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