Connection of Gradient, Jacobian and Hessian matrix in Newton method

Suppose $f: \mathbb{R^n} \to \mathbb{R}$, the gradient of $f(\mathbf{x})$ is $$\mathop{\nabla} f(\mathbf{x}) = \begin{bmatrix} \frac{\partial{f}}{\partial{x_1}} \\ \vdots \\ \frac{\partial{f}}{\partial{x_n}} \end{bmatrix}$$

The Jacobian matrix of $\mathop{\nabla} f(\mathbf{x})$ is \begin{align} \mathbf{D} (\mathop{\nabla} f(\mathbf{x})) &= \begin{bmatrix} \frac{{\partial^2}f}{{\partial}x_1^2} & \frac{{\partial^2}f}{{\partial}x_2{\partial}x_1} & \cdots & \frac{{\partial^2}f}{{\partial}x_n{\partial}x_1}\\ \frac{{\partial^2}f}{{\partial}x_1{\partial}x_2} & \frac{{\partial^2}f}{{\partial}x_2^2} & \cdots & \frac{{\partial^2}f}{{\partial}x_n{\partial}x_2}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{{\partial^2}f}{{\partial}x_1{\partial}x_n} & \frac{{\partial^2}f}{{\partial}x_2{\partial}x_n} & \cdots & \frac{{\partial^2}f}{{\partial}x_n^2}\\ \end{bmatrix} \\ &= \mathbf{H}^T \end{align} where H is the Hessian matrix, which is consistent with the definition in Wikipedia.

The affine approximation of $\mathop{\nabla} f(\mathbf{x})$ around $x_n$ is $$\mathop{\nabla} f(\mathbf{x}) = \mathop{\nabla} f(\mathbf{x_n}) + \mathbf{D} (\mathop{\nabla} f(\mathbf{x_n})) (x - x_n) = \mathop{\nabla} f(\mathbf{x_n}) + \mathbf{H}^T (x - x_n)$$

Setting $\mathop{\nabla} f(\mathbf{x}) = 0$ gives the Newton-Raphson update as $$x_{n+1} := x_n - \mathbf{H}^{-T} \mathop{\nabla} f(\mathbf{x_n})$$

However, in Wikipedia the Newton-Raphson update is given as $x_{n+1} := x_n - \mathbf{H}^{-1} \mathop{\nabla} f(\mathbf{x_n})$. The Hessian matrix is not symmetric if the entry of the matrix is not continuous. Did I do anything wrong with my calculation? If not, does this mean we can generally treat Hessian matrix as symmetric in practice for optimization?

From the first sentence of the Wikipedia article you link - "In optimization, Newton's method is applied to the derivative $f′$ of a twice-differentiable function $f$ to find the roots of the derivative." In other words, the Hessian is symmetric.