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Event Started: 7/1/2013 3:30:00 PM
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>> I have the exams for a couple of you. So [ INDISCERNIBLE, VOLUME TOO LOW. ] .
>> Some of you did well, and obviously not everybody did well, but, you know
there was some really good scores, and some [ INDISCERNIBLE ] if you did not do
well, if you were below a 70, especially, you know, if you were significantly
below 70, then obviously you'll want to, I want to encourage you to come see me
and get some extra help, because, you know, there's just not that much time.
We're already in week four. Can you believe that? Halfway done after Wednesday.
You get fourth of July off this week, so only this class meets three days, first
few days of the week. But anyway, if you are struggling, in particular, like,
you're struggling with the chain rule for example, or like, or some of those
derivatives, you didn't do very well. Even if you got above a 70, you missed
quite a few points on problem number six, then, you know, I want to make sure
that I help did you, because it's really important that you understand what the
chain rule is and how it works.
>> For example, I noticed a little misconception there on some of the problems.
We can certainly go over any of these. If there are particular ones that you
like to look at, I can post the key to Blackboard/, I guess [ INDISCERNIBLE,
[ INDISCERNIBLE, VOLUME TOO LOW. ] I can provide you with a copy of the key or [
INDISCERNIBLE ] version. I haven't gotten it scanned in yet, but with any of the
ones from number six that you'd like to look at, I'd be happy to go over those.
In particular that problem, but really any of the problems. Yes? No?
>> [ INDISCERNIBLE, VOLUME TOO LOW. ] .
>> Okay. I wrote some funny comments. Sometimes I do that. I will pick one to do
because I think probably, I'm trying to think which one was the hardest. Well C
was an interesting one. Why don't we look at C and D together. We'll look at the
[ INDISCERNIBLE ] so C says, does h of t equal negative two times two minus
contingent of t over five see can't of t. Now, some folks cant of t. Now, some
folks I think most people chose to do this as [ INDISCERNIBLE ] and that's fine.
But there's other ways you can do it. I think there was one person who did it
usually the [ INDISCERNIBLE ] rule. By saying I'm going to write this as
negative 2/5 co-sign of t, secant t in the denominator is the name as co-sign t
not in the denominator, and two minus tangent of t. And again, but you actually
don't even need that. So if you, but I think most people did it that way, and
that was fine. But you could have also just multiplied this out and got negative
4/5 co-sign of t plus 2/5 tangent times co-sign, what's that?