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[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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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