BAM!!! Mr. Tarrou. In this lesson we are going
to be introducing and going through lots of
examples of finding derivatives of functions
involving the natural exponential function.
We are going to be talking about the definition.
We will be looking at properties of the Natural
Exponential function. I will do an example
of solving an equation for x that requires
logarithms to do so. That is just a review
of hopefully what you learned in Algebra 2
or PreCalculus. And we will be doing a sketch
of a natural exponential function. Just a
sketch because I want to come about it by
using an reviewing our knowledge about transformations
of functions. Then we will move into five
examples of finding derivatives, three of
which will go in increasing difficulty...
hopefully they all do, and then we will do
an example that involves the Second Fundamental
Theorem of Calculus. The final one will require
some implicit differentiation to review some
skills that we have learned this year. The
Natural Exponential function, the inverse
of the Natural Logarithmic Function f of x
is equal to the natural log of x or log base
e of x is called the natural exponential function.
So if f(x) is equal to ln(x), the inverse
of f(x) is equal to e to the x. They are inverse
functions. How else can we say that? Well
y=e^x if and only if x is equal to the natural
log of y, basically just taking this equation
and going through the steps of solving it
for x instead of y. Another way of looking
at would be to check whether they are inverse
functions by using the definition of the inverse
function. You know, can you do composition
of these two functions f(g(x)) and g(f(x))
and both of those comes out to be equal to
x. Well if so, then f and g are inverses.
So we have the natural log of e to the x is
equal to x. I just wrote that, and it is true,
but if you want to see why it is equal to
just simply x you can take this equation that
is in logarithmic form and convert it into
exponential form. Now remember that when you
take the log of a number you are just getting
an exponent. So I always say in class the
log base 10 of 100 is equal to 2 because ten
squared is equal to a hundred. So what you
get from a log function is an exponent. So
if I want to write this in exponential form,
the base of the natural log is x. What we
get from the log function is an exponent,
so it is e to the x, I have an e to the x
here but that is right here. So log base e
of e^x is equal to x. So this x is my exponent
and what I am taking the log of is sort of
the answer in exponential form. We see that
e^x is equal to e^x. And e to the natural
log of x is equal to x. Well, when you have
a base on your exponential form and if that
equals the base of the logarithm and they
are stacked like this. Maybe if there was
a 2 in front of this natural log I would have
to move it up to be an exponent on the x.
But the base e and the log base e are stacked
with no leading coefficient or any other craziness
going on so they just cancel out and it equals
x. Or you can write this equation that is
in exponential form into log form. We get
log base, the base of the exponent is e, and
the answer that is in this exponential form
is x so it is going to be log base e of x...
then again what you get from a log function
is an exponent. So the exponent that is on
the e is the natural log of x and we have
the log base e which hopefully you remember
is the natural log of x is equal to the natural
log of x. So both of these compositions simplify
to x, thus one more time they are inverses.
Of if that is not enough, you can look at
it graphically which will help me to go over
these properties of natural logarithmic...
excuse me... exponential functions. But we
have an exponential function. The base is
greater than one so it is exponential growth.
The base is e which is approximately 2.718.
It goes through the point of (0,1) unless
there are some kind of transformations that
has been made and it goes off to infinity.
As x goes to infinity the function goes to
infinity. If you reflect that over the line
y=x remember inverse functions, we just talked
about this of course, are a reflection over
the line y=x. You get a graph that is the
function y equals the natural log of x. So
they both look graphically, and would check
out graphically, to be inverse functions as
well. So your properties of natural exponential
functions are the domain of f(x)=e^x is negative
infinity to positive infinity. The range is,
excuse me [voice cracked], the range of course
is the y values goes from zero to infinity
unless you have been messing around with transformations.
These are the parent function properties.
The function f(x) is equal to e^x is continuous,
increasing, and 1 to 1 on its entire domain.
Or as we were talking about in the previous
section it would be monotonic increasing.
That means the first derivative would always
be positive through its entire domain. It
is concave up on its entire domain. Concave
means the second derivative would also be
positive. The end behavior as we would have
called it in Algebra 2 or PreCalculus, the
limit as x approaches negative infinity of
e^x... as you go to the left forever it is
approaching the y value of zero. There is
an asymptotic behavior as x goes to negative
infinity, at least again for the natural exponential
function. Then as x approaches positive infinity,
the function increases without bound and goes
to positive infinity. Let.s get to the first
example, knock some dust off of some old algebra
2 skills or precalculus skills. Then we will
get on to those five derivative examples.
WHOOO! Exciting:D nananananana.... In my first
example we are going to solve this for x.
Now it is a review of some old skills of solving
natural log functions... or solving equations
requiring logarithmic functions. So I am going
to disappear and let you solve this and reveal
the answer step by step. First thing we are
going to want to do is multiply both sides
of the equation by 20 minus e to the x/2 power
to get that x out of the denominator. Then
eventually logarithms will have to be used
to get the variable out of the exponent. Let's
see what the solution looks like. BAM!!! x
is equal to 2 times the natural log of 12.
Now at this point by the way I could have
applied the natural log function to both sides
of the equation. I could have used any base
log. I could do the common log... the log
base of 10, log base 2, log base 5 if for
some reason I wanted to. But I am applying
the natural log to both sides of course because
natural log is log base e and this exponent
has a base of e. We know now that is very
nicely cancel out and just save me a step
in my work before I get my answer that is
in exact form. Of course if you want to check
it, plug it back into your original function
and make sure that it makes the left side
equal to 50. Let's get on to sketching based
on our knowledge of transformations of functions.
For our sketch, the function that we are going
to sketch is y equals negative e to the negative
x plus 2 plus 2. Try to include all of the
transformations that I could in this one equation
just to get, you know, the one to take care
of everything or as much as I can. Now before
talking about transformations, I like to...
because there are two terms up here in the
exponent, I like to take this negative coefficient
away from the x in this exponent and write
it like this -e^-(x-2) + 2. Ok. So I am going
to look at this, the function written this
way to identify what transformations have
been done here that we need to worry about
as we make our graph. Ok. Well the first thing
is when you have a negative leading coefficient
that is going to reflect the graph over the
x axis. So this is going to be a reflection
over the x axis. Just so you don't have to
watch me write. That is a little faster, right.
So negative leading coefficient is a reflection
over the x axis. If there was a value here
that was... let's say like 2. That would be
a vertical stretch of 2. If there was a value
that was between -1 and 1, or I guess if you
want to look at the negative separately, a
coefficient that was between 0 and 1 that
would be a vertical compression. Here in front
of the x that was in our exponent which I
have factored out and wrote negative parenthesis
x minus 2 parenthesis, like to do that because
it is easier to determine does the graph move
to the left or to the right. At least that
works better for me and my teaching. So this
negative coefficient of x in the exponent...
now that is going to be reflection over the
y axis. Now with this negative factored out
and the coefficient of x being positive, I
find it easier to teach and understand what
direction the graph moves. I have a minus
2 that is in my parenthesis. When I put in
a value of x, the very first thing before
I ever let it enter the actual... become an
exponent if you will of e to the something,
it is going to be subtracted by two. So this
minus two with our positive leading coefficient
and the negative factored out in front of
the parenthesis shows that our graph is going
to move to the right two places. So if this
were x+2 with the negative of course factored
out, now there is a plus two here but with
the negative right there in front of the x
you might think that the graph shifts to the
left which.... depending on how you work the
order of operations and how you deal with
this reflection you might still get this right.
I just think this is simpler. Reflection over
the y axis and now it is going to move to
the right. Then finally with our plus two
outside the main function that is in our function,
the e to the something with our variable up
here in the exponent, this plus two is just
a constant at the end of the equation. It
could be at the beginning too. It is just
all about this constant alone outside of or
away from the main function in the equation,
that is going to be a vertical shift. It does
not move in the opposite direction like horizontal
movement. Plus 2 is a shift up. It is not
going to be a shift down like you may think
like with a horizontal movements that always
seem to be in the reverse of what the sign
indicates. This graph is going to up two.
I am going to step out, write that statement,
and then put an x y axis in here and draw
my final sketch by going through each of these
single transformations as we go until I am
done with the final representation or sketch
of this function... As soon as I get the video
stopped. So we have the original parent function
going through the point of (0,1). And I ordered
these as I applied them. You can apply them
in slightly different order. I like to apply
the reflections first, then the horizontal,
and then the vertical movement obviously as
I shown. So we have the reflection over the
x axis. So this graph reflects down and we
get the green one. Then we apply the negative
leading coefficient that we have in the exponent
which is our reflection over the y axis. I
take this green function and reflect it still
having it pass through that point of (0,-1)
instead of what was originally (0,1) because
of the reflection down over the x axis. We
took that function and slid it to the right
two places. I just tried to take these points
as best I could with this sketch and moved
them all over two units. Then finally applied
that last vertical shift of two units up.
So what was a horizontal asymptote of y=0,
that asymptotic behavior of the original parent
function as x approaches negative infinity
the function approaches zero. There is now
a horizontal asymptote, because of the vertical
shift, of y=2. So the range of our final function
is going from negative infinity.... excuse
me... up to two. And though this is a sketch
I tried to make it neat, clear, and easy to
read, I did really want to show this one key
point was plotted correctly. It was going
through the point of (0,1), then with the
reflections both about the x and y axis, it
is still going through that point (0,-1).
We moved it to the right two, now it is going
through (2,-1). Then we shifted it up two
units and it is going through the point (2,1)
where it was going through (2,-1). Ok, we
are going to give you the formula, not prove
it, but give you the formula if you will...
or the rule for finding the derivative of
the Natural Exponential Function and go through
the five examples involving derivatives. So
here we have probably the hardest rule for
finding derivatives in all of Calculus. The
derivative with respect to x of e to the x
is e to the x. It is its own derivative. So
that is going to be... You can't get any easier
than that. There should be a proof of this
in your textbook. We are just going to five
examples and I am just giving you the rule.
And of course our problems are not going to
be looking like this because there is nothing
to do. So with our u substitution and chain
rule we have the derivative with respect to
x of e^u is equal to e^u times du over dx,
or the derivative of u with respect to x or
just u prime. Of course u has to be differentiable
function of x. So over here we have as our
first example we have y is equal to e to the
square root of 2x power plus one. Well this
equation does not exactly match this rule.
I mean as we have practice we can skip some
of these steps, as doing our first example
we are going to do that u substitution. We
have u which looks like it is going to be
our exponent, is equal to the square root
of 2x. Write that in exponential form and
then the derivative of u with respect to x
is going to be, we will bring this power down
and reduce it by one, then multiply by the
derivative of the inside function. Of course
we have some cancellation that can go on here.
This 2x, this 2 is again underneath this power
of negative one-half so this 2 and this division
of 2 cannot cancel, but those two do. And
I am just going to write my final answer,
drop this down into the denominator to take
care of my negative exponent. I am going to
go ahead and put the radical back around the
2x. Well I have a lot of work over here. Let's
get this written in terms of u and finish
our u substitution. So y is equal to e^u+1.
Y prime, or dy/dx... derivative of y with
respect to x, is going to be equal to e to
the u power times du/dx. That is a constant
so it is going to come out to be zero. I am
just applying the rule as I have written here.
We are going to get the U substitution back
out and return our references of x. We have
y prime is e to the square root of 2x times
one over the square root of 2x. This gives
us if you want to just write it as a single
fraction e to the square root of 2x over the
square root of 2x. Maybe your teacher is perfectly
fine with this. Maybe they want you to rationalize
the denominator. Write your final answer as
your teacher is requiring you to do so. That
is the end of our first example. BAM!!! For
our second example we have y=3x^3 times e^-x.
We are going to find the derivative. We are
going to do y prime and of course use the
Product Rule, so we have 3x^2 times the derivative
with respect to x... the first factor times
the derivative of the second factor... If
you know the derivative of course just write
it. I am showing this in very small steps
so I am going to write this and do the u substitution
in a second and finish it. So factor times
the derivative of the second plus the second
factor times the derivative of the first.
Ok. And since we are just starting this out
let's do a little u substitution. We are going
to let u equal -x. That means that du/dx is
going to be equal to -1. And we know that...
Let's finish the u substitution. We have y
prime is equal to 3x^2 times the derivative
with respect to x of e^u plus e^-x times the
derivative with respect to x of 3x^2. We have...
well let's see here. 3x^2 and the derivative
with respect to x of e to the u power is e^u
times du/dx plus e^-x... and this is going
to be 6x. Getting the references back out
of here to get it all in terms of x again
of course. We have 3x^2, e to the... well
u is negative x, du/dx is negative one, and
I can just write this as plus 6xe^-x. Now
we can see that this negative one can brought
out front. Our final is going to be -1 times
3 x squared times e to the negative x plus
6x times e to the negative x. This first term
has an x squared and this one has x to the
first. These are not like terms. We can write
this like this. We can notice that just simplifying
this as much as our teacher needs, we have
a factor of 3 and a factor of x so we can
factor that out. We have a factor of x and
this has two x's and both of these terms have
e to the negative x, so we could write this
as 3xe^-x times... -3 divided by 3 is negative
one... we have two x's and we took one of
them out so basically negative x or negative
one x... and 6 divided by 3 is equal to 2.
We took the x and the e to the negative x
out. Finally we can write this, bring this
e to the denominator, 3x/e^x times 2 - x.
These questions, sometimes you get a derivative
of course and you have done everything correctly
and you check it in the back of the book and
maybe it is in a format that you don't have
it written in. So you can have a correct answer
and have done the correct work, and not match
what is in the back. You have a couple of
choices. You can either keep manipulating
the answer to match what is in the back of
your textbook to check your work, you can
maybe stop at a point that you think is as
far as you can go or just enough and you can
graph that function and then graph the answer
in the back of the book and make sure that
they match as well to check your work. So
that is the end of our second example. As
I was trying to not make any mistakes with
my individual steps, you really should be
making sure that your notation is continuous
all the way through your problem. Maybe I
have to take the second derivative and then
somewhere you forget where your work was for
the first derivative and the second derivative,
or what not. So just wanted to add that y
prime back in. Now for the second example...
or excuse me... the third example, Well I
guess really if you count the first two this
is the fifth example. We are going to have
to use the Quotient Rule for finding derivatives.
Now when I reveal this answer step by step
I would like to give you the opportunity to
try this on your own, or at least speed the
video up and just reveal the answer in small
steps. I am not going to show that U substitution
and show that extra... I am not going to write
it in the same way. The reason why is because
it is simply always going to be... what...
e to the u times u prime basically... your
du/dx, the derivative of u with respect to
x. You might just start short cutting this.
So like if we have a simple function of y=3e^(4x),
that y' is going to be 3 times... and if that
u is 4x... e to the 4x and then you multiply
by the derivative of that u which is going
to be the exponent and you get 4. So this
is going to be 12e^(4x). So that is how I
am going to show my work as I reveal it for
this particular question here. Pause the video
and give it a shot. Practice your Quotient
Rule. Well, here we have it. Our final answer
is y prime is 6e^(3x) over (e^(3x)+1) squared.
Just a couple of quick notes again like the
example I did. Here we have 2 times e to the
3x. The derivative with respect to x, that
u is 3x and u prime is 3, so 3 that u prime
times 2 is where the 6 came from. As well
here except our coefficient here is one, so
we just have 3e^(3x). Distribute these items
together and of course do not forget when
you multiply like bases, our like bases here
are e, you add the exponents. So 3 plus 3
is equal to these exponents of 6x. We have
our final answer. Our next example is going
to involve the Second Fundamental Theorem
of Calculus. So our next example, that Second
Fundamental Theorem of Calculus problem, F(x)
is equal to the definite integral with a lower
limit of 0 and an upper limit of ln(2x) of
the tangent of e^t dt. Ok, so that means that
F prime... we are looking for derivatives
of course... is the derivative with respect
to x of this definite integral. And we start
to see with the lower limit being a constant
and the upper limit being a variable that
we are starting to see the application of
where the Second Fundamental Theorem of Calculus
is coming in. If that was simply an x, then
the answer would be the tan(e^x). But is not,
it is the natural log of 2x. We are going
to need to do a u substitution, and that also
means incorporating the Chain Rule with the
Second Fundamental Theorem of Calculus. What
that is going to look like is F'(x) is going
to be equal to the derivative with respect
to u because we are going to do a u substitution
of our original function F(x) and then we
are going to have to multiply by the derivative
of u with respect to x. My mind was saying
u again or u prime. Ok, so we need to use
substitution and that is going to be about
this upper limit. We are going to say let
u equal the natural log of 2x. So we need
u prime, or the derivative of u with respect
to x, and doing a chain rule here and a u
substitution in our head... the derivative
of the natural log of x is 1/x, but it is
not simply x it is 2x. So we have 1/2x and
then multiplying by the derivative, doing
a u substitution in my head or you can think
of it as chain rule, the derivative of 2x
is simply equal to 2. That simplifies to be
1/x. So the derivative of u with respect to
x is 1 over x. Well, let's come back in here
and fill this out. We have the derivative...
or F prime of x is the derivative with respect
to u of this definite integral starting from
here. I will do my u substitution in a second.
Ok, let's do the U substitution. Let's get
that upper limit out of there and plug in
u. Just getting all that notation in there.
We have our upper limit now of just a basic
u like it was with an x with the original
form of the Second Fundament Theorem of Calculus.
So we are going to have F'(x) is equal to
the tangent of e to the u, again du/dx...
the derivative of u with respect to x. Let's
get the references of u out of here and get
back into our variables of x. We have F'(x)
is equal to, the derivative of u with respect
to x is 1/x, now times the tangent of e to
the natural log of 2x. Now you might want
to go BOOM! I am done. It says F prime and
I have got all of my references of x back
in here. But we have the tangent of e to the
natural log of 2x. So you want to look for
those cancellations. We have a base of e and
we have a log base e. And we have a coefficient
of one, so basically that e and log base e
are stacked... or that natural log are just
stacked against each other. That means that
they can cancel. This ends up being, besides
a piece of chalk hitting the ground, equal
to 1/x times the tangent... and with this
base e and this log base e cancelling, the
tangent of 2x. The tan(2x) over x if you want
to write it that way. Alright. I have one
more example involving implicit differentiation.
We are going to use implicit differentiation
to find the derivative of e to the xy power
plus 3 x to the fourth plus y squared is equal
to five. It would be much easier to do this
implicitly than trying to solve this for y
somehow and then take the derivative explicitly.
So we have e to the x times y. So using that
Chain Rule and U substitution in our head
if you will, we have the derivative of e^(xy)
is going to be e^(xy) times the derivative
of that U... that exponent which is xy...
Now plus... bring that power down and we have
12x^3... plus bringing this power down and
we have 2y. But again we are doing this implicitly
and we are taking the derivative with respect
to x. That is why we simply have a 12x^3.
This is going to be 2y and then it is going
to be times y prime. I am applying the Chain
Rule and we are getting dy/dx, or the derivative
of y with respect to x, or just writing that
as y'. The derivative of a constant is equal
to zero. Alright. Over here we have to apply
the differentiation rule for Products, the
Product Rule. We have e^(xy) times the first
times the derivative of the second factor,
so xy', plus the second factor y times the
derivative of the first factor. Now plus 12x^3
plus 2y times y prime or dy/dx is equal to
0. So basically I just need to get this solved
for y prime or dy/dx. We are going to distribute.
Actually just to speed this process, I don't
know if you need to see me do the Algebraic
manipulation. This is not the first time we
have done implicit differentiation. So let's
speed this up a little bit and I will reveal
the answer step by step. You can pause the
video and try to solve for y' on your own
first. Well let's see... We moved the terms
over to the right hand side that did not have
the y' in them. These two terms both have
a y' so I factored it out... divided both
sides by this big ol mess. We have y', or
the derivative of y with respect to x, is
equal to -12x^3-ye^(xy) over xe^(xy)+2y. I
am Mr. Tarrou. BAM! Go Do Your Homework:)
