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CHRISTINE BREINER: Welcome
back to recitation.

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We're going to practice using
some of the tools you

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developed recently on taking
derivatives of exponential

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functions and taking
derivatives

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of logarithmic functions.

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So I have three particular
examples that I

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want us to look at.

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And I'd like us to find
derivatives of

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the following functions.

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The first one is f of x is
equal to x to the pi

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plus pi to the x.

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The second function is g of x
is equal to natural log of

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cosine of x.

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And the third one is--

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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 to take the

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derivative of with
respect to x.

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I'm going to give you a moment
to to work on those and figure

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those out using the the tools
you now have. And then we'll

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come back and I will work them
out for you as well.

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OK, so let's start off
with the derivative

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of the first one.

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Ok, now, the reason in
particular that I did this

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one, it might have seemed simple
to you, but the reason

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I did this one is because
of a common

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mistake that people make.

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So the derivative of x to the pi
is nice and simple because

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that is our rule we know
for powers of x.

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So we can write this as that
derivative is, pi times x to

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the pi minus one.

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OK, but the whole point of this
problem for me, is to

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make sure that you recognize
that pi to the x is not a

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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 to

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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 that we have for

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exponential, 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, when you have natural log

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of a function, its derivative
is going to be 1 over the

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inside function times
then the derivative

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of the inside function.

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So again, what we do
is we take the

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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

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function, which 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 this one is

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that we see, "Oh, if I wanted
to find a function whose

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derivative was tangent x,
a candidate would be the

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negative of the natural log of
cosine of x." That in fact

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gives us a function whose
derivative is tangent x.

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So it's interesting, now
we see that there are

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trigonometric functions that
I can take a derivative of

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something that's not just
trigonometric and get

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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, I'll

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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.
And then I'll show you

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why it's good to kind of pull
back from a problem sometimes,

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see how you can make it a lot
simpler for yourself, and then

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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 have a composition of

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functions here.

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I have natural log of something
and then I have e to

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the something else.

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Right?

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And then that function
actually, is not

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just e to the x.

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So I have some things I
have to, I have to use

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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, the

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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, I

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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 of the next inside

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function, which the next one
inside after natural log, is e

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to the x squared.

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And the derivative of that,
I'm going to do

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another chain rule.

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I get e to the x squared times
the derivative of this x

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squared, which is 2x.

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OK, so again, this part is the
derivative of natural log

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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 to the

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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

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the x squared part.

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So I end up with a product of
three functions, because I

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have a composition of
three functions.

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So I have to do the chain
rule with three

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different pieces 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 that the natural log

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of e to the x squared-- let me
erase the y here-- e to the x

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squared 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 is the inverse of the

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exponential function
with base 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 to anything

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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, if you

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don't like inverse functions,
you can use one of the rules

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of logarithms, which 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

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e is equal to 1.

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So at some point you have to
know a little bit about logs

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and exponentials.

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But the thing to recognize is,
that h of x is just a fancy

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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 can be done to make

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the problem a little easier.

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But that is where we will
stop with these.

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