Total heat flux $Q$ across a surface can be written as
$Q = \displaystyle \oint_S \mathbf{q} \cdot \mathbf{\hat{n}}$,
where $\mathbf{\hat{n}}$ is the unit normal vector orthogonal to the surface $S$, and $\mathbf{q}$ is the (vector) heat flux per unit surface.
Fourier's law. Fourier's law states that the heat flux per unit surface is linear with the temperature gradient $\nabla T$.
Fourier's law in 3D. General form of Fourier's law in 3D physical space reads
$\mathbf{q} = -k \, \nabla T = - k \left[ \dfrac{\partial T}{\partial x} \mathbf{\hat{x}} + \dfrac{\partial T}{\partial y} \mathbf{\hat{y}} + \dfrac{\partial T}{\partial z} \mathbf{\hat{z}} \right]$
for an isotropic medium, i.e. the heat-flux per unit of area $\mathbf{q}$ is locally proportional to the temperature gradient $\nabla T$, through the conductivity $k$.
Fourier's law in 1D. Looking at an elongated element, such as a structural beam/rod with isolated surface or a pipe containing fluid at rest, we can approximately assume that the behavior of this system is approximately 1-dimensional, with the heat conduction occurring only along the axial direction $x$. Fourier's law thus reduce to its $x$-component only
$q_x = - k \dfrac{d T}{d x}$,
where we can use ordinary derivative assuming all the functions depend only on one variable, namely $x$ coordinate.
Fourier's law in 1D steady problem without heat source. Internal energy equation for 1D steady problem without heat source can be written as
$\dfrac{d q_x}{d x} = 0$
and thus $q_x = - k\dfrac{dT}{dx} = \overline{q}_x = \text{const}$, and thus the temperature profile in space is linear
$T(x) = -\dfrac{ \overline{q}_x }{k} x + C = A x + C$.
To complete the differential problem, you need boundary conditions. As an example, if you know the temperature at two locations $T(x_0) = T_0$, $T(x_1) = T_1$, you can write the temperature and the heat flux as
$T(x) = T_0 + (T_1 - T_0 ) \dfrac{x-x_0}{x_1 - x_0}$
$q_x = - k \dfrac{T_1-T_0}{x_1-x_0}$