomes likewife y + Ay = 
y becomes y 7 2. Ap ae ar 
Bod dw ¥.2.3° 
Now p ee ae eas as a fundtion of x, x becoming 
x + Ax, p bec 
d ive A x? 
ot a ena ax 
d' p Ax 
Now fubftitute this value of A p in the equation. for the 
value of y + ; 
dy dp 2 ap Ax 
= — eB 
POR It Fh an'" T a5 Ie’ Tee 
dy Vp Ax 1 (9p d* 
fg ore aa wt 82.4 (Mae SE, 
an + &c.) + &e. 
Then by comparing the co-efficients of A xin the two 
values of y, 
dy 
ds dp’ dx “dp dp. 
dix. 
po 82. Se Sy sey 
a? pdx dz \ds 
dy? dy d? p : 
ie ~ ap‘ Te 
a (52) 
dx 
dy 
Py dx a 
det dp da? 
_ dx 
(a5) 
dy dy ap 
d x? dx dx? 
“2! ay 
(3 x dv 
Example 2d.—Let y =(ae — 2°) required the values of 
ed . J, : 7 , ke. 
dx 
Put —“#z= 
dw da’ CO GE 
ae ag! 
(aay = 
of 7, = “dp + 
sf ax 
dy 1 
—-=-1ipjp-t = ee 
dp of tis ew 
dy _ 2 
dz A a 
pe. dy Up kL 
ds * 
2) = 4% 
dy lai a eee 
ag P= 4(¥ —a’)j 
