$\frac{\partial}{\partial y_i}(f(x-y))=-\frac{\partial f}{\partial y_i}((x-y))$

Question

If I have $f:\mathbb{R^n} \rightarrow \mathbb{R}$, a sufficiently regular function. Then how to prove that:

$\frac{\partial}{\partial y_i}(f(x-y))=-\frac{\partial f}{\partial y_i}((x-y))$,

where $y_i$ is the $i^{th}$ component of vector $y \in \mathbb{R^n}$. And the second term means:the partial derivative of $f$ with respect to $y_i$ (applied) to $x-y$

Answer 1

Differentiating $f$ with respect to $y_i$ at the point $(x-y)$ can be done using the chain rule for higher dimension. It states that $$\frac{\partial}{\partial y_i} (f(x-y)) = \sum_{k=1}^n \frac{\partial f}{\partial y_k}(x-y) \cdot \frac{\partial (x_k-y_k)}{\partial y_i}(y) = -\frac{\partial f}{\partial y_i}(x-y)$$ where the $y_k$ are the variables you can put into $f$. Therefore the $i$-the partial derivative of the function $f \circ (x-y)$ equals the negative of the $i$-th partial derivative of $f$ at the point $(x-y)$ which was to be shown

Answer 2

We have a function $$f:\quad{\mathbb R}^n\to{\mathbb R}, \qquad u\mapsto f(u)$$ of one vector variable, and form the new function $$g(x,y):=f(x-y)$$ of two vector variables. Expanded this means $$g(x_1,\ldots, x_n,y_1,\ldots, y_n):=f(x_1-y_1,\ldots, x_n-y_n)\ .$$

By the chain rule we then have $${\partial g\over\partial y_i}={\partial f\over\partial u_i}\biggr|_{(x-y)}\cdot(-1)=-f_{.i}(x-y)\ ,$$ where $f_{.i}$ denotes the $i^{\rm th}$ component of the gradient.

Note that writing $((\ldots))$ is the same as $(\ldots)$, so that your equation says $\Phi=-\Phi$ for a certain $\Phi$. This cannot generally be true in your situation. The actual problem you have is one of notation: using the same letter $f$ for various purposes.