BAM!!! Mr. Tarrou. In this video we are going
to start looking at finding the derivatives
with functions dealing with the natural log
function. And like all of my videos there
is going to be lots and lots of examples so
hopefully this one video will get you through
pretty much your entire assignment. The natural
log function differentiation. Well here we
have the power rule for integrating. And there
is a type of function that we actually have
not done indefinite integration with yet.
And we are not going to in the video. But
over the next couple of sections in our book
we are going to see the relationship which
is going to allow us to do this. The indefinite
integral of x to the n dx is equal to x to
the n+1 power divided by n+1 plus c. But the
condition for using this was that n could
not equal negative one. Well if you try to
find the indefinite integral of 1/x, that
actually ends up happening. You have the indefinite
integral of x to the negative one. You raise
that by one, you get x to the zero power divided
by -1+1 which is x to the zero power divided
by zero. This is undefined. So we are going
to see how we are going to be able to find
this indefinite integral over the next couple
of sections, but right now we are going to
just deal with definite integral and this
relationship between the natural log function
and the function of basically y equals one
over x. So our definition of the natural log
function is the natural log of x is equal
to the definite integral from 1 to x of 1/t
dt. Now again, I said the second theorem of
calculus... or I think I did... over here
is coming up. So we are going to see how this
variable changes, unless you remember of course
from the Second Fundamental Theorem of Calculus,
but anyway the definite integral from 1 to
x of 1/t dt where x is greater than zero.
What it says, or it shows us here is that
we have... at least as far as the domain of
x greater than 0, we have the function of
y=1/t. And the natural log of x is equal to
the area bound between that function of y=1/t
and the lower bound of 1 and the upper bound
of x. So the area between the x axis and our
function between 1 and upper limit of x is
going to be... that area is going to be the
natural log of x. So beyond the value of 1
that represents... that shows with this definition
that the natural log of x is going to be positive.
If we find the definite integral at 1, where
the lower and upper limit are equal to one,
then... right here... then our natural log
of x is going to take on the value of zero.
And then if we are to the left of 1, we still
have an area. But because our lower bound
is greater than our upper bound, and we talked
about how you could switch those lower and
upper bounds and create an opposite answer,
it shows that the natural log function is
going to be negative when x is between zero
and one. Excellent. So over here we have the
definition, or lets talk about e because we
are talking about natural log. And of course
natural log is a log base e. And if you are
not really familiar with, or comfortable with
logarithms, remember they just give you an
exponent. Like the log base 10 of 100 is equal
to 2, because ten squared is 100. So exponential
functions and log functions are basically
inverses of each other. Hopefully you learned
that in Algebra 2 or PreCalculus. I know that
my students did. So the Natural Log is just
log base e. So when you take the natural log
of a number, what you are getting is an exponent
such that e to that exponent is equal to the
value of which you are taking the natural
log of. So the letter 'e' which is approximately
2.718. There is a lot of derivations of where
it comes from. It is a constant. It is kind
of like pi, but that is a lot easier to understand
and explain. Just pi is the ratio between
the circumference of the circle and the diameter.
But one way of defining e is the letter 'e'
denotes the positive real number such that
the natural log of e is equal to the definite
integral from 1 to e of 1/t dt... and that
is equal to 1. So basically if you start at
your lower bound of 1, you go out to your
value of e which is 2.718, the area bound
between the x axis and the function y=1/t
is equal to 1. Alright! I started to go off
on a bit of a tangent. I want to take care
of this definition of the natural log function
first. We have, let's see here, the natural
log of x again equal to this definite integral
of 1/t dt. So if I want to find the derivative
with respect to x of the natural log of x,
we have the derivative with respect to x...
of course again from.... the definite integral
from 1 to x of 1/t dt. Now that is equal to...
Is this looking familiar? It is the Second
Fundamental Theorem of Calculus. We have a
lower bound which is, or limit, which is...
bound.. which is a constant and our upper
bound is a variable. I can take that variable
and simply drop it into our function and rewrite
it as 1 over x. So the derivative with respect
to x of the natural log of x is 1/x. So if
you wanted to, you know, graph the function
y equals the natural log of x this function
of y=1/x is going to give you the slope of
the natural log function at a particular point.
Pretty cool:) Now, if we.... We are also going
to talk about in the next section how the
natural log function is anti derivative of
1/x, getting to this original problem which
I kind of pointed out that we have not dealt
with yet. But we are not going to be dealing
with integration in this video. Not quite
straight forward because these have different
domains. This has a domain of all real numbers
except for x cannot equal zero and this has
a domain where x has to be greater than zero.
You cannot take the natural log of a negative
number, or any log of a negative number. So
any way. Getting the chain rule in here, we
get the, you know if you don't have just a
basic x here and you want to do a u substitution
to make the function look simpler so it is
more easily... you can more easily take the
derivative of it. We have the derivative with
respect to x of the natural log of u. So we
are talking about a u substitution so I will
have a u here. That is equal to, instead of
1/x, 1/u but since we have done a u substitution
then we have to multiply by the derivative
of u with respect to x or basically get u
prime over u. This is really really powerful.
You are going to have some really nasty...
with the condition that u is greater than
zero... You are going to have some pretty
nasty looking problems that you know... this
is going to allow them to be pretty straight
forward when finding the derivative of functions
with the natural log in them. Now of course,
it is Calculus. We are always get some problems
get quite complicated dealing with you know...
power rule... quotient rule... have some radicals
that we have to simplify... trig functions.
But it really does make finding these derivatives
with natural log functions at least a lot
easier than they could be. Relatively easy,
we will call it that. Before we get to some
examples I want to remind you of some properties
of logarithms and such so that you can work
out these examples. So one more page of notes
before our first example. Properties of Natural
Log Functions. You domain for the natural
log function is going to be from zero to infinity,
and don't forget that round parenthesis means
that zero is not part in the domain, it is
not inclusive. The range goes from negative
infinity to positive infinity. So your natural
log function is going to come up against,
just the right side of the y axis pass through
the point of (1,0) and then go off to the
right. It is going to be continuous. It is
going to always be increasing, the parent
function. And it is going to be one to one
which means that function would pass not only
the vertical line test to validate that it
is a function, but a horizontal line test
as well. That means that if you were finding
the inverse function it would also be one
to one at least as far as that original domain
goes as being from zero to infinity. And the
graph is concave downward. I am going to go
through 8 examples. I am not going to work
through every single one of them in very small
individual steps because you will recognize
a lot of the skills and the steps that we
are doing which are just basic finding derivatives
and using the product rule, quotient rule,
and such like that... general power rule.
But you are going to see me use these properties
of logarithms that hopefully you look at this
and recall them from either algebra 2 or Precalculus.
I going to do so many examples. I am not going
to do work with just these properties purely
as if they were in algebra 2 or precalculus.
You will see them as we do these derivative
problems. But the natural log of 1 is equal
to zero. The natural log of ab, so you are
taking these two items and multiplying them
together inside the log function. Well that
is equal to the natural log of a plus the
natural log of b. Again, just like I said
a few minutes ago, when you log a value or
natural log a value, or any base for that
matter, you get an exponent...right? Like
again the log base 10 of a 100 is 2 because
10 squared is equal to 100. So we are dealing
with basically exponents here. So when you
multiply with like bases, that like base the
e which is base of the natural log, you add
those exponents. And of course we have the
power rule. The natural log of a to the n
is n times the natural log of a. We can take
that exponent that is again inside the math
function of the natural log and move it down
out front as a coefficient much like the power
to power rule is for your exponents. Then
when you have the natural log of a divided
by b, well just like when you divide with
like bases you subtract the exponents. We
have... I am loosing my balance there. We
have the natural log of a minus the natural
log of b. Ok, one last sort of rule if you
will for doing derivatives which I probably
should have included in the previous page
is the derivative involving absolute value.
If u is a differentiable function of x, again
we are talking about that u substitution that
we will be needing with most of our examples...
actually basically all of them, such that
u is not equal to zero then the derivative
of the natural log of the absolute value of
u... and we get this somewhat often because
again your domain for the natural log function
is only from zero to infinity so you have
to make sure sometimes that what you are taking
the natural log of is positive because if
it is negative it would be undefined or not
in the domain of the natural log function.
That ends up just being u prime over u. All
right. We have lots of examples. Let's get
to those right now! Find the derivative of
the function. So we are going to do eight
examples and start off with of course easy
and go to harder examples as we go through.
We have f(x) is equal to the natural log of
5x. Well we don't have just a basic little
x in here, we have 5x. It is a polynomial
so it will not be too tough. But let's go
ahead and just practice that u substitution.
We are going to say that we are going to let
u equal to... set it equal to 5x, that inside
function. If that is the case, then u prime...
Well the derivative of 5x with respect to
x is equal to just 5. So if f prime of x,
or the derivative of f(x) is with this u substitution
is equal to u'/u then we just simply have
5 over 5x. And 5 divided by 5 is equal to
1. We just have f'(x) is equal to 1/x. So
here we have another example. It looks a little
bit more complicated. But 5x is a polynomial,
we have 4x^2+7 which is another polynomial,
so I am going to step out and let you practice
this u substitution on your own. So not only
was that very straight forward, but in the
end we did not even have any algebraic simplification
that we could do. So really this was even
less work than this one. So let's get on to
some examples that are a little bit more involved.
Ok, so we have over here the natural log of
2x minus 1 cubed. Pay attention to where this
parenthesis is because if it was the ln(2x-1)
to the third power then we would be able to
use some of those properties of logarithms
that I talked about. Like we could take this
three and move it out front and write it as
3 times the ln(2x-1) which would maybe save
us from having to the chain rule. But we don't
have that option, and actually we are going
to be using the chain rule to find the derivative
of this problem. Now over here I have a lot
more going on inside the natural log function.
I will be able to use some of these properties,
and I will need to use some of those properties...I
don't need to. I could make things more difficult
on myself... but separating what is inside
there to finish that problem. Use some of
the properties to expand that natural log
function before working through it. But here,
we are going to take the 3 and use the chain
rule. It is going to be bring down that power.
Move it out front. We have 3 times the natural
log of 2x-1 squared. You know, we are going
to bring that power down and reduce it by
one. But this is not just x cubed, it is a
lot of something cubed. So we have to follow
through with the chain rule which means now
we have to multiply by the derivative with
respect to x of the ln(2x-1). CHALK DOWN!
Ok. So now we get into a situation where we
look very similar to where we were before.
We need to work out this natural log, the
derivative of this natural log of 2x-1. So
u is equal to 2x-1. u prime is therefore going
to be equal to the derivative with respect
to x of 2x-1. That is just going to be 2.
So we have 3(ln(2x-1))^2 times the derivative
of this is going to be u'/u. This is going
to be 2 over 2x-1. Ok, well there is not much
left that is going to be able to be done except
maybe just put this over one so we can see
the numerator and the denominator of these
two values or fractions that are getting multiplied
together. We have 2 times 3 which is equal
to 6 times the natural log of 2x-1 squared
over just 2x-1. Now don't be tempted to think
that you can cancel out this 2x-1 because
this 2x-1 in the numerator is inside the math
function of the natural log. So that is it
unless I have made a mistake. Just checking,
no. Nothing else can be done to simplify this
and this is equal to...oops... I guess I got
to talking and got a little bit sloppy with
my notes. Starting here is where I started
working out the derivative, I am off the camera,
that is f prime of x all the way done until
we get our final answer. Yes, sometimes when
I am talking, and thinking, and writing at
the same time I sort of make small mistakes
and I get... I try not to get sloppy or careless,
but these are going to get longer and longer
as we go through these examples and you do
need to be very meticulous with your work.
Showing every step is going to reduce the
amount of small mistakes or careless mistakes
that you make. And me forgetting to write
that f'(x), that derivative of f on the line
where I started doing the derivative... working
it out... I could have done many many lines
of work on up, lost track of where I was and
put the wrong notation somewhere and have
the wrong answer. Ok, for our next example
we have f(x) is equal to ln(x^2 times the
square root of x^3-2). Now inside this math
function, this natural log function we have
two factors. Not forgetting of course that
radicals are a grouping symbol. So we have
two factors being multiplied together inside
our natural log so like our property page
said the ln(ab) is equal to ln(a) plus ln(b).
So we are going to expand this out. So we
have f(x) is equal to the natural log of x
squared plus the natural log.... and I don't
like working with radicals inside my math
work so we are going to wrap this in parenthesis
and write x^3-2, and we have no parenthesis
or powers inside here so that is going to
be raised to a power of one and we have an
index of 2... so this is going to be raised
to the 1/2 power. Ok. This particular term,
just to point this out it is not really that
important for this particular question, but
sometimes when you use the power property
to take these exponents and bring them out
you are changing the domain of that function.
The natural log of x squared, I can plug in
any value for x that I want except for zero.
That is because even if I plug in a negative,
remember the domain for the natural log function
is zero to infinity, if I plug in a negative
value the first thing is going to happen is
that is going to be squared which makes it
positive. That would mean I could apply the
natural log function. If I bring the 2 out
front and write it as 2ln(x), that natural
log of x now has a domain of just greater
than zero to infinity...not just all real
numbers except x cannot be equal to zero.
So you do need to be aware that as you drop
these exponents down out front, you could
have some issues with domain like if you are
graphing or something like that. So just be
aware that as you do your work, there is probably
well over 100 problems in your Calculus book
and you know... they throw some twists at
you sometimes. So I am going to leave that
power of 2 up there and I am going to bring
this 1/2 down out front just to make finding
the derivative process a little bit simpler.
We have f(x) is equal to the natural log of
x squared plus 1/2 times the natural log of
x^3-2. Ok. So now what? Well we are going
to just simply have two different u substitutions.
We have an inside function of x squared and
we have an inside function of x^3-2. So we
have u is equal to x^2, and u' is equal to
2x. We have this other inside function. Let's
get a different color chalk here. This is
equal to x cubed minus two, so w is x^3-2
which means w prime is equal to 3x^2. And
I realized I was not doing this with some
of my simpler questions as we are getting
up to some of the harder ones, but if I just
wanted to write function f temporarily with
this temporary u substitution we could say
that this is now the ln(u) plus 1/2ln(w).
I realized I was not actually writing that
line where I put the substitution in and then
reminded you that the derivative now that
we start to work out f prime, or the derivative
of f, the derivative of the natural of u is
u'/u. I just wanted to show that super detailed
substitution or work one more time. Excuse
me. The derivative of the natural log of w
is going to be w prime over w. I am going
to go back to just sort of writing like I
did in this example, where I just verbally
said it and wrote it. So u' is 2x and u is
x^2. w' is 3x^2 and w again is x^3-2. Get
those terms of x back in there. At this point
all we have to do is clean things up. So just
to take a couple of seconds of the video I
am going to step off and algebraically clean
this up and see what our final answer is going
to look like. There we go. What did we do?
We did a little cancellation. We found some
common denominators. We distributed and combined
like terms. We have f'(x) is equal to 7x^3-8
over 2x^4-4x. Next examples! Now with these
two examples we are just continuing the u
substitution, so I am not going to talk through
every single step. These I am going to just
barely start this one for you and then step
out and reveal the solution step by step.
So you can either watch them be revealed or
better yet pause the video and try them on
your own to make sure that you are getting
this. Over here it seems little funny that
we have the natural log of the natural log
of x cubed. But that just simply means...
well that our inside function is going to
be the natural log of x^3. And hopefully at
this point in the video you can just look
at this maybe now with the pattern and identify
what u prime is. If you can't, if you still
need that u substitution, you can just focus
on this as your own separate little problem
and say maybe... well... let my inside function
w be x cubed and then w prime maybe is 3x^2.
Then just work it in really really small incremental
steps that way. Over here we just have another
inside function in the natural log. It is
x plus the square root of x^2+4 this time.
So we are not going to be able to... We are
not going to be able to separate or expand
this natural log function because in the property
for expanding a log, or a natural log, these
have to be factors and not two separate terms.
So you are going to have to deal with all
of this as your inside function. So let's
see what those solutions look like on step
at a time. So, with this example over here
we had a u value that when we found u' we
had to of course use the chain rule. And we
had a one-half from the power moved out front,
reduced that power by 1. This gives us 1 plus
1/2(x^2+4)^-1/2 times 2x the derivative of
the inside function... the derivative of x
squared plus four. The 1/2 and the 2 cancel
out. We need to drop that negative one-half
into the denominator, or we can, and write
it now back into a radical form because at
this point.... besides finding common denominators
and such we are not really going to work with
it much. This one over one, we need to find
a common denominator of course. I multiplied
the top and bottom by the square root of x^2+4.
We have it simplified into a single term.
That way when we write it as u prime over
u, when you have a fraction over a fraction
that of course means a division symbol...
and you can change the division of these two
fractions by multiplying, right here, by the
reciprocal of the second fraction which is
going to be our denominator. When we do that,
at least the way I have it written, we have
the square root plus x and x plus that square
root. Of course addition is commutative so
that does not make any difference. That is
going to come out to be the same answer. So
they cancel each other out to be a value of
one. Rationalize the denominator and badda
bing... badda boom... y prime is right there.
So um... our next two examples, we have one
with an absolute value function and trig functions
mixed in with it, and our last example dealing
with a little bit implicit differentiation.
BAM!!! Last board. We have y is equal to the
natural log of the absolute value of sine
x over cosine of x minus 1. And remember when
we first started doing derivatives... finding
derivatives with the absolute value function
gave a pretty special little case, a set of
challenges that we had to deal with. But like
our notes showed if I want the derivative
of the natural log of the absolute value of
u, it is just going to be u' over u. So it
is going to work out just the same as all
of these other problems. So we are going to
expand this natural logarithm first. Separate
this division and make this a little bit easier.
So we have y is equal to the natural log of
the absolute value of sine of x minus the
natural log of the absolute value of cosine
of x minus one. Well in here our u would be
sine and our u prime, the derivative of sine,
is going to be cosine. So we have y' is...
ok... u'/u and u is sine of x and u'... the
derivative of sine is cosine of x. This minus,
now here our u is cosine of x minus one. Our
u', well the derivative of -1 or constant
is zero, so the derivative of cosine is negative
sine of x. Now we got a negative and then
a negative in our numerator, so subtraction
a negative is addition. So we have y prime
is equal to cosine of x over sine of x plus
sine of x over cosine of x minus 1. If you
would like to write that as a single term,
I don't think it really simplifies much. Yeah,
if you want to write that as a single term
you can find common denominators. But I am
just going to leave it as our two separate
terms. Now over here we have a different,
a slight different direction, just reminding
you how to do implicit differentiation. We
have 2y plus the natural log of x squared
times y is equal to 6. So instead of trying
to isolate y and just doing the regular derivative
process, let's go through this implicitly.
So we are going to find the implicit differentiation
with respect to x. So we have... well let's
see here. Let's make life a little bit easier
and use our properties of logarithms, natural
logarithm... same thing, to separate this
multiplication. We have 2y plus the ln(x^2)
plus ln(y) equals 6. Working through this
now, the derivative of 2y with respect to
x is going to be 2, and normally it would
just be 2 if it was the derivative of 2x it
would be 2, but then again this implicit differentiation
we have 2y' because we are finding the derivative
with respect to x. It would be dy/dx which
we can write as y prime. Over here we have
plus... now u is going to be x^2 and the derivative
of u..... I said 2x but I typed it right:)...
I was thinking of the answer. x squared. The
derivative of x squared with respect to x
is going to be 2x. So we will have some cancellation
going on there. This x in the numerator and
one of these two x's in the denominator will
cancel out. Then over here we have plus the
natural log of y. Ok, well that means that
for this u substitution, or it you could call
it w since we are working in the same equation
if you are actually writing this out now.
Our u substitution, u or w or whatever variable
you want to assign, is y. The numerator is
that u prime which is going to be... well...
the derivative if y with respect to x is dy/dx
or we can just write that as y'. Then over
here the derivative of a constant is equal
to zero. So I just really need to work this
problem, getting it solved for y'. So I am
going to first take the 2/x and move it over
to the other side. We got 2y' plus y'/y is
equal to that 2/x which is positive and we
are going to move it over with subtraction,
so minus 2/x. I can factor the y' out now
but I am going to go ahead and get this y
out of the denominator. I am going to multiply
both sides of the equation by y. That is going
to give me 2yy'. Over here in the second term
the y's are going to cancel out just leaving
me with y'. Then over here we have -2y/x.
Each of these terms have a y' and that is
what I am trying to isolate. I am trying to
solve for y prime. So I am going to factor
that out and we get... I am probably going
to run out of room here. We have y' times
2y+1 is equal to -2y/x. And our last step
is to divide both sides of our equation, we
will just use purple, by 2y+1. When do that
of course this cancels out and our final answer
is going to be y', I know you can see it but,
negative 2y over... Yes I break my chalk fairly
often... x times 2y+1. Just to make sure since
I am talking, writing, and thinking all at
the same time.... YES. I am right. I am Mr.
Tarrou. Now it is time for you to GO DO YOUR
HOMEWORK!
