so if you're also a calculus teacher then I'm pretty sure that you have also experienced the following
and that is whenever you're trying to teach the definition of derivative
let's say we're trying to find the derivative of sin(x) by using the definition
you know there's always going to be some students in the back yelling out the answer cos(x) because they know it already
however when you ask them why they wouldn't be able to explain
they just don't want to do the definition of derivative I think the best way to make the
students learn the Interventional derivative is by forcing them
so today I would like to show you guys the classic example that you really have to use the definition of derivative in order to find the derivative okay this is a very classy question every calculus teacher and student should
know here we have the function f of x is defined to be x squared times sine of 1 over x if x is not
equal to zero and if x is equal to 0 then we get zero and our goal is to find F Prime of zero first
off I would like to just tell you f is indeed continuous as zero you can go ahead and check that
out later on your own now let's just focus only derivative first you might be wondering why can't
we just take it derivative maybe we just have to use the product rule and also a chain rule well
let me show you if you focus on this right here that only works for axis 9.0 so I will have to say
for X is not equal to zero then we look at F Prime of X okay use the product rule we keep the first
function which is x squared times D the root of the second which is cosine of 1 over X and don't
forget the chain rule multiply by the derivative 1 of X which is negative 1 over x squared
and then we add the second function then that's sine of 1 over x times the derivative of the first
and now give us 2X and now this and that cancel so we see that F Prime of X is just negative
cosine of 1 over X plus 2X sine of 1 over X it looks pretty cool right but here's the thing
right here X is not equal to zero I cannot plug this 0 in here all right
in fact like it's not defining either and we do have the second piece when X is equal to 0 we get
zero but that's just a point you cannot really just differentiate that and say the answer is
zero I know the answer to this question is 0 but no we cannot just do that well let's take a look
right here and I'll show you guys this if we take the limit as X approaching 0 of this expression
negative cosine of 1 over X plus 2X times sine of 1 over X I will tell you though this right here
will give us 0 you will see that later on we will have to use the squeeze theorem to prove that and
we will do that so just kind of wait for it but here is the part that will give us the trouble
here when we put zero here this is cosine of 1 over 0 it's like infinity and this right here
just means that this is going to give us you know between negative one and one oscillating forever
so the limit of the derivative when X is approaching 0 doesn't exist
so the stamping this doesn't exist no it's really weird no no let's go ahead and just check this out
to get F Prime of 0 well as I mentioned it we'll be using the definition of derivative and let me
just take a look right here ah I'll use this one the first one all right this right here I will say
this is the limit as X approaching zero and then we have f of x minus F of zero over x minus zero
keep in mind X is approaching zero f of x is that because when we have X is approaching
zero we don't care about X is the second gas zero so this is limit as X approaching zero
this right here is just x squared times sine of 1 over X and this right here by
this we get 0 so we don't need to write that down and then on the bottom we have X
well cancel cancel so we are just going to observe the limit as X approaching 0 of x
times sine of 1 over X which is the similar thing right here and now I will give you guys the detail
unfortunately we cannot just plug this student to here and here it does look that we have zero times
sine of 1 over 0 this right here again it's like sine of infinity is D and E
but you cannot just say 0 times DN is zero no don't do that that's bad
whenever these kind of things happen sine cosine we will use the square zero for this
kind of situation let me show you how to write that down first we can say sign of any angle
so I'll just put on Theta for this this is in between of negative one and one
so for whip one of X into the angle here we can say sine of 1 over X
this right here must also be in between of negative one and one
now I want to have the X let's just multiply Everybody by X so we can say Negative X
and right here I will have X this right here we will have x times sine of 1 of X in the middle
and this must be between of this and that at the end let's go ahead and take the limit I'll just
write down as X approaching 0 this we can do is just zero now we can also do it's also just
zero in the middle I will write down the limit as X approaching 0 of x times sine of 1 over X
this is what we're trying to find and we know that this right here has to be in between of
this zero and also that's the row of course the answer to this will be what it has to be equal to
zero and when you do this this right here it's the squeeze theorem so let me write this down
and some other books let me also call this the sandwich theorem that yeah you just pick your
favorite name for this so as we can see F Prime of 0 is just equal to zero and let me just write that
down for you guys that's it yes all the work right here for zero the value of f Prime of 0 we do have
a value of zero but if you take the limit for an X is approaching 0 of f Prime of X it doesn't exist
so if you liked today's video then you must really like calculus and as a calculus teacher
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