Prove that $\frac{\partial}{\partial x}f(x-x')=-\left(\frac{\partial}{\partial x'}\right)f(x-x')$

Question

I haven't formally studied partial derivatives.

My attempt:

$\displaystyle\frac{\partial}{\partial x}f(x-x')=\frac{\partial f}{\partial x}(x-x')\times\left(1-\frac{d x'}{d x}\right)$

$\displaystyle\frac{\partial}{\partial x'}f(x-x')=\frac{\partial f}{\partial x'}(x-x')\times\left(\frac{d x}{d x'}-1\right)$

Also,

$\displaystyle\frac{\partial f}{\partial x'}(x-x')=\frac{\partial f}{\partial x}(x-x')\times\left(\frac{d x}{d x'}\right)$

How do I proceed?

Answer 1

The most straight forward approach is to utilise a variable transform (explicitly) $$ u = x - x' $$ then you have $$ \frac{\partial}{\partial x} = \frac{\partial u}{\partial x}\frac{\partial}{\partial u} = \frac{\partial}{\partial u}\\ \frac{\partial}{\partial x'} = \frac{\partial u}{\partial x'}\frac{\partial}{\partial u} = -\frac{\partial}{\partial u} $$ then we have $$ \frac{\partial}{\partial x}f(x-x') = \frac{\partial f(u)}{\partial u} $$ and the lhs $$ -\frac{\partial}{\partial x'}f(x-x') = -\left(-\frac{\partial}{\partial u}\right)f(u) = \frac{\partial f(u)}{\partial u} $$ both sides line up correctly.