What's the Difference Between dy/dx and d/dx in Calculus?

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The discussion clarifies the difference between dy/dx and d/dx in calculus, emphasizing that dy/dx denotes the derivative of y with respect to x, while d/dx indicates the derivative of whatever follows it with respect to x. It explains that when differentiating y as a function of x, dy/dx is appropriate, whereas d/dx is used for general differentiation. The conversation also touches on implicit versus explicit differentiation, noting that dy/dx is used when the relationship between y and x is not explicitly defined. Participants express confusion over notation but ultimately gain a clearer understanding of how these derivatives relate to the functions being analyzed. The thread concludes with a sense of resolution regarding the initial misunderstandings.
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I do not understand the difference between taking the derivative, and taking the derivative with respect to x, or taking the derivative with respect to y (or any other variable).

If I take the derivative of y = x^2, I get y' = 2x. What if I use the dy/dx or just the d/dx notation?

so

dy/dx y = dy/dx x^2
vs
d/dx y = d/dx x^2

another example I don't understand would be

dy/dx = 2x
vs
d/dx = 2x
vs
f'(x) = 2x

I know that the d refers to an infinitesimally small number, but I just don't understand the difference between the stuff I mentioned before.

Someone enlighten me?
 
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if you differentiate y(x^2 ) with respect to x, you get 2xy. if you differentiate y(x^2) with respect to y, you get x^2. What is going on is that one is seen as a function of x and the other is seen as a function of y. That is, when y(x^2) is a function of y, after fixing some x, x^2 is just a constant, so that differentiating y(x^2) is like differentiating cx with respect to x, resulting in c.
 
Sorry, I still don't understand.
 
Ok, so let's assume you have a function y = f(x). There's a lot of overlap in notation, as you'll see:

f'(x) means to take the derivative of y with respect to x. (same with y')

d/dx means to take the derivative of whatever's after it with respect to x. For example:

d/dx (y), would mean to take the derivative of y with respect to x.

dy/dx means to take the derivative of y with respect to x. The "numerator" indicates what function you're taking the derivative of. The "denominator" indicates what you're differentiating with respect to.
 
I actually hate the d/dx notations and similar...

guss said:
I do not understand the difference between taking the derivative, and taking the derivative with respect to x, or taking the derivative with respect to y (or any other variable).

If I take the derivative of y = x^2, I get y' = 2x. What if I use the dy/dx or just the d/dx notation?

The point is that y is actually a function, so it would be better to write y(x)=x^2. Then dy/dx just means the derivative of y with respect to x. So

\frac{dy}{dx}=y'

If you want to evaluate this in the point 2, then you write

\frac{dy}{dx}(2).

Sometimes, if y=x^2, for example, people will write

\frac{dx^2}{dx} instead of \frac{dy}{dx}

But I consider that to be very bad notation...

so

dy/dx y = dy/dx x^2
vs
d/dx y = d/dx x^2

The first notation doesn't really makes sense to me. The second would be

\frac{d}{dx}y:=\frac{dy}{dx}=y'

another example I don't understand would be

dy/dx = 2x
vs
d/dx = 2x
vs
f'(x) = 2x

The second notation doesn't make sense to me. The first does, but I think it's bad notation and I would never use it...

I know that the d refers to an infinitesimally small number, but I just don't understand the difference between the stuff I mentioned before.

Not everybody will agree with me, but don't think of d as infinitesimal number. Just think of d as a notation. Thinking of d as a number causes you to make mistakes, and in (standard) real numbers, there are no such things as infinitesimals...
 
Thanks guys, I think I'm starting to understand it.

gb7nash said:
The "denominator" indicates what you're differentiating with respect to.

I still don't understand what this means, though. What does "with respect to" really mean?
 
It's nothing spectacular, "with respect to" simply indicates the variable.

For example, if f(x)=2x, then f'(x)=2, and the notation would be df/dx
But we can also write f(z)=2z (this is the same function), then we would write df/dz.

This notation is useful for functions like f(x)=2a+x, where a is just a number. If we do not know what our variable is (x in this case), then we could both have df/dx or df/da. The dx in the bottom just serves as a reminder to what the variable of f is called...
 
Ahh, I understand now. Thanks!

But, last question. In explicit differentiation, d/dx is usually used to represent the change of the function with respect to x. However, in implicit differentiation, why is dy/dx used to represent the change of a function with respect to x?
 
When you do implicit differentiation, y is a function of x so when you take the derivative of y with respect to x you write it as a derivative of the function.

When you differentiate an explicit function of x you know how the function is dependent on x so you can explicitly take the derivative. You don't know how y depends on x, so you must leave it as dy/dx.
 
  • #10
I'm not really following, sorry. I think we have a misunderstanding in your second paragraph. I am just referring to an equation like y = 5x^2 or f(x) = 5x^2. Not a multivariable expression.
 
  • #11
Ah, I thought you meant implicitly differentiating a function like xy^2 = 2x/y or similar.

I'm not quite sure what you mean then by explicit and implicit differentiation.

As far as the notation does, d/dx is just a differential operator, meaning take the derivative w.r.t. x, where as dy/dx applies the operator to some function y.
 
  • #12
That is what I mean.

You said
When you differentiate an explicit function of x you know how the function is dependent on x so you can explicitly take the derivative. You don't know how y depends on x, so you must leave it as dy/dx.
It seems to me that you are talking about something like f(x) = x^2 + 6y.

Could you rephrase what you said before? Sorry for being unclear I am very new to this stuff haha.
 
  • #13
Yay, I finally understand. I was just overthinking it.

It's funny how the solution to something so simple can seem so amazing after finally understanding it.
 

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