Differenzialformel. 
282 
Differenzialformel. 
e die Basis des natürlichen Systems =2,7182 81828... 
m den Modul des Briggschen Systems = log br e — 0,43429448 
1 1 
so ist 
81. y = e z 
82. y = a~ 
83. y — 10* 
logn 10 2,3025 8509.. 
dy = e : 0s 
dy = a z logn a 0s 
Oy = aZ ÜZ ' ln 10 = 2 ’ 3025 bo09 •••"'■ 05 = 0,43429448... 
1 
II 
Si 
oo 
dy = — = s- 1 05 
85. y = Zoy 6r s 
JV cs, 02 02 
J * 2 ln 10 2,3024 85 .. 2 
= log br e • — = 0,43429 ... — 
* 7» 
86. y = Zn (a ± 6s) 
c, ± 6 02 
Ö?/ = —¡r 
a. ± 62 
87. y = ln (as ± 6z. 2 ) 
(a ± 262) 02 
8 »=- 
88. y = Zn (s -f ] a 2 s 2 ) 
9,= * 
]/a 2 + 2 2 
89. y = ln (2 + ]/jj* — a 2 ) dy = 
j/2 2 — a 2 
90. y = Zn (2 — ]/s 2 — a 2 ) 
dy = 
]/2 2 — a 2 
91. y = Zn — (a-fj/a 2 +2 2 ) 
7» 
^ «05 
Oy = 
2 J/a 2 + 2 3 
92. y = ln — (a—]/ö 2 — s 2 ) dy = -——— 
5 2|/a 2 —z 2 
93. y = ln (a + 2) (6 + 2) 
„ 02 02 a + b -f 2z- _ 
!J ~ a + 2 b + 2 - (in -f 2) (6 + z.) C ' 
94. y = ln(a — 2) (6 — 2) 
95. y — l?i —■ — Z 
a — s 
, a — 2 
96. y = ln ;— 
J a-f 2 
. 2 — a 
97. 1/ = Zn —■— 
■ 2 + a 
» 2 -f a 
98. y = Zn 
2 — a 
02 02 «4-6 — 22 „ 
a — 2 6 — 2 (a — 2) (6 — s) 
~ 2a 02 
2a 02 
8 » 
„ 2a 02 
8 »=iw 
» 2a 02 
8 »= ä »-„> 
99. y = (Zn 2)« 
100. y = z m ln 2 
101. y =-- 2"' (ln 2 ] 
m \ m / 
0y = (Zn 2) 1 • — 
0y = [m Zn 2 + 1] 2 1 02 
( 0y = 2 ff| - 1 Zn 2 02 
_s 
II 
5* 
cd 
0 
05 
oy = —i— 
2 Zn 2 
103. y = l • br (l • br • 2) 
(log-bre) 2 
8 ^ s * /öj - Ar s 8 " 
104. y = sin 2 
0y = COS 2 • 02 
0y = — SIW 2 02 
dy = sec 2 s 0s = (1 + z# 2 2) 0* 
104. y = sin s 
105. y = cos s 
106. y = tg z>