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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.