What is the partial derivative of f with respect to w?

[slr77]

Homework Statement

Define f(x,y) = x+2y, w = x+y. What is ∂f / ∂w?

Homework Equations

The Attempt at a Solution

f = w+y so:

∂f/∂w = ∂(w+y)/∂w = ∂w/∂w + ∂y/∂w = 1 + ∂y/∂w. But I'm really not sure if this is right and if it right so far, I can't figure out what ∂y/∂w should be...

 

[RyanH42]

∂f/∂w = ∂(w+y)/∂w

This is right.

Now partial derivative means the derivative of given parameter.The other parameters will be assumed constant.Cause they are not changing respet to that parameter.I mean If I set two parameter a and b and make a fuction using them (It can be anything).Lets call it f(a,b)=a+b then ∂f/∂a means derivative of function respet to a not b.So we will assume b is constant and ∂f/∂a=1 so.Do the same thing.

 

[mfb]

@RyanH42: the problem here: which other variable is supposed to be constant? Why should it be y (as you seem to suggest), not x, or x-y? Those would lead to different answers.

 

[slr77]

This is right.

Now partial derivative means the derivative of given parameter.The other parameters will be assumed constant.Cause they are not changing respet to that parameter.I mean If I set two parameter a and b and make a fuction using them (It can be anything).Lets call it f(a,b)=a+b then ∂f/∂a means derivative of function respet to a not b.So we will assume b is constant and ∂f/∂a=1 so.Do the same thing.

But w depends implicitly on y so can I really take y as constant? If I get y in terms w (y = w-x) and continue this way (so 1 + ∂y/∂w = 1 + ∂(w-x)/∂w). I just get an endless chain of 1+1+1+1+1... That's why I think what I'm doing is not right.

 

[RyanH42]

I don't know but he answer might be this ∂f/∂w=∂f/∂y.∂y/∂w+∂f/∂x.∂x/∂w look this rule.

 

[slr77]

I don't know but he answer might be this ∂f/∂w=∂f/∂y.∂y/∂w+∂f/∂x.∂x/∂w look this rule.

I think this is the chain rule but we haven't learned that yet. I'll read ahead and come back to this and make sense of it but apparently there should be a way to do this without directly making use of the chain rule.

 

[RyanH42]

I have an idea ∂f/∂w=1 + ∂y/∂w now ∂f/∂w=1 + 1/∂w/∂y then ∂f/∂w=2 Its a trick but I don't know its ture or not.I know it was stupid idea

 

[verty]

This problem can't be solved, it is underspecified.

I have an idea ∂f/∂w=1 + ∂y/∂w now ∂f/∂w=1 + 1/∂w/∂y then ∂f/∂w=2 Its a trick but I don't know its ture or not.I know it was stupid idea

So ∂y/∂w + 1/∂w/∂y = 0? It wasn't actually a stupid idea, you just made an algebra mistake.

 

[slr77]

This problem can't be solved, it is underspecified.

Hmm, ok. I think I should have posted the full problem because I think it's more open ended than what my original post conveys:

I'm just treating w as the variable and going from there but maybe that's not the right definition? So is there some way to do this problem that makes sense?

 

[verty]

Hmm, ok. I think I should have posted the full problem because I think it's more open ended than what my original post conveys:

View attachment 85560

I'm just treating w as the variable and going from there but maybe that's not the right definition? So is there some way to do this problem that makes sense?

I still maintain this is a flawed question and you should move on. It doesn't contain enough information to answer it.

 

[RyanH42]

So ∂y/∂w + 1/∂w/∂y = 0?

Is this true ? or You mean ∂w/∂w + 1/∂w/∂y ? I am confused

 

[RyanH42]

If you use chain rule you get 2 again.I think answer is 2.
$∂f/∂w=∂f/∂y.∂y/∂w+∂f/∂x.∂x/∂w$
$∂f/∂y=2$
$∂y/∂w=1/2$
$∂f/∂x=1$
$∂x/∂w=1$
So answer is 2 I guess.

Whats your's idea ?

 

[verty]

$f(x,y) = x + 2y$
${∂f \over ∂w} = {∂x \over ∂w} + 2 {∂y \over ∂w}$

The problem happens because we don't know what ${∂x \over ∂w}$ and ${∂y \over ∂w}$ are, we don't have enough information to determine them. If this isn't clear, be sure to look again at partial derivatives and what they mean.

 

[mfb]

$∂y/∂w=1/2$
$∂x/∂w=1$

w=x+y. Why should the two derivatives be different?

Also, the rule doesn't work like that with partial derivatives. You could introduce arbitrary new parameters (e. g. z=x) and add more and more terms that would change the result.

 

[RyanH42]

Here my last idea then.The question ask's us for general solution and we do the general solution : ${∂f \over ∂w} = {∂x \over ∂w} + 2 {∂y \over ∂w}$ or
${∂f \over ∂w} = {∂f \over ∂x}{∂x \over ∂w} +{∂f \over ∂w}{∂y \over ∂w}$
Part b asks solve these equation with spesific parameters.I think there must be some difference between question a and b so I thought we can think x and y like numbers or actually constant parameters.So I mean f=x+2y is a constant cause x and y is contant so the answer is zero.

 

[RyanH42]

Forgive me but Why you guys stop answering the question.Theres a problem and you are avoiding to answer.If one of you find the answer he can tell here cause I am curios.

 

[ChrisVer]

The question (a) asks for the definition of the derivative of a general function f(x,y)...The definition of the derivatives is with limits I guess...even with or without limits, you can write your 2nd expression in p#15, and try to find the dx/dw, dy/dw.

(b) asks for the given function: f=x+2y

 

[mfb]

Maybe we are overthinking this. There is no explicit dependence on w in f...