I know this will be another simple question of mine, but I just can't help it.

Often in integration, we use substitution like

\(u=2x\)

Then \(du=2 dx\)

Okay, I understand it in intuitive sense. I also understand why 2x became 2.

Now my question is

1. Why \(du\) is no fraction like \(\frac{du}{dx}\)?

2. Why \(2 dx\) has dx on it? When we differentiate things like: We have a function \(y=3x^2\), then the notation and result would be \(\frac{dy}{dx}3x^2=6x\), right? There isn't \(dx\) on the right side of the equation.

My guess to my question is this:

Actually, suppose we have \(u=2x\), then we differentiate it \(\frac{du}{dx}2x=2\) Then we multiply \(dx\) (I guess this is very informal because I treat as if dx is a number when in fact \(\frac{du}{dx}\) is just a notation.) to both sides of the equation.

\(\frac{du}{dx}dx=2dx\)

\(du=2dx\)

I know I have removed the 2x in \(\frac{du}{dx}2x=2\). But as of now, that is how I understand it.

I know my guess is very informal so I hope the geniuses here will help me.

Thank you.

 

I don't have the full answer for you and I'm away from my reference, but treating differentials like dy and dx separately works OK if the functions are well behaved. In this context it means continuous and integrable. When functions get pathological many informal methods breakdown.

 

When functions get pathological many informal methods breakdown.

That's serious PB!

 

1) Because you've separated the variables into differentials.

2) No.

\( \frac{d}{dx} 3x^2 = 6x \)

\( (d)(3x^2) = (6x)(dx) \)

 

If you let u = 3x^2, then you'll see:

\( du = f'(x)dx \)

The change in u accelerates the change in x.

 

First you select the quantity that you want to be u.

In your case you want 2x to be u. So, u=2x.

Then you take the DERIVATIVE of u with respect to x. Get it? du=2dx is not something that you pull out of air or magic. You have to take DERIVATIVE of u with respect to x. So, derivative of u with respect to x is derivative of 2x with respect to x:
\(\frac{d}{dx}u=\frac{d}{dx}2x=2\frac{d}{dx}x=2(1)=2\)

\(\frac{d}{dx}u=2\)

\(\frac{du}{dx}=2\)

\(du=2dx\)

The only thing YOU choose is which "thing" will be u. Afterward everything is mathematically derived.

 

Keep in mind that the notation dy/dx isn't arbitrary.

It is the ratio of a small change in y, Δy, to a small change in x, Δx, as the changes get arbitrarily small. So fundamentally you are working with something like:

y = 2x²

Δy/Δx ≈ 4x

It then makes perfect sense to multiply both sides by Δx to get

Δy ≈ 4x*Δx

What this says is just that if we change x by a small amount, the change in y is 4 time that amount times the value of x. If x is twice as big, then the change in y will be twice as big.

The approximation becomes equality in the limit that we are talking about infinitesimal changes.

dy = 4xdx

but the interpretation is essentially the same.

 

You are being too abstract and forgetting what derivatives and differentiation mean.

Differentiation means taking the slope or gradient, i.e. Δy/Δx

Let us start with a simple example:

y = 4x

Graphically, this can be represented by a straight line.
What is the slope of this line? We know the slope of a straight line is a constant value.

Hence we can use Δy/Δx.

y = 0 when x = 0
y = 4 when x = 1
Δy/Δx = (4 - 0)/(1 - 0) = 4/1 = 4

Δy/Δx = 4 for all values of x

Now let us take

y = 4x²

We know this is not a straight line.
What is the slope of the line? It is not constant. It varies at every position of x.

Can use Δy/Δx to find the slope? Yes, we can find the slope at any point x as long as we make Δx very small at the point x.

For y = 4x²

dy/dx = 8x

This is the same as saying

Δy/Δx = 8x in the limit as Δx approaches 0.

Thus dy/dx is simply a way of expressing the slope or gradient of a function at any value of x.


dy = 8x . dx is the same equation as dy/dx = 8x

just as writing

Δy/Δx = 8x is the same as Δy = 8x . Δx

Note that neither dx nor Δx is zero.

 

Thank you everyone!

Keep in mind that the notation dy/dx isn't arbitrary.

It is the ratio of a small change in y, Δy, to a small change in x, Δx, as the changes get arbitrarily small. So fundamentally you are working with something like:

y = 2x²

Δy/Δx ≈ 4x

It then makes perfect sense to multiply both sides by Δx to get

Δy ≈ 4x*Δx

What this says is just that if we change x by a small amount, the change in y is 4 time that amount times the value of x. If x is twice as big, then the change in y will be twice as big.

The approximation becomes equality in the limit that we are talking about infinitesimal changes.

dy = 4xdx

but the interpretation is essentially the same.

I can't help myself but can you give an example? Please?



y = 2x²

Δy/Δx ≈ 4x

Let's say that we choose x to be 0.001 so the y will be 0.000002.

So we have Δy/Δx as 0.000002/0.001 = 0.002 but that's not equal to 0.004 (or 4*0.001)

Isn't it equal because im wrong or because it can't be equal because that's just an approximation?

Or my interpration is wrong?

P.S.: If it would be okay pls help me with http://forum.allaboutcircuits.com/showpost.php?p=699874&postcount=19 ... I'd appreciate everyone!

Thanks!

 

y = 2x²

Δy/Δx ≈ 4x

At x = 0.001, y = 0.000002

Δy is not 0.000002
Δx is not 0.001

At x = 0.001, y = 0.000002
At x = 0.0011, y = 0.00000242

Δx = 0.0001, Δy = 0.00000042
Δy/Δx = 0.00000042/0.0001 = 0.0042

dy/dx at x = 0.001 = 4 times 0.001 = 0.004

 

y = 2x²

Δy/Δx ≈ 4x

At x = 0.001, y = 0.000002

Δy is not 0.000002
Δx is not 0.001

At x = 0.001, y = 0.000002
At x = 0.0011, y = 0.00000242

Δx = 0.0001, Δy = 0.00000042
Δy/Δx = 0.00000042/0.0001 = 0.0042

dy/dx at x = 0.001 = 4 times 0.001 = 0.004

touché!

That's why earlier, I think there was an error in my post. I thought Δy/Δx must be like change in over change in x....

I really learn many things here. The things I think I already know, experts here falsify that! That's why I love asking here, eh. People's not bully.

Thanks...

Can someone help me with my Fourier problem? I'd like to understand it in mathematical sense, at least... thanks!

 

What's the Fourier problem?

 

touché!

That's why earlier, I think there was an error in my post. I thought Δy/Δx must be like change in over change in x....

I really learn many things here. The things I think I already know, experts here falsify that! That's why I love asking here, eh. People's not bully.

Thanks...

Can someone help me with my Fourier problem? I'd like to understand it in mathematical sense, at least... thanks!

Δy/Δx IS change in y over change in x.

Let's again consider

y = 2x²

Pick any x, say x = 10

y = 200

Now pick a small change in x, say Δx = 0.01

y at x=10.01 = 200.4002

So Δy = 0.4002

Δy/Δx = 0.4002/0.01 = 40.02

dy/dx = 4x = 4*10 = 40

 

What's the Fourier problem?

http://forum.allaboutcircuits.com/sh...4&postcount=19

I hope you'll help me.

 

http://forum.allaboutcircuits.com/sh...4&postcount=19

I hope you'll help me.

You need the full link. The "sh...4" indicates that it was shorted for display purposes.

 

Don't paste link into your reply box. Use the Insert Link button (icon) at the top of the message box.

 

Fourier Series Problem

Okay. Please help me...