Linearization Differentials: trying to wrap my head around how a differential is defined

[jpanknin]

Hi, I'm trying to wrap my head around how a differential is defined. We were given the function [imath]f(x) = x^3+x^2-2x+1[/imath] and asked to calculate [imath]dy[/imath] from 2 to 2.05 with [imath]dy[/imath] defined as [imath]dy = f'(x)dx[/imath]. I got the derivative of [imath]f'(x) = 3x^2+2x-2[/imath] and then calculated [imath]f'(x)[/imath] for both 2 and 2.05, subtracting [imath]f'(2) from f'(2.05)[/imath] and got the correct answer.

The book then says that, in general, [imath]dy[/imath] can be calculated using the formula [imath]dy = f'(x)dx[/imath], which in this case would be [imath]dy = (3x^2+2x-2)(.05)[/imath]. I'm thinking of this as a kind of factoring where the long way would be: [imath]dy = [3(2.05)^2-2(2.05)-2] - [3(2)^2+2(2)-2][/imath] and the short way [imath]dy=(3x^2+2x-2)(.05)[/imath], but for the life of me I can't see how this would be done. It's not a typical factoring formula like [imath](5sinx-3sinx)[/imath] where [imath]sinx[/imath] can be factored out because the 2 and 2.05 must be placed in the f'(x) expression. We're not multiplying f'(x) by 2 and 2.05, we're placing 2 and 2.05 inside as the variable, so I can't figure out how f'(x) can be isolated algebraically. Hope this is clear. Any help would be appreciated.

 

[Dr.Peterson]

Hi, I'm trying to wrap my head around how a differential is defined. We were given the function [imath]f(x) = x^3+x^2-2x+1[/imath] and asked to calculate [imath]dy[/imath] from 2 to 2.05 with [imath]dy[/imath] defined as [imath]dy = f'(x)dx[/imath]. I got the derivative of [imath]f'(x) = 3x^2+2x-2[/imath] and then calculated [imath]f'(x)[/imath] for both 2 and 2.05, subtracting [imath]f'(2) from f'(2.05)[/imath] and got the correct answer.

You were told how the differential is defined: it says that a small change in y is approximately equal to f'(x) times a small change in x. It linearizes the function, and uses that line to approximate the curve. You seem to be overthinking.

The book then says that, in general, [imath]dy[/imath] can be calculated using the formula [imath]dy = f'(x)dx[/imath], which in this case would be [imath]dy = (3x^2+2x-2)(.05)[/imath]. I'm thinking of this as a kind of factoring where the long way would be: [imath]dy = [3(2.05)^2-2(2.05)-2] - [3(2)^2+2(2)-2][/imath] and the short way [imath]dy=(3x^2+2x-2)(.05)[/imath], but for the life of me I can't see how this would be done. It's not a typical factoring formula like [imath](5\sinx-3\sinx)[/imath] where [imath]\sinx[/imath] can be factored out because the 2 and 2.05 must be placed in the f'(x) expression. We're not multiplying f'(x) by 2 and 2.05, we're placing 2 and 2.05 inside as the variable, so I can't figure out how f'(x) can be isolated algebraically. Hope this is clear. Any help would be appreciated.

No, it is not factoring. But you could, in this particular case, do the same work you do to find the derivative from the definition in order to show what is happening here, and that will look a little like some of what you have done.

The actual change in y is given by [math]\Delta y = [(x+\Delta x)^3+(x+\Delta x)^2-2(x+\Delta x)+1] - [x^3+x^2-2x+1]\\=[(x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3)+(x^2+2x\Delta x+\Delta x^2)-(2x+2\Delta x)+1]-[x^3+x^2-2x+1]\\=3x^2\Delta x+3x^2\Delta x^2+\Delta x^3+2x\Delta x+\Delta x^2-2\Delta x\\=(3x^2+3x^2\Delta x+\Delta x^2+2x+\Delta x-2)\Delta x [/math]
Since [imath]\Delta x^2[/imath] is considerably smaller than [imath]\Delta x[/imath], we can ignore terms containing it in an approximation, and use [imath](3x^2+2x-2)\Delta x[/imath]. And that is [imath]dy[/imath], when you replace [imath]\Delta x[/imath] with [imath]dx[/imath]. So [imath]dy[/imath] is an approximation of [imath]\Delta y[/imath].

Does your textbook not explain where this approximation comes from?

 

[jpanknin]

You were told how the differential is defined: it says that a small change in y is approximately equal to f'(x) times a small change in x. It linearizes the function, and uses that line to approximate the curve. You seem to be overthinking.


No, it is not factoring. But you could, in this particular case, do the same work you do to find the derivative from the definition in order to show what is happening here, and that will look a little like some of what you have done.

The actual change in y is given by [math]\Delta y = [(x+\Delta x)^3+(x+\Delta x)^2-2(x+\Delta x)+1] - [x^3+x^2-2x+1]\\=[(x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3)+(x^2+2x\Delta x+\Delta x^2)-(2x+2\Delta x)+1]-[x^3+x^2-2x+1]\\=3x^2\Delta x+3x^2\Delta x^2+\Delta x^3+2x\Delta x+\Delta x^2-2\Delta x\\=(3x^2+3x^2\Delta x+\Delta x^2+2x+\Delta x-2)\Delta x [/math]
Since [imath]\Delta x^2[/imath] is considerably smaller than [imath]\Delta x[/imath], we can ignore terms containing it in an approximation, and use [imath](3x^2+2x-2)\Delta x[/imath]. And that is [imath]dy[/imath], when you replace [imath]\Delta x[/imath] with [imath]dx[/imath]. So [imath]dy[/imath] is an approximation of [imath]\Delta y[/imath].

Does your textbook not explain where this approximation comes from?

Thank you. The book does a good job of explaining the concept of linearization and how to calculate [imath]dy[/imath] using [imath]f'(2.05) - f'(2)[/imath], so I get the idea. I'm just having trouble seeing how that can be generalized as [imath]dy = f'(x)dx[/imath]. Below is the section of the book where it's generalized.

 

[Dr.Peterson]

Thank you. The book does a good job of explaining the concept of linearization and how to calculate [imath]dy[/imath] using [imath]f'(2.05) - f'(2)[/imath], so I get the idea. I'm just having trouble seeing how that can be generalized as [imath]dy = f'(x)dx[/imath]. Below is the section of the book where it's generalized. View attachment 36194

I don't understand what you don't understand!

It doesn't need to be generalized! It is defined as [imath]dy = f'(x)dx[/imath].

Please show how they explain the idea -- not an example like this, but what they say dy is.

The basic idea, which they surely tell you, is that for small changes in x, [imath]\frac{\Delta y}{\Delta x}\approx\frac{dy}{dx}[/imath], so [imath]\Delta y\approx\frac{dy}{dx}\Delta x[/imath].

 

[jpanknin]

I don't understand what you don't understand!

It doesn't need to be generalized! It is defined as [imath]dy = f'(x)dx[/imath].

Please show how they explain the idea -- not an example like this, but what they say dy is.

The basic idea, which they surely tell you, is that for small changes in x, [imath]\frac{\Delta y}{\Delta x}\approx\frac{dy}{dx}[/imath], so [imath]\Delta y\approx\frac{dy}{dx}\Delta x[/imath].

Below is the first place [imath]dy=f'(x)dx[/imath] is mentioned. The previous screenshot immediately followed this definition.

Defined to me is kind of like my dad always saying, "Because I said so." The part I'm trying to figure out is how they went from [imath]f'(2.05) -f'(2)[/imath] to [imath]dy=[3(2.05)^2−2(2.05)−2]−[3(2)^2+2(2)−2][/imath] to [imath]dy=f'(x)dx[/imath]. There is no proof in the book showing how this can be generalized.

 

[Steven G]

Defined to me is kind of like my dad always saying, "Because I said so." Good, you know understand perfectly what defined means.

Consider when you were in grammar school and your teacher said + means to add. But why couldn't * or & be the symbol to add? Answer: It could have one of those symbols or any other symbol but someone defined + to mean add.

Never ever question a definition. One can define whatever they want. I define, for this post, that a**b means a+2b where a and b are numbers-so 7**2=11. You can't say that I am wrong or the definition is wrong! Have you ever said that the definition of a word in a dictionary is wrong? No, because the word was defined exactly as it says in the dictionary.

In your case, dy = f'(2)(0.05) = (3*4+2*2-2)(0.05)= 14(0.05) = 0.7

Recall that dy/dx = f'(x). (a little hand waving follows) Multiplying both sides by dx gives us that dy=f'(x)dx.

Alternatively f'(x) = dy/dx =[f(2.05)-f(2)]/dx. This means that f(2.05)-f(2) = f'(x)dx.

 

[Dr.Peterson]

Below is the first place [imath]dy=f'(x)dx[/imath] is mentioned. The previous screenshot immediately followed this definition.

Defined to me is kind of like my dad always saying, "Because I said so." The part I'm trying to figure out is how they went from [imath]f'(2.05) -f'(2)[/imath] to [imath]dy=[3(2.05)^2−2(2.05)−2]−[3(2)^2+2(2)−2][/imath] to [imath]dy=f'(x)dx[/imath]. There is no proof in the book showing how this can be generalized.

View attachment 36195

Okay, this is a definition, and tells you what they mean by the word differential and the symbol dy.

Unfortunately, it doesn't tell you why it can be used to approximate the result of a small change.

That's what my last line was meant to fill in, in case they didn't show that. Does it mean anything to you?

The basic idea, which they surely tell you, is that for small changes in x, [imath]\frac{\Delta y}{\Delta x}\approx\frac{dy}{dx}[/imath], so [imath]\Delta y\approx\frac{dy}{dx}\Delta x[/imath].

Now, they don't go from [imath]f'(2.05) -f'(2)[/imath] (which is not mentioned at all) to [imath]dy=[3(2.05)^2−2(2.05)−2]−[3(2)^2+2(2)−2][/imath] (which is the actual change [imath]\Delta x[/imath]) to [imath]dy=f'(x)dx[/imath]. The latter is the definition of what we mean by dy; that's the starting point, as far as notation is concerned. In terms of meaning, here is the thinking they are apparently expecting you to see for yourself:

Since [imath]\Delta y\approx\frac{dy}{dx}\Delta x[/imath], and [imath]\Delta x = x-x_0[/imath] and [imath]\Delta y = y-y_0[/imath], we have [math]y-y_0\approx\frac{dy}{dx}(x-x_0),[/math] so that [math]f(x)-f(x_0)\approx f'(x_0)(x-x_0).[/math] In the example, [math]f(2.05)-f(2)\approx f'(2)(0.05).[/math]
Perhaps you need to find a better source. Here is one textbook's explanation of the differential approximation:

4.2: Linear Approximations and Differentials

In this section, we examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions are the easiest functions with which to work, so …

math.libretexts.org

This does as I would do, showing how and why the approximation can be found, before even mentioning differentials.