If v'(x) = u(x), and y denotes some function, does

>>8061752
Sorry, meant:
Does v'(y) = y*u(y)?

>>8061752
What is v? What u? What is x?

Does v' mean the derivative?

>>8061759
Yes.

https://en.wikipedia.org/wiki/Notation_for_differentiation#Lagrange.27s_notation

v and u are both functions

>>8061752
>>8061756
No
>>8061766
functions of?

>>8061766
>v and u are both functions

I meant what were they exactly.

If you mean that this is general case for any functions v,u and y then I am sorry to tell you that even a toddler could construct a counter example.

That is why I supposed v and u were actually functions constrained in some way so that this general case would make sense.

>>8061752
>hamburger
>actually chicken sandwich
fug off

>>8061780
Those are any functions which satisfy the equalities. For example, in the OP, we can clearly see that u(x) is a function which is equal to the derivative of v(x).

>>8061752

If y is a function that uses x then no. It also needs to be derivated according to derivation rules.
If it's just a number or not a function that includes x then yes

File: 1355592808571.jpg (113KB, 500x500px) Image search: [Google]

113KB, 500x500px

this conversation has me wondering, what is the most general case in which a function can be said to be differentiable?

do the domain and codomain need to be uncountably infinite? or is it something simpler?

>>8061789
Yeah but so what?

That is not enough of a restriction to make any of this a true statement.

Also, v' might aswell just be another function because we get no information about what v is so who cares.

>>8061804
>what is the most general case in which a function can be said to be differentiable?
https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_higher_dimensions

>>8061756
No.

If you differentiate with respect to y:
v'(y) = u(y)

If you differentiate with respect to x:
v'(y(x)) = y'(x) u(y(x))

>>8061905
I also want to add to my post that Lagrange's notation isn't always clear. I prefer Euler' notation (or Leibniz's), since those notations allow you to mention the variable you differentiate to.
[math]\text{D}_{y}^1v(y)=u(y) \text{ and }
\text{D}_{x}^1v(y)=u(y)\text{D}_{x}^1y[/math]