Partial Derivatives of Multi-Variable Functions

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!

• 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 – Calvin Khor Jan 30 at 22:13
• Probably the second derivative is with respect to y. Try to use the chain rule! – Thomas Jan 30 at 22:13
• Chain rule is the key word. – callculus Jan 30 at 22:13
• 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 – joseph Jan 30 at 23:25

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.
• 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. – joseph Jan 31 at 13:09
• 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 $\mathbf{R}^q$). – Will M. Jan 31 at 17:45