Calculating Partial Derivative of F(u,v) w.r.t u

[-EquinoX-]

Homework Statement

Let F(u,v) be a function of two variables. Find f '(x) for f(x) = F(x, 6).

Homework Equations

The Attempt at a Solution

I need to find the answer in terms of F_u, how can I do this?

 

[Dick]

Remember what the definition of F_u is.

 

[-EquinoX-]

I am confused what are we supposed to find the derivative of F_u with respect to what? x or y?

 

[Dick]

F_u is the derivative of F(u,v) with respect to the first variable with the second variable held constant, right? f'(x) looks like pretty much the same thing.

 

[-EquinoX-]

my guess is that it will be something like:

F_u * v(6)

is this right?

 

[Dick]

What does that mean? Try an example f(u,v)=u^2-u*v. What's F_u? What is f'(x)?

 

[-EquinoX-]

I assume that F_u is the derivative of f(u,v)

f(u,v)=u^2-u*v

is just:

(u -v) is this correct?

 

[Mark44]

I assume that F_u is the derivative of f(u,v)

No, F_u is the partial derivative of F(u, v). You calculate it by assuming that v is constant and that only u is changing. F_u can also be written as \partial F/ \partial u

f(u,v)=u^2-u*v

is just:

(u -v) is this correct?

No.
One way to to calculate this is to take the limit:
\lim_{h \rightarrow 0} \frac{F(u + h, v) - F(u, v)}{h}

 

[Dick]

You mean F_u(u,v)=2u-v, yes? Now what's are f(x) and f'(x)?

 

[-EquinoX-]

f(x) is therefore 2x-6 right? and f'(x) is 2 ?

 

[Mark44]

Equinox,
Don't forget that you're answering Dick's question about a specific example (post 6), not the question you first posted. He's trying to get you to think about this the right way.

 

[-EquinoX-]

Yes I am aware of that.. I am supposed to find the relation between the example he's given and the real answer to my question. I believe so the answer is then F_u(x,6) ?

 

[Dick]

If you mean f'(x)=F_u(x,6), yes. f(x)=2x-6, and f'(x)=2 are NOT right.

 

[-EquinoX-]

so f'(x)=F_u(x,6) is not correct?

 

[Dick]

so f'(x)=F_u(x,6) is not correct?

Yes it is. The other two things aren't.

 

[-EquinoX-]

what if a variation of the question asks for f(x) = F(x, x), is it just then 0?

 

[Dick]

No, why would you say that? Wouldn't you need to think about F_v as well? v isn't fixed anymore.

 

[-EquinoX-]

Hmm..is x here actually a number or is x another function?

 

[Dick]

Actually, sorry, I meant yes. For the specific function F(u,v)=u^2-u*v, sure f(x)=F(x,x)=0. f'(x)=0. Sorry, I misspoke. But you can't express f'(x) purely in terms of F_u.

 

[-EquinoX-]

hmm.. I tried to input 0 as the answer and it didn't accept it..

 

[Dick]

Why did you input 0? I said that was only for that particular function that I made up in my head. For a general function F(u,v), f'(x) is going to depend on F_u AND F_v.

 

[-EquinoX-]

so it is F_u + F_v ?

 

[Dick]

Yes it is. But why do you say that? It would be really helpful if you would give reasons or show a calculation instead of just pulling the answer out of a hat.

 

[-EquinoX-]

I think it's just because of the definition of the chain rule. As now u and v varies with x and both are not constants therefore I have that conclusion in mind

 

[Dick]

Good. Yes, it's the chain rule. More specifically the answer is F_u(x,x)+F_v(x,x), right?

 

[-EquinoX-]

that is correct! thanks for helping me out Dick I appreciate it