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What does it mean to take $F(x, y)$ as $F(x, y(x))$? I just do, not, get the difference, despite numerous explanations.

To put it into context, I see this done when applying the chain rule to $F$ to find $\frac{\mathrm{d}F}{\mathrm{d}x}$.

So what does it mean? What is the difference between $F(x, y)$ and $F(x, y(x))$?

i.e. take $F(x, y) = x^2 + y^3 + 2y$. What would $F(x, y(x))$ be equal to?

Please dumb explanations down, and don't skip steps, as I've heard numerous explanations and don't get any of them!

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    $\begingroup$ Of course $F(x, y(x))$ is just $x^2+(y(x))^3+2y(x)$. The point here is that you can define a function of single variable this way. Specifically, $f(x)=F(x,y(x))$. If, say, $y(x)=x$, then $f(x)=x^2+x^3+2x$, as a specific example. $\endgroup$
    – lulu
    Sep 28 '20 at 23:24
  • $\begingroup$ @lulu When why bother representing it as two variables in the first place, if we can just reduce it to a single variable? $\endgroup$
    – kentrid
    Sep 29 '20 at 0:20
  • $\begingroup$ Because you may not know in advance $\textit {which}$ function of one variable you'll want. $\endgroup$
    – lulu
    Sep 29 '20 at 10:03
  • $\begingroup$ In real world situations, things you wish to measure are, generally, functions of many variables. If you want to understand them, it often helps to restrict the variables. Maybe you hold all but one of them constant and vary the other. Maybe you make all the variables a simple function of a single variable. And so on. In these cases, you "want" to understand the function of many variables and these specialized functions are ways to get a handle on it. $\endgroup$
    – lulu
    Sep 29 '20 at 10:06
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$y(x)$ seems to denote a function in the variable $x$. So if $F(x,y)$ is given for $(x,y) \in \mathbb{R}^2$, $F(x,y(x))$ means the evaluation of $F$ at $(x,y(x))$ : you get a function for the variable $x$.

Your confusion must be the double use of $y$ as a variable and as a function. Rather, call $\varphi(x)$ instead of $y(x)$. Then you are actually just looking at $F(x,\varphi(x))=x^2+\varphi(x)^3+2\varphi(x)$ (you replace the variable $y$ by the value $\varphi(x)$) and the chain rule explain how to get the derivative of $x \mapsto F(x,\varphi(x))$ in terms of the partial derivatives of $(x,y) \mapsto F(x,y)$

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  • $\begingroup$ This makes a bit more sense. But how do we know $y$ is a function of $x$ (or can be represented as one)? $\endgroup$
    – kentrid
    Sep 29 '20 at 0:19
  • $\begingroup$ The variable $y$ and the function $x \mapsto y(x)$ are different things. The variable $y$ is not a function. It is just that in your exemple you replace the variable $y$ by a function $x \mapsto y(x)$. But I don't know much about the context of your problem $\endgroup$
    – user598294
    Sep 29 '20 at 0:30
  • $\begingroup$ So I understand we are changing the variable $y$ to some arbitrary function of $x$. But why bother? $\endgroup$
    – kentrid
    Sep 29 '20 at 0:41

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