What is the meaning of dx and how does it relate to dy/dx?

[FS98]

Often I have to solve problems using dx or dq. I always don’t quite understand what’s going on.

I understand what dy/dx is but not just dx. Can someone walk me through in plain language a somewhat rigorous definition of differential like dx?

 

[jedishrfu]

There’s some YouTube videos online by 3blue1brown on the Essence of Calculus where they go over this very topic.

Here’s the first of the sequence

 

[FS98]

There’s some YouTube videos online by 3blue1brown on the Essence of Calculus where they go over this very topic.

Here’s the first of the sequence

I notice that he and some other sources sometimes use delta x and dx interchangeably. He also says that dx is not infitesimally small.

He sometimes also refers to dx as a small change in x.

Is dx literally just a small change in x? Not infinitely small, but sufficiently small?

 

[mathman]

In elementary calculus \frac{dy}{dx} is defined as the limit as \Delta x \to 0 \ of \ \frac{\Delta y}{\Delta x}. Therefore dx is infinitely small. However in elementary calculus, dx by itself has no meaning.

 

[FS98]

In elementary calculus \frac{dy}{dx} is defined as the limit as \Delta x \to 0 \ of \ \frac{\Delta y}{\Delta x}. Therefore dx is infinitely small. However in elementary calculus, dx by itself has no meaning.

In calculus 1 dx and du are used alone for u substitution.

 

[jedishrfu]

In elementary calculus \frac{dy}{dx} is defined as the limit as \Delta x \to 0 \ of \ \frac{\Delta y}{\Delta x}. Therefore dx is infinitely small. However in elementary calculus, dx by itself has no meaning.

This depends on the calculus teacher or book used. However infinitesimals like $dx$ can be treated like real numbers with all the same algebraic properties as real numbers when Calculus is cast in the framework of hyperreals.

https://en.m.wikipedia.org/wiki/Hyperreal_number

You can find Kreisler’s Elementary Calculus copyright free online here:

http://www.math.wisc.edu/~keisler/calc.html

 

[fresh_42]

In calculus 1 dx and du are used alone for u substitution.

This sounds as if you are talking about integrals. Here we have something like $\int f(x)\,dx$, whereas in the differentiation process we have $f\,'(x)=\dfrac{df}{dx}$. The name of the variables, $x$ or $u$, $f(x)$ or $y$, are not important. It's just a label to work with and a convention which are used when. In both cases it basically means something like a small amount of. Technically it stands for the limit where this small amount gets smaller and smaller. In differentiation, it is a quotient, a slope of secants narrowing down to one of a tangent, and in integration, it is height $f(x)$ times width $dx$, i.e. the limit of Riemannian sums, see e.g.
https://www.physicsforums.com/insights/omissions-mathematics-education-gauge-integration/

For all practical purposes of integration, it is enough to consider $dx$ as just a notation which determines what the integration variable is, because all others, which don't change with $x$ are constants.

 

[FS98]

This sounds as if you are talking about integrals. Here we have something like $\int f(x)\,dx$, whereas in the differentiation process we have $f\,'(x)=\dfrac{df}{dx}$. The name of the variables, $x$ or $u$, $f(x)$ or $y$, are not important. It's just a label to work with and a convention which are used when. In both cases it basically means something like a small amount of. Technically it stands for the limit where this small amount gets smaller and smaller. In differentiation, it is a quotient, a slope of secants narrowing down to one of a tangent, and in integration, it is height $f(x)$ times width $dx$, i.e. the limit of Riemannian sums, see e.g.
https://www.physicsforums.com/insights/omissions-mathematics-education-gauge-integration/

For all practical purposes of integration, it is enough to consider $dx$ as just a notation which determines what the integration variable is, because all others, which don't change with $x$ are constants.

But u substitution uses dx beyond that.

If we want to integrate something like

int2x/(x^2+1)dx

you set up a u substitution

u = x^2 + 1
du = 2xdx

And then you replace the 2xdx in the original integral with du.

So dx and du are being used in ways other than just to define what we’re integrating with respect to. All of calculus makes sense to me up until the point where we start using differentials in ways that aren’t to define what we’re integrating with respect to or in the form of something like dy/dx.

 

[rootone]

'd' is a difference in some measurement, 'x' just means 'something'.
You could be unconventional and call it d, but you won't be thanked for that.

 

[fresh_42]

But u substitution uses dx beyond that.

If we want to integrate something like

int2x/(x^2+1)dx

you set up a u substitution

u = x^2 + 1

Which is indeed $u(x)=x^2+1$, that is a function, that depends on the variable $x$.

du = 2xdx

And then you replace the 2xdx in the original integral with du.

Which is due to the chain rule. We have $[f(g(x))]' = f\,'(g(x))\cdot g'(x)$. Now write $u(x)=u$ instead of $g(x)$.

So dx and du are being used in ways other than just to define what we’re integrating with respect to.

No. We just changed horses and adjusted the way to ride to the kind of horse we have. In both cases they signal the integration variable, but as we didn't simply wrote $u$ for $x$ but substituted an entire expression in $x$ by just a $u$, we had to adjust $dx$ as well. Maybe the better comparison is, that we changed the scaling on our $x-$ axis and this affects the measurement of our width, which had been in $x-$ units and is now in $u-$units. This is different from just renaming it. Afterwards, we end up with an expression, where $du$ denotes the integration variable again, now along the $u-$axis.

All of calculus makes sense to me up until the point where we start using differentials in ways that aren’t to define what we’re integrating with respect to or in the form of something like dy/dx.

Edit: Let's take an example. We want to calculate the distance traveled in time $0$ to $60\,sec$ at a velocity of $2\,m$ per second. Then we have
$$
distance \,= \int_{t=0}^{t=60} v(t)\,dt = \int_{0\,sec}^{60\,sec} 2\,\frac{m}{sec} dt = 2\,m \cdot (60 - 0) = 120\,m
$$
Now change the seconds to minutes. You cannot go ahead with $dt$ measured in seconds anymore, you'll have to rescale it to minutes, too, for otherwise the units wouldn't fit anymore.

 

[FS98]

Which is indeed $u(x)=x^2+1$, that is a function, that depends on the variable $x$.

Which is due to the chain rule. We have $[f(g(x))]' = f\,'(g(x))\cdot g'(x)$. Now write $u(x)=u$ instead of $g(x)$.

No. We just changed horses and adjusted the way to ride to the kind of horse we have. In both cases they signal the integration variable, but as we didn't simply wrote $u$ for $x$ but substituted an entire expression in $x$ by just a $u$, we had to adjust $dx$ as well. Maybe the better comparison is, that we changed the scaling on our $x-$ axis and this affects the measurement of our width, which had been in $x-$ units and is now in $u-$units. This is different from just renaming it. Afterwards, we end up with an expression, where $du$ denotes the integration variable again, now along the $u-$axis.

Edit: Let's take an example. We want to calculate the distance traveled in time $0$ to $60\,sec$ at a velocity of $2\,m$ per second. Then we have
$$
distance \,= \int_{t=0}^{t=60} v(t)\,dt = \int_{0\,sec}^{60\,sec} 2\,\frac{m}{sec} dt = 2\,m \cdot (60 - 0) = 120\,m
$$
Now change the seconds to minutes. You cannot go ahead with $dt$ measured in seconds anymore, you'll have to rescale it to minutes, too, for otherwise the units wouldn't fit anymore.

It still feels like those differentials a being used in ways that I’m totally unfamiliar with.

I understand that something like dy/dx is shorthand for the limit definition of a derivative.

I also understand that something followed by dx with an integral sign in front of it is shorthand for a Riemann sum as the value of n approaches infinity.

I have no idea what dx means outside of these 2 things. I’m able to use it in some situations such as u substitution because I’ve seen examples done, but I have no idea what it actually means outside of knowing that a couple different expressions involving dx are shorthand for something that I do understand.

 

[fresh_42]

I have no idea what dx means outside of these 2 things. I’m able to use it in some situations such as u substitution because I’ve seen examples done, but I have no idea what it actually means outside of knowing that a couple different expressions involving dx are shorthand for something that I do understand.

If you want the full dose, then you'll find it here:
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
You can consider $dx$ as a basis vector of the tangent space. Of course for functions $f \, : \,\mathbb{R} \longrightarrow \mathbb{R}$ this will be the only basis vector, as tangents in this case are always lines.