BAM! Mr. Tarrou. In this lesson we are going
to expand on the knowledge that we already
know about finding derivatives of natural
logarithms. We are going to expand it to find
the derivative of logarithms regardless of
what base we have, base 'a'. Now we are going
to have a new rule for this. And if you memorize
it, then you can do these problems very very
quickly. If you don't have this memorized
then it is pretty close to the derivative
rule we have for natural logarithms. But if
you don't have this new rule memorized, I
am looking at the u substitution one here,
then you can remember what you learned back
in algebra 2 or PreCalculus about changing
the base of the logarithms. So the first one
example I am going to of 3, I am going to
do twice. Once using this new rule and once
where we change the base from base 'a' to
base e, or natural logarithm. And the next
two examples I am going to convert them into
natural logarithms. I am going to change the
base so we can use those skills we have already
learned in the previous section. That way
you can see how to do the problems, again
if you don't have this memorized. So definition
of a logarithmic function with base 'a', if
'a' a positive real number... again not equal
to one... and x is any positive real value,
because we cannot take the log of a value
unless it is positive... then the logarithmic
function to base 'a' is denoted by log base
a of x and is defined by log base 'a' of x
is equal to 1 over the natural log of a times
the natural log of x. Basically what they
are showing here again, and I am going to
show you where this comes from... hopefully
you remember the change of base formula from
Algebra 2 or PreCalculus, is they are kind
of... with this definition showing you that
you can use derivative rule that you have
already learned in your previous section...
our previous lesson. Derivatives of natural
logarithms for base 'a', let 'a' be a positive
real number and not equal to 1 and let u for
the u substitution be a differentiable function
of x. Let's just go to the u substitution
here. The derivative with respect of x of
log base 'a' of u is equal to 1 over ln(a)
times u... times the derivative of u with
respect to x. We can see that 1/u, or that
u prime over u. That should look familiar
from us finding derivatives of natural log
functions. We see the natural log of a here
in the base, so they have with this rule that
you can again memorize, changed that base
from base 'a' to base e. I will show you how
to do, or remind actually. Because it should
not be new. If I have like say y is equal
to log base a of x, now there is a change
base formula if you remember it you are going
to know this answer immediately and you can
see where this comes from, but to .... If
you don't remember that change of base formula
and just to practice our skills for properties
of logarithms, I can changes this base by
going from... I can take this function and
go into exponential form. So I can say...
Well, let's see here. Log... Base... Taking
the logarithm of something and what you get
out of a logarithm is an exponent. Hopefully
this is all unnecessary because of all the
work we have done with logarithms recently.
But just in case, or maybe you are not watching
my whole series of videos. So here we have
an equation in logarithmic form and here it
is in exponential form. Base. Exponent. And
then in exponential form this is actually
the answer, what you are taking the log of.
So we have 'a' to the y is equal to x. Now
if I want to solve this equation for y, just
like again for years now right... you need
logarithms to solve equations when the variable
is the exponent. I want to solve for y. But
I want to get into the natural log function,
so I am going to take the natural log of a^y
is equal to the natural log of x. Bring this
power down out front. Divide both sides by
ln(a). And voila, we have our definition of
a logarithmic function. The base 'a' right
here. So that is where this comes from. It
is just simply changing the base from base
'a' to base e, natural log. Let's get on to
that first example. BAM! [nananananana] Alright.
So here is the change of base formula. This
is what I just showed you how to work through
the steps of in the previous screen. But,
I thought maybe just having the formula would
be nice for your notes if you needed it to
remind you of what it was. I know you have
seen that before. So I am not going to show
all of those steps for changing this base.
We have a log base 5 and I want to go into
the natural log, or a log base e. So we can
see that it is going to be this expression
that you are taking the log of goes into the
numerator, and the old base ended up in the
denominator. We are going to have y is equal
to the natural log of 4x^2+x over then natural
log of 5. If you want that broken apart even
more to see in small steps, we have y is equal
to 1/ln(5) times ln(4x^2+x). And you are going
to start seeing very very quickly again, where
this new rule comes into play. Honestly again,
if you can memorize it then use it. It is
going to be a little bit less work. But I
have not added too many steps in here. I am
already at a point where I can use the rules
and the skills that we practiced when doing
the derivative of natural logs because I have
a natural log right there. My u is going to
be 4x^2+x. And my derivative of u with respect
to x is equal to 8x+1. And if I cover up that
natural log of a that is in the denominator,
that is 1/u times the derivative of u with
respect to x. That is the rule that we just
learned like I said in the previous section.
So our final answer y prime, excuse me, is
equal to 1 over the ln(5) times u prime which
is 8x+1 over u which is equal to 4x^2+x. And
voila. I mean, other than writing this as
one fraction, which maybe I should, we are
done finding our derivative. Now of course
there it is. 1 over.... The derivative of
log base 'a', not natural log, of u is 1/ln(a)
which we got from converting into our log
base e or natural log and there it is in the
denominator... 'a'.. ln(a) in the denominator...
du/dx... or just simply u prime over u. There
is my u prime, the derivative of my inside
function, over u. That is the end of our first
example. The next one I am going to let you
do on your own. I will bring it up here and
just reveal the solution one step at a time
to allow you to practice it as well. BAM!
Voila:) Y prime is equal to 6x over 2ln(3)(3x^2+1)
If you... for students... or if you have a
little bit of trouble with fractions, I like
to let this natural log of 3 come out and
show that 1 over ln(3) and then bring down
that power of 1/2. That way you can more easily
see how the numerators and denominators line
up. And just mixed up the notation a little
bit showing you u prime here instead of du/dx
and instead of y prime, just to get you used
to both of those notations. Our last example
is of course going to be involving logarithms
and we will have to do some expansion with
these before, or we should, before we try
to find the derivative of the original function
given. Let's find the derivative of log base
3 of x squared times the square root of 3x+1
divided by 5. And again, this will be best
I think tackled if you do the expansion of
the logarithm first. So we are going to have
y is equal to log base 3... now the numerator,
we have basically what... We have the product
of... Well we have two factors. We have multiplication.
So that is going to be addition. So we have
log base 3 of x^2 plus log base 3 of the square
root of 3x+1 minus, and then in the denominator
we just have a constant, so it is going to
be minus the log base 3 of 5. Ok. And I am
going to go ahead and erase this right now.
Normally we want to deal with these radicals
and fractional exponents. We are going to
take that square root and of course change
that to a power of 1/2. Ok. Now let's get
these converted into natural logarithms. We
are going to have y is equal to the natural
log of x^2 over ln(3)... plus the ln((3x+1)^1/2,
I am trying not to skip any steps so you guys
can follow along. Some of you maybe saying,
"why don't you move these powers up front
and right now?" I am going to next. I don't
really care about converting the log base
3 of 5 to a natural logarithm because it is
a constant. So when we find a derivative it
is going to be zero. We are going to move
these powers out front. We get y is equal
to 2ln(x)/ln(3)... You can write it like that,
or if you really want to separate the coefficient
away from the variable factor, you can write
that coefficient out front and write that
as 2/ln(3) time ln(x)... that is how I am
going to do the rest of these terms. We are
going to, like before here, this is 1 over
ln(3) and we are going to bring that 1/2 out
front. It is going to be 1/2 times 1/ln(3),
that's our natural log coefficient with the
variable in it. So we are going to have 1/2ln(3)
times ln(3x+1) minus again the log base 3
of 5 which is going to become zero when we
find the derivative. So our derivative is
going to be y', kind of running out of room
here. Well this is just 1/x. That is going
to be... Or the derivative of ln(x) is 1/x.
We have 2/ln(3)*x plus... here with a u substitution,
our u is going to be 3x+1 and the derivative
of this inside function is going to be 3.
So just doing that u'/u a little bit here,
and skipping at least one step. We have 3
in the numerator and in the denominator we
have 2ln(3)*(3x+1)... Minus, and of course
this is zero. Oops, I forgot my derivative
mark. So y prime is equal to 2 over natural
log of 3 times x plus 3 over 2 times natural
log of 3 times 3x+1. And you don't need the
zero there. This is... The derivative is done.
So this can be your final answer, or you can
find common denominators and write this as
one single fraction. But, I know that on the
test... Like again, I have said this in the
previous video, don't over simplify. Simplify
to where your teacher expects you to simplify
to, or just simply do the problem until you
get to a point where that would be acceptable...
say if you are going to take a standardized
test, like maybe the AP Calculus test. If
you make.... If this is correct and then you
make a mistake in the simplification process
trying to clean up your answer, and maybe
they are not expecting to have do that on
the say AP Calculus test, you could lose points
even though this is correct and you did the
Calculus right but you made a small mistake
in your simplification of your answer. So
don't over simplify and possible lose some
points that you would otherwise had. So, I
am Mr. Tarrou. BAM! Go do your homework!
