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Chapter 3 94

# Chapter 3 94 - /—— 244 CHAPTER 3 DIFFERENTIATION RULES...

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Unformatted text preview: /—— 244 CHAPTER 3 DIFFERENTIATION RULES 58.:E2+4:cy+y2:13 2 2m+4(my'+y'1)+2yy’:0 => x+2my’+2y+yy’=0 : 2xy’+yy’=—x—2y => y’(2:c+y)=—x—2y => ’=———'\$“2y.A:21 ':#2_2=_é~ an equation of the tangent line is y — 1 = -§(:c — 2). ory : —%:1: + 1—53 59. y = (2 + we” => 3/ = (2 + sex—e”) + e” -1: e_\$[—(2 + a?) + 1] = e‘x(—m — 1). At (0,2). y' : 1(—1)= —1. so an equation of the tangent line is y — 2 = !1(:n — 0). or y = —;v + 2. 60. f(:r) = we‘d“ => f’(m) = xIeSI”(cos '13)] + e“"1(1) = eSI”(at cosa: + 1). As a check on our work. we notice from the graphs that f’(a:) > 0 when f is increasing. Also. we see in the larger viewing rectangle a certain similarity in the graphs of f and f’: the sizes of the oscillations of f and f’ are linked. _ + _———————_———— 2t/5—m 2 5—w 2\/—5—m 2x/5—z (d) The graphs look reasonable. since f ’ is positive where f has tangents with positive slope. and f ’ is negative where f has tangents with negative slope. 62. (a) f(ac) : 4m # tang: => f'(a:) : 4 , sec2 :3 :> f”(\$) : —2 seem (seem tartan) : —2 see2 an tanw. (b) We can see that our answers are reasonable. since the graph of f' is 0 where f has a horizontal tangent. and the graph of f ’ is positive where f has tangents with positive slope and negative where f has tangents with negative slope. The same correspondence holds between the graphs of f ’ and f ", ...
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