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In Professor Leonard's lectures:

https://www.youtube.com/watch?v=aiBD9aI69C8&list=PLF797E961509B4EB5&index=25

... we are starting to integrate. He keeps using 'dx', our old friend from 'dy/dx' back in differentiation. What is 'dx'? If 'dy/dx' is a ratio, which it sure seems to be, then 'dx' = one:

f(x) = x^2
f'(x) = dy/dx = 2x = 2x/1 (obviously).  
So 'dy' = 2x and 'dx' = 1.

Solving this:

(integral) x^2(x^3-4)^5 dx

... he is careful to hang on to 'dx' through the whole chain of calculations but is seems to me that 'dx' is always one and can be ignored, no?

I've read 'similar' threads here, that ended up as long fights about 'dy/dx' being or not being a ratio, so I'm clearly not the first person to wonder about what it is. Maybe if I can get clear on what 'dx' by itself is, that will help. He treats it as a variable holding a value on which one performs algebra but we never see it's value. What is it's value?

Ray Andrews
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    https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio/21209#21209 – Arturo Magidin Dec 02 '19 at 22:54
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    Wait wait wait wait wait. How did you get from $\mathrm dy/\mathrm dx = 2x/1$ to $\mathrm dy=2x$ and $\mathrm dx=1$? Do you think that if $a/b=c/d$ then automatically $a=c$ and $b=d$? –  Dec 02 '19 at 23:10
  • I think that if any fraction is reduced to 'n' then that is implicitly 'n/1'. If 'dy/dx = 2x' then that is identical to 'dy/dx = 2x/1' and thus 'dx = 1'. – Ray Andrews Dec 02 '19 at 23:55
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    @RayAndrews: $\frac{6}{3}=2=\frac{2}{1}$, but one does not say $6=2$ and $3=1$. –  Dec 03 '19 at 01:10
  • No, but we do say that 1 = 1. If 2/1 = 2, then if we write '2/x', 'x' = 1. – Ray Andrews Dec 03 '19 at 01:24

1 Answers1

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In many contexts, you can read $dx$ as "With respect to $x$".

When you get the derivative of a function or attempt to find the anti-derivative (by integration), you need to know what variable is to be used.

In a function such as $y=x^2$, $y'(x)=\frac{dy}{dx}=\frac{d}{dx}x^2=2x$.

In the above case we formulated the derivative in "with respect to $x$".

When you write an integral you must specify that you want to perform the integration with respect to a specific variable also.

$$\int \:2x\:dx=x^2+c$$

This says "find the anti-derivative of the function $2x$ with respect to the variable $x$".

Note that if $x$ was not a variable and was a constant instead, the result would differ. Maybe you are familiar with the constant $e=2.71828182846...$

if you write $y=e^2$ then:

$\frac{dy}{dx}=\frac{d}{dx}e^2=zero$.

There are functions of more than one variable such as $y=zx$ where both $z$ and $x$ are variables. Again you need specify which variable you want to use to find the derivative (called partial derivative in this case). The result will differ based on the variable chosen. for example to take the derivative of $y(x,z)$ with respect to $x$ we could write:

$$\frac{\partial }{\partial x}y\left(x,z\right)=\frac{\partial }{\partial x}\left(xz\right)=z$$

The partial derivative with respect to $z$ is:

$$\frac{\partial }{\partial z}y\left(x,z\right)=\frac{\partial }{\partial z}\left(xz\right)=x$$

Additional Note: When given an equation like: $\frac{dy}{dx}=2x$

You could re-write it as:

$dy=2x dx$

If you integrate the left hand side with respect to $y$ and the right hand side with respect to $x$, you get $y=x^2+c$.

In this process one may be lead to think that $dx$ is a variable.

In calculus, the differential represents the principal part of the change in a function $y = f(x)$ with respect to changes in the independent variable. The differential $dy$ is defined by: $$dy=f'(x).dx $$

A more detailed explanation (that is beyond basic Calculus) is Differential of a function.

I don't have enough knowledge to rule out the usage of the symbols as independent variables in the general case though.

NoChance
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  • "There are functions of more than one variable such as y=zx where both z and x are variables."

    Ok, that helps. So it is just information then? A reminder of the variable to be processed? But if that is so, how come Prof. L does algebra with/on it? It treats it exactly like a variable holding a value.

     – Ray Andrews Dec 02 '19 at 23:47
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    @RayAndrews He should not treat dx as a variable, if he does, he is sloppy. – Moishe Kohan Dec 03 '19 at 00:35
  • Treating $dx$ as a variable shifts the discussion to the concept of "function differential" - Please see my edit above. – NoChance Dec 03 '19 at 08:07
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    @Ray Andrews Note also that in physical situations, dx carries units of the variable x and is required for all of the units to balance out. It is an entity, which in most ways follows the rules of algebra, except you can't assign a value to it. You may want to look up the term "differential". – electronpusher Dec 03 '19 at 08:18