Unformatted text preview: Math 1A — Quiz 7 — March 21, 2008 Quiz/7 lNC/S Total/10 NAME ll Find the value of sec (tan‘ 12) Le: tan’ ‘ 2 = 9 then 2 = mine so we can draw a triangle: l 0 V“—
with hwotenuse \/§ since mane =7
1 _ _;_i_£_ﬁ_
Thensec(mn 2)_Sece’cose_£—A_ 1 _\/5 ﬂ 1
H
2‘ Find the derivative of each of the following functions.
a) y=cos“(e7") y’z— , INF] 321.2
i e)
b) y=ln(cosh3x) y’= 1 ~sinh3x3
cosh 3x
1 . _ r I
o) y= . _, =(sm 1)6) so
5111 x
 2
y’—— 1(sin'1x)
3, Find the limit 3) lim tim“(lnx) since lim lnx=— oothen
X—o 0' x—H’J' lint}. tan‘ 1(In x) = — 3% (See the graph of tan‘ ’ in the book) . wshx . e‘+e"‘ . 1+e‘7” 1
11m =11m =12m = b) x—vzo ex x—om 23: pm, 2 ’2— 0 lim 4 asxHoowegetg...indeterminate
X~°°ln(1+2e‘) 00 we better use I ’ Hopital ’ s: so thelimit becomes lim 1 = lim “28 was 1 ‘22,, mm gex =lim 1+1=1 xq “’26" NOTECARD BONUS QUESTION: (1 point) Evaluate the limit
. x“ — ax + a ~ 1 0 . .
11m —~——2— as x —> 1 weget — . . l mdetermmate
x —v 1 (x _ l) 0
we better use I ’ Hopital ’ s:
11v 1 _ so the limit became: lim —a and asx —> 1 x A» 1 2x — 2 . 0 , . we still get — . . . Indeterminate 0
so we better use] ’ Hopital ’ s again: . _ _ Z _
so the limit becomes lim1 5% = gig—1) ...
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 Spring '08
 WILKENING
 Calculus

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