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Let $f(x)$ be a differential function and $F(x,y)=f(x-y)$. Show that $$ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} = 0 $$

I have no idea where to even begin with this problem. A small hint to start would be a great help!

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    $\begingroup$ for starters you probably don't mean that $2\frac{\partial F}{\partial x} = 0$. Once you copy the correct problem, try writing $F = f \circ g$ and use chain rule $\endgroup$ – Calvin Khor Jan 30 at 22:13
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    $\begingroup$ Probably the second derivative is with respect to y. Try to use the chain rule! $\endgroup$ – Thomas Jan 30 at 22:13
  • $\begingroup$ Chain rule is the key word. $\endgroup$ – callculus Jan 30 at 22:13
  • $\begingroup$ Yes sorry for the mistake I just fixed it. I'm not sure what you mean by the chain rule as it isn't stated what f(x) is $\endgroup$ – joseph Jan 30 at 23:25
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If $T(x, y) = x - y,$ then $T$ is linear and $T'(x,y) = T.$ By the chain rule, $F'(x, y) = f'(x - y) \circ T'(x, y) = f'(x - y) T$ (the latter being a scalar multiple of a linear function). Identifying $T$ with its canonical matrix $[1, -1],$ we reach $\partial_x F + \partial_y F = f'(x - y)(1 - 1) = f'(x - y) (0) = 0.$ Q.E.D.

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  • $\begingroup$ How did you identify T with its matrix and how did you use it? It looks like you just plugged it in to cancel out the $f'(x-y)$ term. $\endgroup$ – joseph Jan 31 at 13:09
  • $\begingroup$ You have so many gaps, it is difficult to help. But I'll try. The canonical matrix of a linear function $u:\mathbf{R}^p \to \mathbf{R}^q$ is the matrix whose columns are $u(e_{p,i})$ (where $e_{p,i}$ is the canonical basis of $\mathbf{R}^p$). Then, $T(1, 0) = 1$ and $T(0, 1) = -1,$ making the matrix of $T$ to be $[1, -1]$ as claimed. Now, if the linear function $u:\mathbf{R}^p \to \mathbf{R}^q$ is the derivative of $F$ at the vector $x$ then $\mathbf{D}_i F_j(x) = u(e_{p,i})_j$ (the $j$th entry of the vector $u(e_{p,i})$ of $\mathbf{R}^q$). $\endgroup$ – Will M. Jan 31 at 17:45

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