0
$\begingroup$

Suppose that we have a function $f(x)$ of one variable, like $f(x)=x$.

And I want to derive this function with respect to another multivariate function $g(x,y)=x-y$.

My hypothesis is that the derivative is equal to $0$:

$$\frac{{\partial f(x)}}{{\partial g(x,y)}}=0$$

My reason to think that is that $f(x)$ can never be expressed in terms of $g(x,y)$, so $g(x,y)$ does not impact directly $f(x)$ in any way. Also $x$ and $y$ are independent, meaning they are not related, there is no function between them.

Or maybe is undetermined?

But if that is true, I would need to prove it. I would appreciate if you could help me.

1 Answers 1

-1

Basically it's a normal derivative, but the fact that it does operate on the variable $(x-y)$.

For example:

$$\frac{\partial (x-y)^2}{\partial (x-y)}$$

By calling $(x-y) = z$, you can find the usual derivative:

$$\frac{\text{d} z^2}{\text{d}z} = 2z \longrightarrow 2(x-y)$$

as it shall be.

In your case your function is simply $x$.

By a change of variable like $x-y = z$ you find that $x = z-y$ hence your derivative is like

$$\frac{\text{d} (z-y)}{\text{d}z} = \frac{\text{d}}{\text{d}z}(z-y) = 1$$

In any case, let me tell you that that is a very ill notation.

  • 0
    If I see it with out a change of variable $\frac{\text{d}(x)}{\text{d}(x-y)}$ what we consider as a new variable does not exist in the function, so that is why I think the derivative would be zero, – 2017-01-24
  • 0
    I also thought that with they change of variable $\frac{\text{d}(z+y)}{\text{d}z} = \frac{\text{d}z}{\text{d}z} + \frac{\text{d}y}{\text{d}z} = 1-1=0$. – 2017-01-24
  • 0
    You're wrong: if you make that change of variable, then $dy/dz$ is zero, not $-1$. – 2017-01-24
  • 0
    Also, you can always write $x$ as $x - y + y$ that is: you can always find that the derivative is $1$ in this case. – 2017-01-24
  • 0
    Finally, your $g(x, y)$ is somehow related to $f$, because $x-y$ is just $x$ minus $y$. It's not like $dx/dz$; there is a hidden connection between $x$ and $x-y$. – 2017-01-24
  • 0
    Is weird to think that if $d(x-y+y)/d(x-y)=1$, then by separating $d(x-y)/d(x-y)+dy/d(x-y)$ would imply that $dy/d(x-y)$ is zero, and would be exactly the same case as $dx/d(x-y)$. – 2017-01-24
  • 0
    It's not zero. Can't you really see the indirect connection between the functions? – 2017-01-25
  • 0
    As I said, the best way to visualize it is to perform a change of variable like $x-y = z$ and run the derivative in the usual way. – 2017-01-25
  • 1
    Ok, I see it now, thank you for your patience and time. – 2017-01-25
  • 1
    Just because the joint effect can't be isolated does not mean that it does not affect the function. I think I get it now. – 2017-01-25