= Solution
Write the <Lorentz transformation> as $x'^\mu=\Lambda^\mu{}_{\nu}x^\nu$. Acting on a <scalar field>, the <chain rule> gives the <four-gradient>
$$
\partial'_\mu=\frac{\partial x^\nu}{\partial x'^\mu}\partial_\nu
=(\Lambda^{-1})^\nu{}_{\mu}\partial_\nu.
$$
For the specified <Lorentz boost>, the inverse is obtained by changing $\beta$ to $-\beta$, so
$$
\partial'_0=\gamma(\partial_0+\beta\partial_1),\qquad
\partial'_1=\gamma(\partial_1+\beta\partial_0),\qquad
\partial'_2=\partial_2,\quad\partial'_3=\partial_3.
$$
These are precisely the <covector> component rules for the lowered coordinates. Raising the index with the <Minkowski metric> $\eta^{\mu\nu}=\operatorname{diag}(-1,1,1,1)$ reverses the time component, yielding
$$
\boxed{\partial_\mu=(c^{-1}\partial_t,\partial_x,\partial_y,\partial_z),\qquad
\partial^\mu=(-c^{-1}\partial_t,\partial_x,\partial_y,\partial_z).}
$$
Indeed $\partial'^0=\gamma(\partial^0-\beta\partial^1)$ and $\partial'^1=\gamma(\partial^1-\beta\partial^0)$, the contravariant <four-vector> rules. Thus the index on a <partial derivative> with respect to an upper coordinate is lower.
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