Learn How To Use the Tangent Line Approximation (by Example)

Video Series: Calculus 1 (Explore, Discover, Learn) Series

(D4M) — Here is the video transcript for this video.

00:00
[Music] in this video we explore what differentials are and how to use them
we also discover what linearization is and then see how to
linearize a function at a point so you will understand the so-called
tangent line or linear approximation with these examples here we go
okay everyone welcome back this is episode seven
in the calculus one explore discover learn series this episode is

00:01
linearization and differentials by example so we’re going to
cover differentials first i’m going to talk about differentials
and then after that we’re going to get into the the main part of the talk
or the episode which is going to be about linear approximation
and i’m going to make some nice drawings you can see it
and understand it and hopefully get a lot out of this
and then we’ll work some examples we’re approximating real numbers
using linearization and then we’ll linearize functions
so sometimes this is called tangent line approximation
but i usually just call it linearization then at the end we’ll talk about some
exercises we’ll talk about the exercises at the
end all right so let’s get started okay so up first our differentials and

00:02
we’re going to say that we have a differentiable
function that’s we’re going to start off with that’s our starting point
we have a differentiable function i’m just gonna
you know name it f of x x is our variable we have a differential d y and it’s
going to be defined by this equation right here
now for the for the purposes of this introduction here
we’re just going to say that dx is an independent variable
and then we’re going to look at why we use this notation here
at d y and a dx for our differentials of course you’ve seen the derivative
notation where you have y prime and you also see in leibniz notation
where you have d y over dx but at this point you know maybe
everything you’ve seen about d y over d x that was just notation
and you were wondering was there something more because the notation was just so
um intuitive in any case we’ll go over that here and this

00:03
but first here’s what a differential is d y and so let’s look at some examples
so here’s our first one we have our function fourth root of x and let’s
you know find the differential so the way to do that
is just to simply take the derivative right so we have fourth root of x
or think about that as x to the one fourth and so
we can say y prime is one-fourth x to the fourth minus one
so minus three fourths or we can write it as one over four
and then the fourth root of x to the third [Music]
so there’s our derivative and if you want to call this f of x
and we can just call this f prime here in any case d y is f prime times dx

00:04
where dx is just a variable here so our differential g
y is just f prime right here so one over four
and then we have the fourth root of x to the third and then times d
x so there’s our d y right here there’s our d y there’s our differential
d y is just this right here the derivative times d x
there’s really not much more to it other than you know use find a derivative
okay and we’ll see why this is useful in a minute like i said
but here we go here’s the differential first we find the derivative
and then the differential is just multiply d y is f prime times dx

00:05
all right let’s do one more quick example so here’s our y prime here is 10
times x squared minus two x minus three to the ninth power
and then times two x minus two [Music] because that’s the chain rule all right
so there’s y prime and so d y is just this right here however you want to
write it you can bring out a 2 and multiply if we like
but that’s not really much of a change there so i’m just gonna leave it
the way it is and then times dx so there’s our differential dy

00:06
so it’s just the derivative times dx all right let’s do one more
so there’s a derivative and there’s the derivative times dx
let’s look at one more ah let’s do something a little bit more fun
so let’s see what the differential is there
y prime is all right well first maybe let’s look at y
y is let’s say here we have x plus a 2x minus one
so just to see what you’re seeing in my head
this is way i’m looking at it in my head all right so y prime is one half
times all of this to the um minus one half power

00:07
so one half minus one and then now times the derivative of this inside part here
which is one plus and now derivative of this part so that’s one half
and then two x minus one to the minus one half
and then times the derivative of the inside part here times two
so there’s our derivative there let’s check that one half leave all this alone
to the one half minus one now times the derivative of all this
derivative of x is one derivative of this part is one half
leave all this alone subtract one and then now times the derivative of
this inside part here all right and so then now d y is
well you know d y is really nothing more than the derivative times d x
that’s just that’s just what d y is and then you can

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make this look as nice as you want to by writing it in radical form [Music]
so let’s do that there’s y and there’s the derivative
removing the negative exponents and so the differential d y
is just y prime times dx so that’s just formally what differentials are
and we’re going to see the relationship between the d y over dx
which is all one entity what is that entity it’s just the limit
as delta x goes to zero of delta y over delta x
so so this is just notation for all of this and then we just said what dx is an

00:09
independent variable and d y is f prime times d x so
you know you might think of d y here’s a two variable entity
where x is independent and dx is independent but we’re gonna make the connection
between all three of these concepts here we’re gonna do that right now
so notation variable defined quantity here and let’s kind of
put them all together so this notation makes a little bit more sense all right
[Music] all right so we’re going to start off with a
differentiable function f is a differentiable function
which means the derivative exists and we’re going to understand linear

00:10
approximation now where does it come from well it only comes from
that definition of what a derivative is nothing more nothing less
and we’re going to understand it hopefully better than you understand it
before and you’re getting value out of this but we need to look at a
diagram here to help us so this is kind of a review but it’s a little
bit more meatier than what we did on differentiation in that episode
so what we’re going to start off with here is a
diagram and i’m going to try to sketch a function through here
just a a a differentiable function so let’s look at something like this

00:11
and let’s look at the point about right here let’s say we’re about right there
and let’s look at a tangent line through that x
okay so tangent line through that x looks something about right there
now let me make that a little bit darker perhaps
now let’s give that x a little bit of a bump [Music]
let’s say that’s x plus delta x so that’s x so this distance right here
is delta x just a little small little distance there
um now to make the diagram look nice let’s actually
exaggerate because delta x is supposed to go to zero right

00:12
so you know we’re looking at this um derivative d y over dx is the limit
as delta x goes to zero of delta y over delta x and so if we come up here
and we come up here we’re going to hit right there
and we hit right here we have a horizontal line right here
we know this distance right here that’s delta x and this distance right here is
is called uh d y and the distance right here [Music]
is the delta y remember the delta y is so this height let’s come over here
and get a height and a height over here for this x so this x is right here
and we come over here we get a height that’s f of x

00:13
and we come over here and get a height so that’s x plus delta x
in the function it hits the function in red and so we come over here and get a
height let’s call that height here f of x plus delta x
okay so let’s put that in red this is the graph of y equals f of x
and the delta y here remember the delta y is the total height which is found by
that right there f of x plus delta x minus right so this height right here is
the whole height take away this height right here so that’s f of x
minus f of x so that’s the delta y so you can see the difference between
dx and delta delta y if i was to say graph the secant line through those two

00:14
points right there so we would have a secant line right through there
let’s see if we can do that in blue okay so in blue let’s make the graph of the
secant line through there i’ll dash that in there and how do you
find the slope of the secant it’s going to be the delta y over the delta x
that’s the slope of the secant delta y over delta x now when these are
when delta x is going to zero right that’s the whole point but when it’s not

00:15
zero then so let’s say delta x is approximately 0 then
d y over dx is approximately delta y over delta x now we can look at that
and say okay if this is if we have the limit we pass the limit then we’re equals
because that’s the definition but if we don’t have the limit
then the delta x could be some positive value there
and then these are only approximately this is the slope of the tangent line
right here that’s the slope of the line in green
and this is the slope of the secant line the slope of the line in blue now when
delta x becomes closer and closer to zero so when delta x is close to zero
then those two lines are the slope of those two lines are almost the same so
that’s what we mean by this right here now we’re going to take dx to be

00:16
delta x and [Music] what what then we’re going to have
d y is approximately delta y so d y is approximately delta y and
what can we get from that well let’s just go over here and do that
all right we’ll put it down here so d delta d y is approximately
and the delta y remember the delta y was the absolute change here um from
from the uh f of x plus delta x minus the f of x right
so that’s approximately so this is a delta y here f of x plus delta x
minus f of x just plugging in what that is right there
and then now moving this to the other side what we’re coming up with is

00:17
d y plus f of x is approximately this number right here this value right here
[Music] so let’s go put that over here let’s say we have
f of x plus delta x is approximately d y now remember d y is our differential
which is f prime times d x but we’re using dx as a delta x let’s
just say delta x right now and then so and then plus the f of x
right so that value right there is approximately d y there’s our d y plus f of x
and this holds when delta x is close to zero
and this is what i’m going to call the linear approximation formula
and it just comes from the definition of the derivative
so what it says is this value right here this height

00:18
is the derivative the slope of the tangent line times the delta x
plus this value right here so this value right here plus this d y right here
so the you can see this part right in here is the error
right in there but as the delta x gets closer and slower
closer and closer to zero then that error becomes smaller and smaller
so this is our linear approximation formula [Music]
so there’s the function there and this diagram right here this triangle
right here this right triangle describes what’s going on we have a change in x
and what what are the consequences of making that change in x
knowing that we have a differentiable function the consequence is we have this
linear approximation formula here which we’re going to practice using now

00:19
okay so let’s look at all this [Music] we have a differentiable function so d y
uh the differential represents the amount the tangent line rises
or falls right so the d y represents the amount the
tangent line rises or falls right this one’s rising because the tangent line is
is increasing whereas the delta y represents the absolute change in the
in the function f so [Music] when it changes by an amount uh dx
which we’re using to be our delta x so that’s why we’re using a delta x
so the f is differentiable so that means the derivative exists which
means that limit exists and so that equality is is relaxed to an approximation

00:20
so the further you’re away from delta x being close to zero
um the less that approximation holds most likely depending upon your function
um so if you take dx to be delta x then we get the delta y to be
approximately the d y and that will give us our linear approximation formula
so that’s we did right here so d y is approximately delta y
so if we just substitute in here what the delta y is
and then we substitute in here what is the d y
right move that over and then substitute in what the d y
is f prime times delta x then this is our linear approximation formula right
here so it all stems from this right here
d y is approximately delta y and you can see that from the graph right here

00:21
this is d y right here and there’s delta y right there and the
closer this is to zero the closer these are to each other right here okay so
[Music] um let’s look at the equation of the tangent line
which we’re looking at before in the previous videos
um so the derivative here f prime at a is the slope of the tangent line and so
this is another way now that we’re talking about differentials d y this is
just d y evaluated and so we have y equals f of a
plus d y so when using differentials to approximate
we’re going to use this formula right here and
here’s the here’s the main idea when you’re looking at

00:22
this linear approximation formula right here so this part right here is easy
when i say easy i mean usually easier and this part right here is also easier
so these are easy so these are often easier and then it’s
usually easy to add things and this is usually hard
so we’re going to find something that’s hard but we’re only going to find it
by doing two things that are easier and by approximating
so we’re going to see situation we’re going to see some examples where these
are easy to find and they will help us approximate something that’s hard to find
and so that’s kind of the idea behind the linear approximation formula
using d y’s approximately delta y and then the d y is easy to compute
the f of x is easy to compute but if we make a small

00:23
change that could be hard to compute okay so okay so now we’re going to use our
linear approximation formula that we just developed and we’re going to
approximate this number right here the cube root of
218 without using a calculator of course of course the calculator can do this
very quickly we’re doing it without a calculator
so first thing is we need a function right because to talk about
differentials we need a function and it needs to be a differentiable function so
what should our function be so i’m going to start off with a function

00:24
and since i’m trying to approximate the cube root of something
i’m going to look at the cube root function and so there’s our function there
now it’s just to use differentials so i’m going to find the differential v y
now remember this is x to the one third right so this will be
one third x to the minus two thirds power so in other words one over three times
the cube root of x squared and then times dx so there’s our differential
now we’re going to use the linear approximation formula that we just developed
so the linear approximation formula says x
plus delta x is approximately equal to the derivative at x times r

00:25
delta x plus the function value at x so how does
all this fit here with the f of x and the d
y how does all this fit together well when we’re looking at cube root of
218 we need to know a value that is easy to find but it’s close to that
so if you look at cube root of 216 that’s easy to find right what is that
cube root of um 216 is what is this easy to find what is um 8 times 8 times 8
64 times 8 so that’s not it so what is the cube root of 216

00:26
well do some thinking about it it’s six all right not six point something it’s
just six and so what if i’m looking at cube root of
218 now and a different light i think it’s something close to six right but
can we quantify that better can we say six point something seven point
something what is it so you could for example
you know multiply 7 what’s 7 to the third that’s too far so
i’m going to use my delta x to be um 218 minus 216 which is two
now you might say delta x is supposed to be close to zero
right so this is when delta x is close to zero
is too close to zero well it all depends upon your point of view

00:27
if you made a two percent or two out of a hundred on an exam
then you would say two is well almost zero [Music]
but if if you’re thinking of decimals like
point zero zero zero one well then two is pretty far away from zero
but um in terms of 216 for example we have here
let’s say two is pretty pretty small there so we’re going to start off with
the cube root of 218 that’s what we’re trying to find 18
and we’re going to say that is our function right here if plugged in at 218
right if you plug into it to 18 you plug in 218 so that’s the cube root of 218

00:28
and this is 216 plus 2 right so we have x plus delta x
and this is going to be approximately equal to the derivative at the x
so the x is the 216 times the delta x which is 2
and then plus f at the x this is the x so f at 216 here
so the question is we know this one is easy to compute that’s six that’s
that was our you know idea here that this is six this is two
what about this one right here if that one’s easy to find also then we got a
good approximation well if we put 216 in here and
let’s say we cube root it first we’re going to get six so that’s going to be

00:29
36 right there so it’s 3 times 36. and so we can go to equals here now
so let’s just write this as 1 over 3 times cube root of 216
that’s our x and then squared and then we have times the two still
i’ll put up here two and then plus the six so then we can go and calculate this
right here two over three times the cube root of
216 squared and i would take the cube root of that first right that’s six
that’s 36 right there so that’s very doable right there
and so that’s how we would use the linear approximation
to approximate um a real number cube root of 218 well we need a function
the function i’m going to use to plug in a number

00:30
to get the value that we’re trying to approximate this is going to be the
function right here cube root of x because if i plug in 218
that’s the number i’m trying to approximate
and then i try to think of 218 as being close to something
that is easier to work with so cube root of 216 we see
i have to do a little bit of thinking that’s easy to work with that’s just six
so i’m going to write it as 216 plus what’s the difference 2. so
218 is 216 plus 2 and this is an x plus a delta x so this is my x right here
and this is my delta x so then i just go with the right hand side
f prime of the x x is here times the delta x which is 2
plus and then f of x which is 216. this one we said was easy that’s why
we’re doing it all the derivative is easy to find
216 into the derivative is easy to find and when we calculate all this right

00:31
here we’re going to get 325 over 54 which is approximately 6.01852
and so that’s what the cube root of 218 is approximately 6.01852
now if you go to your calculator you can check out how accurate it is
it’s pretty pretty good okay so here we go first thing is i need a function and
then i go find ui your y is easy you find the derivative
you find dx now the reason why i do dy next is because
i know that i need to go i’m going to need to plug something into the function
and something into the derivative and they’re going to be need to be easy to
find i’m going to plug in 216 into my function and i’m going to plug in

00:32
216 into my derivative so i want to have them available so i can say yeah those
are easy to find and then the difference between the easy one to 216
and the 218 that’s the two so now we have everything we need
to do the approximation so i always start with what i’m trying to approximate
that way so we can easily see that hey i approximated it
that’s f of 216 plus two so identified my x plus my delta x
so then we use our differential a linear approximation formula
remember the d y is just f prime times d x everything is easy to compute
we do a little bit of simplification there and we get the decimal there
so let’s do one more let’s do one more here what’s the function pause the video

00:33
whenever you want and try it yourself i’m going to go ahead here
f of x is what do you think it should be cube root of x plus
fourth root of x now the reason why i’m doing this
is because the one point two there are the same that’s one point two one
point one point zero two so those are the same so i’m just going to
use one function on this all right so what’s our derivative
so this will be 1 over 3 cube root of x squared plus 1 over 4
and then we have the fourth root of x to the third right that’s cube root
so i’m always thinking about these with rational exponents here
when i’m using the power rule there so it’s 1 4 and then 1 4 minus 1

00:34
so minus 3 4. anyways there’s our derivative and now let’s do our linear
approximation so our linear approximation formula is
x plus delta x this is what is difficult to find is approximately f prime of x
times the dx or the delta x and then f of f of x
so i need to know this easy and i need to know this easy
um we know how to take the cube root of one and the fourth root of one so
i’m going to use um i’m going to approximate this with the
one so my delta x here will be um so my change in x is 0.02

00:35
and so that’s just going to be 0.02 and now i’m going to start trying to
approximate it so the cube root of 1.02 plus the fourth root of 1.02
is approximately or is equal to the function at 1.02
right so if i plug in 1.02 into the function
i get this number out right so i plug in 1.02 here
1.02 here we get the function out so that’s the left hand side right well
let’s double check this is f of 1 plus .02 right 1.02 is the same thing as 1
plus 0.02 so identified my x and my delta x
so now i’ll go to approximately and now i got the right hand side here

00:36
the derivative at 1 times the change .02 plus f of one
and the idea that these are all easy now this is this should be easy
that’s easy that’s easy so now i can go to equals because i know this is exactly
what’s one when you put it into the derivative here’s the derivative
it’s going to be one-third plus one-fourth so i’ll just write that
one-third plus one-fourth times what’s 0.02 so that’s what
you know 2 over 100 and then what happens when you plug in 1 here
you plug in 1 to the derivative well that’s just a two right
so plus two when you plug in one here and one here we get one plus one
so if we calculate all that up we get 12 1207 over 600

00:37
and that’s um 2.011 and then the six start repeating
so it’s about what you thought it would be the cube root of 1.02
plus the fourth root of 1.02 is approximately two well
this is pretty good here you know if you go use the calculator on this
you’ll see how accurate that is it’s pretty good
all right so there’s our approximation right there and let’s write that up so
first off i need a function so i choose x to the one third plus
x to the one fourth it needs to be differentiable so we found the derivative now
that derivative doesn’t exist everywhere but we need it to be
um around one we’re interested in the point around one

00:38
so the linear approximation is um one plus zero point zero two
is approximately f of one plus d y and we found the d y the differential
it’s the derivative at one times the change in x
and there’s the fraction and there’s the approximation all right [Music]
so here we go now now we’re going to start approximating functions
so before we were approximating real numbers like the cube root of 218
and now we’re going to approximate functions which is more powerful
and it actually doesn’t even take very much more work in fact students actually

00:39
like this a lot um approximate functions and it gives you more power because you
can do more than just approximate one single number at a time
so we’re going to start by talking about the linear approximation right there
and that is basically a um tangent line approximation because it uses the
tangent line equation to do the approximating and so
um you know sometimes called the tangent line approximation
and we’re interested at a point a so i use i’m using this a
at this point so it says that f of x which is the height of the function is
approximately f of a plus the derivative at a times x minus a
so the one on the left is you know in terms of x and it could it’s

00:40
potentially non-linear the one on the right is linear f of a is a number
the derivative at a is a number it’s the slope and then x minus a so this is a
linearization and here’s the actual linearization so i used an
l for linearization so the linearization is equal to
and then we’re using that right hand side because it’s approximately f of x
and this is certainly a line again we have f of a that’s a number
f prime of a that’s a number and then we have the variable x here so x minus a
so this is the linearization of f at a and just for short it’s called the
linearization okay so let’s look at our first example here
we have this function and as you can see it’s not linear

00:41
and we’re interested in this function at zero so my a is zero
or you can call it x1 whatever you want but we’re at the number of the value
zero that’s where we’re interested in linearizing it
so here we go the linearization is just f at the number which is zero
plus the derivative at the number and then x minus the number [Music]
now we obviously need to go find the derivative of f
but this is the linearization right here it’s the function at zero plus the
derivative at zero times x minus the number there so let’s
go over here and find the derivative so let’s think of the function first as

00:42
two plus x to the minus one half power right and so the derivative is simply
um minus one over two and then two plus x and then to the minus
minus one half minus one so minus three halves
um in fact i’ll just write it like that minus one half
two plus x and then minus three halves and then times the derivative
of the inside part here but that’s just one
so i need to plug in zero into both of these
so i wouldn’t worry about simplifying them too much
we need their we need these numbers so the linearization is
what happens when you substitute 0 into this function right here
so we’re going to get 2 to the minus 1 half power
and then plug in 0 into the derivative and we’re going to get

00:43
zero right there and so we’re going to get minus one half two to the
minus three halves and then x y zero so that’s just an x
so we just need to worry about what is two to the minus one half power
plus minus one half and then two to the minus three halves power
and that’ll be the slope right there so there’s the linearization [Music]
so writing that up linearization is so i used an x1 there x1 is zero
there’s the definition right there and we plug in zero
and when we plug in zero into the function
we plug in zero into the derivative and then x minus zero
so then we calculate that up two to the minus one half

00:44
and then calculate that up and then of course you can rationalize
if you want but there’s our linearization right there so what that means is
the function f of x is approximately equal to l of x
whenever x is near zero that’s the point where we’re linearizing
at zero so as long as x is near zero we have this nice linearization which i
went ahead and rationalized for us so which would you rather use if you wanted
to know [Music] what is so we have 1 over square root of 2 plus
x is approximately square root of 2 over 2 plus minus square root of 2 over 8

00:45
times x so this is our function right here and this is our linearization at x
equals zero so if i wanted to know what is f of x of zero point one i could say
oh that’s just one over square root of two plus point one or two point one
right so that involves finding the square root and taking reciprocal though
those operations aren’t the funnest in the world
so instead i could say because the change right here is point one that’s
pretty close to zero so like instead i could say that’s
square root of two over two which actually you may know of right off
the top of your head plus and we have minus square root of
two over two eight over eight sorry minus square root of two over
eight and then times so what’s the x we’re inputting zero point one

00:46
or one over ten so these operations over here on the right are much simpler
and you can get these right here just simply by
you know doing some scratch work where of course you can do this right here
scratch work two but these operations are more complex than these operations
over here um well both sides involve ratios and square roots but
i think you get the feeling for why this could be very useful
um let’s look at the next example [Music]
so now our function is the fourth power down there
so let’s look at the linearization so first i’m going to take the derivative
so i’m thinking about this to the minus four power

00:47
i’m gonna have minus four one plus two x to the minus four minus one so
minus fifth power and then times two derivative of the inside
and we’re looking at zero so what is f prime at zero rather than simplify that
looks like it’s minus eight times what um just minus eight right
and so our linearization is going to be f of zero plus the derivative at zero
times x minus zero that’s our linearization
so again this problem is looking at zero so the linearization
l of x is f of 0 plus the derivative at 0 times x minus 0. so l of x
is if i plug in zero into my function right we get out one so that’s one

00:48
and this right here is minus eight and then times x so minus eight x
okay so this is approximately equal to one over one plus two x to the fourth
so that’s when um x is close to zero when x is close to zero these are
approximately equal to each other so now in this example i think it’s much
more clear if i choose the value close to zero
for example point zero one or point zero three or
you know something small close to zero i’d rather
plug it in over here on this and multiply it by eight
and then do a negative and then add one this is much much easier
whereas if i plugged in a point zero one here
multiplied it by two added a one and then i gotta do the fourth power and
then i gotta do reciprocal so definitely in this example right here

00:49
the left hand side is much easier to use than the right hand side right here
all right let’s see that written up so there’s our function and we’re
linearizing at zero so there’s our linearization formula
we plug in zero everywhere we calculate what is f of zero we calculate
f prime and zero and we get our linearization what that means is the function is
approximately equal to the linearization when x is near zero so think about this
differently if you were to go say use a graphing
calculator and you were to look at that function right there
f of x is 1 over 1 plus two x to the fourth and you say
sketch the graph go ahead and sketch it and then you zoom in
on the window around zero say you’re zooming in
say point one to or minus point one to point one
and you keep zooming in you keep zooming in and it looks like a line what’s that

00:50
line that it looks like one minus eight x okay so um let’s look at some
exercises now and um here we got um some exercises for us
so the first one is look at the um difference between the
delta f and the differential on those first two examples there
first three examples and then on the next uh couple there four and five
we’re going to be writing differential formulas um
so we’re given some kind of formula there v and s
and see on the next uh six and seven here is purely just find the differential

00:51
there find the d y so that’s basically practice right uh finding more
derivatives remember d y is just derivative times d x
so practice you finding those derivatives um number two there is also implicit
differentiation so these are some fun exercises that
i like for you to try and if you get stuck or want to see
your your work versus mine let me know in the comments below
and i can make a video and i can show you how i work them out
all right so here’s some approximating real numbers for example 2.99 to the
third approximate that number using the linear approximation formula
um and then 9 and 10 are find the linearization of those formulas there
and so then there’s some good exercises for you to practice on
remember math is more than just knowing it’s also about your skill um

00:52
you know how fast or how accurate are you when you do your mathematics
all right so at this point i want to say um thank you for watching and if you
enjoyed this video please remember to hit that subscribe button um
also if you want to find me on social media the links are below
so i would appreciate it uh say thank you for watching
and um well the next video is going to be over
related rates um this can be exciting we’re going to work out lots of word
problems and get a better feeling for how we can use these rates of change in in
the world around us um and so we’ll see that next time
and um i really look forward to seeing you thanks for watching
if you like this video please press this button
and subscribe to my channel now i want to turn it over to you
math can be difficult because it requires time and energy to become skills

00:53
i want you to tell everyone what you do to succeed in your studies
either way let us know what you think in the comments

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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