## Differentiating second order term of Taylor polynomial (multivariable)

Why not work in small but certain steps? Simply start without the vector notation: $$f(x+\Delta x,y+\Delta y,z + \Delta z) = f(x,y,z) + \Delta x \frac{\partial f}{\partial x} + \Delta y \frac{\partial f}{\partial y} + \Delta z \frac{\partial f}{\partial z}\\ + \frac{1}{2}(\Delta x)^2 \frac{\partial^2 f}{\partial x^2} + \frac{1}{2}(\Delta y)^2 \frac{\partial^2 f}{\partial y^2} + \frac{1}{2}(\Delta z)^2 \frac{\partial^2 f}{\partial z^2}\\ + (\Delta x)(\Delta y) \frac{\partial^2 f}{\partial x \partial y} + (\Delta y)(\Delta z) \frac{\partial^2 f}{\partial y \partial z} + (\Delta z)(\Delta x) \frac{\partial^2 f}{\partial z \partial x}$$ Then convert this to the vector notation and check out if you can reproduce the above result: $$f(x+\Delta x,y+\Delta y),z + \Delta z) = f(x,y,z) + \left[ \Large \begin{array}{ccc} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{array} \right] \left[ \begin{array}{c} \Delta x \\ \Delta y \\ \Delta z \end{array} \right] \\ + \frac{1}{2} \left[ \begin{array}{ccc} \Delta x & \Delta y & \Delta z \end{array} \right] \left[ \Large \begin{array}{ccc} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^2 f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z^2} \end{array}\right] \left[ \begin{array}{c} \Delta x \\ \Delta y \\ \Delta z \end{array} \right]$$