Oh in this video we will take the definition that natural logarithm of x is equal to the
integral from 1 to X of 1 over T dT and we'll prove some logarithm properties based on this
definition so here we go but first of all let me just show you guys for the derivative of natural
log of X it is first and keep in mind we are using this right here for the definition so to
differentiate natural log of X we will just put this right here and as we can see we just have
to differentiate the integral going from 1 to X of 1 over T DT and this is so nice because we can use
the first fundamental theorem of calculus all we have to do is to put the X right here because it
pretty much says derivative and integral pretty much cancelled so we just have 1 over X like that
and the derivative X is just 1 so chandu doesn't really matter this right here see it don't put
on the plus C because this right here it's just the derivative right and now let's take a look
at the properties perhaps the first one to show you is this what if we have natural log of X to
the hearts power as we know from our algebra class we can take the R to the front but don't minus 1
because we're not doing that the rule of D yet but this right here should be equal to R times
Ln of X right how can we do this though well we will actually do it by taking the derivative of
this function and you'll see how we are going to do it here we go of course you can first put the
X to the our power right here but I have already show you that natural log of X has a derivative
of 1 of X right here right so here is the deal let's make some observation observation today if
we have to differentiate let's say I just want to differentiate natural log of X to the ars
power because what this already so what we can do first is well put inside in the denominator
so we have 1 over X to the first power and then by the chandu we have to multiply by the derivative
things I and as we all know the derivative of x to our power is equal to our X to the R minus
one power here you do - the one wait for the power and that's pretty much it and now we just have to
simplify this a little bit as you can see though this is they're saying X to the R times X to the
negative one and we can just cancel all the X to R and that's very nice on the top we will have R
over on the bottom this is extra negative one put it on the bottom it's just X like this now it's
very nice so this is our first observation and the second observation is that if today we are going
to just differentiate our x natural log of X well we can use the properties of derivative a constant
multiple you can just take that to the front so this is the same thing R and you multiply by the
derivative of natural log of X and because we have seen that this right here is just one of X all all
we just get R times 1 of X of course that's the same as R all / X now here is the deal notice this
and that have the same derivatives so that means this in that they are not necessary equal yet we
just know they are you know off by a constant so now we can say ln of x to the our it's equal
to r times Ln of x but we don't know if it's off by what constant so I will just put down plus C
so that's pretty much what we have hmm how can we figure this out oh don't worry just plug in
some number for X and we'll see what I see is okay before we do that we will have to figure
out a value of natural log that we actually know so let me actually do this one right here number 3
let's figure out what natural log of 1 is because I would like to use one right here right so let's
do that one first okay to figure out natural log of 1 using this for the definition we will just
have to compute the integral one two one and we have one over T here and entity and this is so
nice because the integral from one to one this right here is nice the equal to zero so that's
very very nice so now we can come back here and we'll just say that X equal to one pick an easy
for you and just go ahead plug die so we get the left hand side natural log of X which is one now
and then to the ars power and this is equal to our times natural log of one and this right here we
have plus C we're way out one to the Arts power is just one so here we have Ln of one and this right
here is our one of course plus C now do we know what Ln of one is we do zero likewise this would
be zero so as you can see this is zero is equal to R times 0 is 0 plus C so of course we can say
that C is equal to 0 therefore we do have the nice property that this is indeed equal to that because
this you go through anyway this right here it's equal to R times natural log of X very good and
that's the second property and in fact we have to store one right here as well very good right
ok now let's see what if we have the famous ones okay well if we have two things multiplying in
South natural log and let me just call them to be X&Y hmm what's natural log of x times y as we all
know from our algebra class this right here it's equal to Ln X plus y but we are prove it this time
using this for the definition though so here we go put x times y in here so we have the integral
from 1 to X Y and we have 1 over T DT like this hmm how can we handle this next oh this is the
integral going from 1 to X Y the do is that we are going to make a cut somewhere and a nice cut
is we will just go from one to X first and we have the one over here this is so nice because
this is exactly Ln of X but don't forget the other piece we have to pick it up from where we left off
which is going from X and we have to finish it so this is X Y and we still have the one over T
DT right here so that's pretty much the idea this is good it's just Ln of X this right here we don't
know if it's the end of something yet because we don't have this enough whoa it's okay just do some
substitution this is how we're going to do it let u equal to something ideally we want to see a 1
for the limit of integration right one somewhere okay this right here is T goes from X 2 T is equal
to X Y well I want to have one somewhere so I have more to have the X and let me use another variable
let me just say you like that this is actually very nice okay look at this keeping my T and you
are our variables X is just like a number right so I'm with the friendship both sides we get DT
equals x times T you that's very nice and we'll take this from the T world to the new world check
this out when T goes from X to X Y let's see this right here will give us the integral in terms of U
when T is equal to X put it here we have X being equal to x times u that means egg that means you
have to be 1 that means U has to be 1 so you have to go from one very nice and if he goes to x times
y we put here that has to be Xu and you will have to be what you will have to be why isn't it yes so
you will have to go to Y like that very nice and now check this out we have 1 over T and he is Xu
so here we have 1 over X you DT is x tu this is so nice because the X and X cancel each other out
very nicely and now let's just put this down as well we have the integral going from 1 to X and
we have 1 over T DT and we add this with that one as you can see this right here is nice to
equal to Ln of X very nice and for the second one as you can see we go from 1 to Y 1 over u D U and
keep in mind use just a dummy variable you can use that P whatever letter that you want right here as
long as you have the integral going from 1 to X of 1 over Y upper and this is the same wafer you
can natural log of X I in this case we have the U so for the blue portion we have you know and
of Y very very nice and of course we do have the plus in between so this is the property that we
know now n of X Y is equal to Ln of X plus Ln of Y very very interesting huh now of course we have
the other one what if we have x over Y so let's do that quickly number 5 Ln of X over Y and you
guys can pause the video and try to think about this first but it's pretty much the same step it's
just a substitution part slightly tricky anyway we are going to integrate this from 1 to X over Y and
we will have to have the 1 over T DT and just a poverty over here we will have to break your part
the first one is the integral going from 1 to X and we have the 1 over T DT I know you want to put
on subtraction but no we cannot yet this is the integral going from 1 to X we have to pick it up
by putting down plus and the integral going from X to x over Y right that's the integral property
right and then we have the 1 over T DT so this is what we will have to deal with this is done
this is not yet again do a substitution so let's do that right here let he equal to something well
if you let equal 2x you sure you will work but you had to do it another substitution so I will
eat at you I am going to that he equal to x over Y pans you you'll see why first of all X&Y they are
just numbers so if I differentiate both sides this right here was turkey me x over Y and we will have
the tu right here and now take a look right here here we have the integral going from X T is equal
to X 2 T is equal to x over T Y right now in a new world when T is equal to X we will have to put the
X right here so we have X being equal to x over Y times new so in that case you will have to be
what you have to be equal to Y so that they cancel you get X of X X is equal to X so you get u equals
y right when he is equal to X u will be equal to Y and when T is equal to x over Y let me just put
that right here X over Y that's equal to x over Y times u so you will have to be 1 very good U is
equal to 1 very interesting and again we'll just do the usual business this is 1 over T which is
that let me just keep it as how the years x over Y times U and DT is that which is x over y tu they
cancel each other very nicely now it's very good and now of course we still have this right here
namely the integral going from 1 to X 1 over T DT Plus this integral in blue hmm something is
right something is wrong I don't know this right here it's actually totally right because we're
actually going to switch the limit of integration and to do that we just have to make the plus naked
here that's all so in fact this right here is think that we're going from 1 to X and we
have 1 over T DT but I'm just going to make that negative so that legitimately switched the limit
of integration and we will just go from 1 to Y and here we have the 1 over u the U and as you can see
for the first one this right here is nicely equal to natural log of X minus for the second one we
have what this right here is pretty much the same as that it is going to be natural log of Y that's
it natural log of Y so another property that we used to know and we proved it with integral
anyway as always take care so much watching and that's it