0

I have a function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$

and a function $\phi: \mathbb{R}^{n-1} \rightarrow \mathbb{R}$

$f$ is defined as $f(x_1,...,x_n) = f(x_{1},...x_{i-1},\phi_{i}(x_{1},...,x_{i-1},x_{i+1},...,x_{n}),x_{i+1},...,x_{n})$ and differentiable.

for any $x_j\;$ such that $i\neq j\;$ why is $\frac{\partial f}{\partial x_{j}}=\frac{\partial f(x_{1},...x_{i-1},\phi_{i}(x_{1},...,x_{i-1},x_{i+1},...,x_{n}),x_{i+1},...,x_{n})}{\partial x_{j}}=\frac{\partial f}{\partial x_{j}}+\frac{\partial f}{\partial x_{i}}\cdot\frac{\partial\phi_{i}}{x_{j}}\;$

How is the chain rule applied here?

GoodWilly
  • 905
  • Your ‘definition’ of $f$ is inconsistent. If $f$ is given as a function of $n$ variables, how can it be defined as a function of $n-1$ variables? – Bernard Jun 20 '20 at 17:42
  • @Bernard, You're correct It should be $\mathbb{R}^{n-1}$ – GoodWilly Jun 20 '20 at 17:45
  • But in this case, you can't write $f(x_1,\dots,, x_n)$. – Bernard Jun 20 '20 at 17:50
  • @Bernard I got confused. why it doesn't have n variables? it just happens the one variable is a function of the others. – GoodWilly Jun 20 '20 at 17:52
  • I suspect you're given a function of $n$ variables , and they associate to this function a function of $n-1$ variables through $\phi$. – Bernard Jun 20 '20 at 17:56
  • @Bernard Correct, using the implicit function theorem. Could you explain the chain rule application though? – GoodWilly Jun 20 '20 at 17:58
  • It might be easier to understand this if you just let $i = n$ without loss of generality, define another function $g$ such that $g(x_1, x_2, \ldots, x_{n-1}) = f(x_1, x_2, \ldots, x_{n-1}, \phi(x_1, x_2, \ldots, x_{n-1}))$, and ask about $\frac{\partial g}{\partial x_j}$ for $1 \leq j < n$. (Assuming I understand the question.) – Brian Tung Jun 20 '20 at 18:02
  • In short: Given an infinitesimal change in $x_j$, there are two effects on the value of $g$. One comes from the direct dependence of $f$ on $x_j$, and one comes from the dependence of $f$ on $x_n$, which in turn depends on $x_j$ through $\phi$. In other words, only the second term of the addition on the right-hand side is an application of the chain rule. – Brian Tung Jun 20 '20 at 18:05
  • @BrianTung I understand it. Although the formal proof of the chainrule doesn't relate to such situations, thats why it feels a bit not formal to me – GoodWilly Jun 20 '20 at 18:07
  • Whatever proof you saw can, I'm sure, be straightforwardly extended to cover the present situation, provided appropriate constraints on $f$ and $\phi$ are set. – Brian Tung Jun 20 '20 at 18:21

0 Answers0