WEBVTT
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CHRISTINE BREINER: Welcome
back to recitation.
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We're going to practice
using some of the tools
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you developed recently
on taking derivatives
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of exponential functions
and taking derivatives
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of logarithmic functions.
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So I have three
particular examples
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that I want us to look at.
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And I'd like us to find
derivatives of the following
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functions.
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The first one is f
of x is equal to x
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to the pi plus pi to the x.
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The second function is g of x is
equal to natural log of cosine
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of x.
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And the third one is-- that's
an h not a natural log--
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h of x is equal to natural
log of e to the x squared.
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So you have three
functions you want
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to take the derivative
of with respect to x.
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I'm going to give you a
moment to to work on those
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and figure those out using
the the tools you now have.
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And then we'll come back and
I will work them out for you
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as well.
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OK, so let's start off with the
derivative of the first one.
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OK, now, the reason
in particular
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that I did this one-- it might
have seemed simple to you,
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but the reason I did this one
is because of a common mistake
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that people make.
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So the derivative of x to
the pi is nice and simple
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because that is our rule
we know for powers of x.
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So we can write this
as that derivative is,
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pi times x to the pi minus one.
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OK, but the whole point
of this problem for me,
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is to make sure that you
recognize that pi to the x
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is not a power of x rule
that needs to be applied.
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It's actually an exponential
function right, with base pi.
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So if you wrote the derivative
of this term was x times pi
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to the x minus one, you would
not be alone in the world.
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But that is not the correct
answer, all the same.
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Because this is not a power
of x, this is x is the power.
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So this is an
exponential function.
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So the derivative of
this, we need the rule
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that we have for derivatives
of exponential functions.
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So that's natural log
of pi times pi to the x.
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That's the derivative
of pi to the x.
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So that's the answer
to number one.
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OK.
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Number two, I did
for another reason.
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I think it's an interesting
function once you find out
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what the derivative is.
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So, this is going to require
us to do the chain rule.
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Because we have a
function of a function.
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But you have seen
many times now,
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when you have natural
log of a function,
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its derivative is going to
be 1 over the inside function
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times then the derivative
of the inside function.
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So again, what we do is we take
the derivative of natural log.
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Which is 1 over cosine x.
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So we take the derivative
of the natural log function,
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evaluate it at cosine x.
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And then we take the derivative
of the inside function, which
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is the derivative of cosine x.
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So you get negative sine x.
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So you get this whole thing
is negative sine over cosine.
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So this is negative tangent x.
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So the reason I,
in particular, like
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this one is that we
see, "Oh, if I wanted
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to find a function whose
derivative was tangent x,
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a candidate would be the
negative of the natural log
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of cosine of x."
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That in fact gives us a function
whose derivative is tangent x.
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So it's interesting,
now we see that there
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are trigonometric
functions that I
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can take a derivative
of something that's
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not just trigonometric and get
something that's trigonometric.
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So that's kind of
a nice thing there.
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And then the last
one, example three,
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I'll work out to the right.
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There's a fast way and
there's a slow way to do this.
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So I will do the slow way first.
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And then I'll show you why
it's good to kind of pull back
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from a problem
sometimes, see how
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you can make it a lot
simpler for yourself,
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and then solve the problem.
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So, I'll even write down
this is the slow way.
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OK, the slow way
would be, well I
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have a composition
of functions here.
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I have natural log
of something and then
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I have e to the something else.
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Right?
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And then that function actually,
is not just e to the x.
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So I have some
things I have to, I
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have to use the chain rule here.
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OK, so let's use the chain rule.
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So I'll work from
the outside in.
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So the derivative of the
natural log function,
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the derivative of the
natural log of x is 1 over x.
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So I take the derivative of
the natural log function,
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I evaluate it here.
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So the first part gives me
1 over e to the x squared.
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And then I have to
take the derivative
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of the next inside
function, which
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the next one inside
after natural log,
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is e to the x squared.
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And the derivative of that, I'm
going to do another chain rule.
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I get e to the x squared
times the derivative
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of this x squared, which is 2x.
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OK, so again, this
part is the derivative
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of natural log evaluated
at e to the x squared.
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This part is the derivative
of e the x squared.
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This one comes just
from the derivative of e
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to the x is e to the x.
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And so I evaluate
it at x squared.
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And then this is the derivative
of the x squared part.
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So I end up with a product
of three functions,
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because I have a composition
of three functions.
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So I have to do the chain rule
with three different pieces
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basically.
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So, but this simplifies, right?
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e to the x squared divided
by e to the x squared is 1.
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So I get 2x.
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OK, so what's the fast way?
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That's our answer: 2x.
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But what's the fast way?
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Well, the fast way
is to recognize
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that the natural log
of e to the x squared--
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let me erase the y
here-- e to the x squared
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is equal to x squared.
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OK?
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Why is that?
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That's because
natural log function
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is the inverse of the
exponential function with base
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e.
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Right?
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This is something you've
talked about before.
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So this means that if
I take natural log of e
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to anything here, I'm going
to get that thing right there.
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Whatever that function is.
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So natural log of e the
x squared is x squared.
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OK?
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If you don't like to
talk about it that way,
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if you don't like
inverse functions,
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you can use one of the
rules of logarithms, which
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says that this
expression is equal to x
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squared times natural log of e.
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That's another way to
think about this problem.
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And then you should remember
that natural log of e
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is equal to 1.
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So at some point you
have to know a little bit
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about logs and exponentials.
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But the thing to
recognize is, that h of x
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is just a fancy way
of writing x squared.
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And so the derivative
of x squared is 2x.
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So sometimes it's
better to see what
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can be done to make the
problem a little easier.
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But that is where we
will stop with these.