WEBVTT
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PROFESSOR: Welcome
back to recitation.
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In this segment I'm going
to actually show-- well,
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you're actually going to show--
the derivative of tangent
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x using the quotient rule.
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So what I'd like you to
do, I wanted to remind you
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of what the quotient rule is.
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So u and v are functions of x.
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We want to take the
derivative of u divided
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by v. I've written
the formula that you
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were given in class for this.
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And I'm asking for you
to take d/dx of tangent x
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using the quotient rule.
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And the hint I will give you
is the reason we can obviously
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use the quotient rule
is because tangent x is
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equal to a quotient
of two functions of x.
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It's sine x divided
by cosine of x.
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So I'm going to give you
a minute to work this out
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for yourself, and then when we
come back I will do it for you.
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OK.
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So we want to find the
derivative of tangent x.
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So let me-- let me work
on this side of the board.
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So I'm actually going to
take d/dx of sine x divided
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by cosine of x.
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OK.
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So in this case, sine x is u.
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Cosine x is v. So using my
quotient rule I know that first
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I have to take the
derivative of sine
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x-- that's cosine x--
and then I multiply it
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by the denominator, the
v, which is cosine x.
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So my first term in the
numerator is cosine squared x.
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Again, one cosine x comes
from the derivative of sine x,
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one cosine x is the v. It's the
cosine x in the denominator.
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Then I have to
subtract v prime u.
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The derivative of cosine
x is negative sine x.
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I'll actually just
write that one down.
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And then I bring the
u along for the ride.
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So I multiply by sine x here.
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And then I take v squared in the
denominator from the formula.
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v, again, is cosine
x, so I take cosine
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squared x in the denominator.
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Now this at this point
is a little bit messy,
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but the nice thing
is that we can
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use some trigonometric
identities to simplify this.
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So let me first write out what
it is a little more clearly.
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Minus a negative
gives you a positive.
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And then here I get
sine x times sine
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x, so I get the sine squared x.
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And then I keep divided
by cosine squared x.
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Now at this point
some of you might
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have divided by
cosine squared x here
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and gotten 1, and divided
by cosine squared x here
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and gotten tangent squared x.
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And then from there
you could simplify
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to another
trigonometric function.
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I'm going to go straight a
different way to show you
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what that actually also equals.
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So there, at this
point I want to stress
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there are sort of two ways
you can get to the same place.
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But I'm going to use the
fact that the numerator is
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a very nice trigonometric
identity that we know.
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We know cosine squared x plus
sine squared x always equals 1.
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So this is quite lovely, the
numerator simplifies to 1,
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the denominator stays
cosine squared x.
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What is this function?
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1 over cosine x is
actually secant x.
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So if you need, at this point,
to rewrite the whole thing
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like this, right?
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1 squared is 1 and in
the denominator we still
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get cosine squared x-- this
tells you that 1 over cosine
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squared x is actually just
equal to secant squared x.
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So again, what I
want to point out
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is we've now taken the
derivative of tangent x
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and we got that that's
secant squared x.
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Now using this
quotient rule, you
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can do the same kind of
thing with cotangent x,
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with cosecant x, with secant x.
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You can find all
these derivatives
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of these trigonometric functions
using the quotient rule.
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So if you want to know what
the derivative of secant x is,
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you should take d(dx of 1
divided by cosine x and use
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the quotient rule.
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Or, in fact, the chain rule
would work well there also,
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to find that derivative.
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So we are building up
the number of derivatives
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we can find using
these different rules.
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So we'll stop there.