Transformation of differential operators Consider the variable transformation $(t,x) \mapsto (\xi,\eta)$
$$ \xi = t - \frac{x}{c}, \hspace{1cm} \eta = t + \frac{x}{c}.$$
How to transform then the operators ($\frac{\partial}{\partial x},\frac{\partial}{\partial t}$) to ($\frac{\partial}{\partial \xi}, \frac{\partial}{\partial \eta}$) ? They are
$$ \frac{\partial}{\partial \xi} = \frac12 \left( \frac{\partial}{\partial t} - c \frac{\partial}{\partial x} \right) , \hspace{1cm} \frac{\partial}{\partial \eta}  = \frac12 \left( \frac{\partial}{\partial t} + c \frac{\partial}{\partial x} \right),$$
but I cannot derive them.
 A: Using the standard change of variable for derivatives 
$$
\partial_\zeta = \partial_\zeta x \partial_x + \partial_\zeta t \partial_t \\
\partial_\eta= \partial_\eta x \partial_x + \partial_\eta t \partial_t 
$$
we also have
$$
2t = \eta + \zeta \to t = \frac{\eta + \zeta }{2}\\
2\frac{x}{c} = \eta - \zeta \to x = c \frac{\eta - \zeta }{2}
$$
so 
$$
\partial_\zeta x = -\frac{c}{2},\,\, \partial_\zeta t = \frac{1}{2}\\
\partial_\eta x = \frac{c}{2},\,\, \partial_\eta t = \frac{1}{2}
$$
Place in the equations above.
A: Looks like homework for special relativity to me! So I'll only give a hint / fill-in-the-gaps.
Use the multivariable chain rule on an arbitrary (appropriately differentiable) function $f$:
$$
\frac{\partial}{\partial t} f(\xi(t,x), \eta(t,x)) = \frac{\partial}{\partial \xi} f \cdot \frac{\partial \xi}{\partial t} + \frac{\partial}{\partial \eta} f \cdot \frac{\partial \eta}{\partial t}.
$$
Do something similar for $\frac{\partial}{\partial x}$, and thus set up a pair of simultaneous equations on your derivatives (treating $\frac{\partial}{\partial t}$ and $\frac{\partial}{\partial x}$ as known). You can re-arrange to get the required equations.
