General

[quote=Chrysalids]I have gone through 3 pages of comments, so we all know dy/dx in the given example is 5.

Your goal here is to find the equation of the tangent line at a given point, (2, 3).

By using point-slope form, y = y1 + m*(x - x1), m is the slope which is dy/dx

Equation of the tangent line will be y = y1 + (dy/dx)*(x - x1) = 3 + 5*(x - 2) --> y = 5x - 7.

I hope this will clear things up.[/quote]
The goal for the TS here was to show an appropriate way to express dy/dx = x + y at the point given (2,3).

@ Whoever mentioned partial derivatives:
This isn't necessarily that advanced. My guess is it is some sort of implicit function.

[quote=cchpm]ok first of all, what class are you taking?
If you are trying to find a tangent line of a function f(x,y), you would need to do a partial derivative of x and the partial derivative of y.[/quote]

Although I only assumed, it definitely can't be about partial derivatives class, cause if it is you would have been given the original function, instead of dy/dx.

If you'd argue you could just integrate with respect to x to get the original function, what if the original function was f(x,y) = (x^2)/2 + xy + y^3? Then it would have been more troublesome

Just say it's the definition of a derivative: the slope of a tangent line.

[quote=esporteen]@WinterMyth: Alright if you are forced to use that, I would define:
y = y(x,y) <- yes it seems a bit confusing but the y's in y(x,y) are two separate y's if that makes sense. The first being y being the function itself, and y being a coordinate. Or if you want to know y(x,y) means "a function y evaluated in terms of x and y"

Then you have dy/dx right? This might seem tedious but its the same as writing:
(d/dx)*y <- so the other poster was correct BUT I would DEFINITELY SUGGEST you to define y as I did from above so as to avoid potential "oh you weren't precise on what y is" that the teacher might tell you.

[Note: (d/dx) is just a notation which means derivative with respect to x, and you usually can sort of isolate your function, say make dy/dx as (d/dx)*y which will still have the same meaning]

So you have:
dy/dx = (d/dx)*y = (d/dx)*y(x,y) = (d/dx)*y(2,3)

which you can solve from here I believe.[/quote]

Okay, thanks again!

ok first of all, what class are you taking?
If you are trying to find a tangent line of a function f(x,y), you would need to do a partial derivative of x and the partial derivative of y.

@WinterMyth: Alright if you are forced to use that, I would define:
y = y(x,y) <- yes it seems a bit confusing but the y's in y(x,y) are two separate y's if that makes sense. The first being y being the function itself, and y being a coordinate. Or if you want to know y(x,y) means "a function y evaluated in terms of x and y"

Then you have dy/dx right? This might seem tedious but its the same as writing:
(d/dx)*y <- so the other poster was correct BUT I would DEFINITELY SUGGEST you to define y as I did from above so as to avoid potential "oh you weren't precise on what y is" that the teacher might tell you.

[Note: (d/dx) is just a notation which means derivative with respect to x, and you usually can sort of isolate your function, say make dy/dx as (d/dx)*y which will still have the same meaning]

So you have:
dy/dx = (d/dx)*y = (d/dx)*y(x,y) = (d/dx)*y(2,3)

which you can solve from here I believe.

[quote=WinterMyth]Alright, thanks. But let's just say that you were forced to express it using dy/dx and strictly only dy/dx, how would you have done it?[/quote]
I'm not entirely sure it's possible since in d(y)/dx, y is supposed to be a function of x and dx is a infinitely small change in the value of x.... it'd be hard to do without knowing the original function.... honestly I would just write down dy/dx=2+3=5 and be like, "screw this, the teacher knows exactly what I mean"
In this question you're solving for dy/dx for a specific point, and therefore the point needs to be specified (though I think it already is in the original problem?)

[quote=esporteen]Alright then let's define this as follows:

Let f = f(x,y) <- which just simply means f is evaluated in terms of x and y
Then obviously f'(x,y) is just the first derivative but in terms of what?
It could either be:
f'(x,y)|x <- derivative in terms of x (I don't know how you are thought to denote such things, but usually you write f'(x,y) with a long vertical bar next to it then at the lower right corner of the bar write x)
OR
f'(x,y)|y <- derivative in terms of y

Since you are given dy/dx which means derivative of y with respect to x, then you use f'(x,y)|x

so you have dy/dx = f'(x,y) = x + y
f'(2,3) = 2 + 3[/quote]

Alright, thanks. But let's just say that you were forced to express it using dy/dx and strictly only dy/dx, how would you have done it?

ugh I hate differential calculus
edit: ooh wait it's not that bad! gimme a sec...
just assign values to x and y.
Before saying dx/dy=somethingoranother, say x=2 and y=3
then plug them in to the equation to get dx/dy
done!

[quote=WinterMyth]dy/dx is the equation.

Okay, let me try to explain it from another view.

let's say I was given x=2 and f'(x)= 3x

to show that you substituted the work in, you would show it like this: f'(2)= 6

I'm looking for a way to show the substitution work with a set of coordinates in a dy/dx expression*, because if you just said dy/dx, you're just stating the statement and not the substitution work.[/quote]

Alright then let's define this as follows:

Let f = f(x,y) <- which just simply means f is evaluated in terms of x and y
Then obviously f'(x,y) is just the first derivative but in terms of what?
It could either be:
f'(x,y)|x <- derivative in terms of x (I don't know how you are thought to denote such things, but usually you write f'(x,y) with a long vertical bar next to it then at the lower right corner of the bar write x)
OR
f'(x,y)|y <- derivative in terms of y

Since you are given dy/dx which means derivative of y with respect to x, then you use f'(x,y)|x

so you have dy/dx = f'(x,y)|x = x + y
f'(2,3)|x = 2 + 3

[quote=esporteen]You are given dy/dx = x + y right? You plug in the coordinates on this equation to get:

dy/dx = 2 + 3 , this gives you the slope of the tangent line AT THIS GIVEN coordinate. I don't know what you're really asking but its rather straightforward from here on out[/quote]

dy/dx is the equation.

Okay, let me try to explain it from another view.

let's say I was given x=2 and f'(x)= 3x

to show that you substituted the work in, you would show it like this: f'(2)= 6

I'm looking for a way to show the substitution work with a set of coordinates in a dy/dx expression*, because if you just said dy/dx, you're just stating the statement and not the substitution work.

@LowWillpower

Is that an accurate way to display the substitution work? I have to be 100% sure about this.

You are finding the slope of the function y at the point (2,3), aka y(2,3), so I would write it as

d/dx ( y(2,3)) = (2) + (3)
or
d(y(2,3))/dx = (2) + (3)

[quote=WinterMyth]I know how to find the tangent line, but how do you show in plugging a set of coordinates into dy/dx with (2,3)?[/quote]

I don't think there's any particularly fancy way of doing it. All I've ever done is:

dy/dx= x+y
dy/dx = (2) + (3)
dy/dx = 5

[quote=WinterMyth]I know how to find the tangent line, but how do you show in plugging a set of coordinates into dy/dx with (2,3)?[/quote]

You are given dy/dx = x + y right? You plug in the coordinates on this equation to get:

dy/dx = 2 + 3 , this gives you the slope of the tangent line AT THIS GIVEN coordinate. I don't know what you're really asking but its rather straightforward from here on out

[quote=esporteen]Read what I said on my comment [/quote]

I know how to find the tangent line, but how do you show in plugging a set of coordinates into dy/dx with (2,3)?

I know the equation of the tangent line and everything, I know how to get the answer, but I can't show the work with the substituting because I don't know how to write out how I'm showing that I'm substituting.

[quote=WinterMyth]Sorry, I should have been more clear. I forgot to mention "find the tangent line"

If I were given this set of data, and I had to find the tangent showing work, how do I display plugging in the set of coordinates into dy/dx to find the slope?[/quote]

Read what I said on my comment

Sorry, I should have been more clear. I forgot to mention "find the tangent line"

If I were given this set of data, and I had to find the tangent showing work, how do I display plugging in the set of coordinates into dy/dx to find the slope?

I don't really know what you're trying to show in the end, but I'll tell you a few stuff:
(1) dy/dx stands for the derivative of y with respect to x, which alternatively can be known as the slope of the line
(2) If you are looking for the final "equation of the line" you can use the formula:
" y - y1 = m*(x-x1)" where x1 and y1 stands for the coordinates you are given, and m stands for the slope of the line which you can easily calculate to be 5

@all

What if I was given dy/dx=x+y and a set of coordinates (2,3) and my goal was to find the tangent line?

I would have to plug in the coordinates to show the work to find the slope, but how do I show that properly?

are you confusing it with u-substitution maybe?

substitution with derivatives is only used to represent a complex variable with a simple one for example 3x^2 = u

[quote=swass]...Dy/dx is the derivative of y. You don't plug in the values like what you did with -> d(3)/d(2).[/quote]

And if you're given a derivative and a set of coordinates to solve for a tangent line?

"d" is neither a variable nor a constant, d stands for derivative
This isn't your every day math; welcome to the realm of calculus

EDIT: Ninja'd