Two-dimensional Taylor linearisation

Question

I have to perform a first order taylor expansion of a function $f(\vec{x}) = f(x+u,y+1)$ at the point $\vec{a} =(x,y)$. My solution reads $$ f(\vec{x}) \approx f(x,y) + \left( \begin{matrix} x + u - x \\ y + 1 - y \end{matrix} \right)^\top \left( \begin{matrix} \partial_{(x+u)} f(x,y) \\ \partial_{(y+1)} f(x,y) \end{matrix} \right) \\ = f(x,y) + \partial_{(x+u)} f(x,y) \cdot u + \partial_{(y+1)} f(x,y) $$ but the example solution is given by $$ f(\vec{x}) \approx f(x,y) + \left( \begin{matrix} x + u - x \\ y + 1 - y \end{matrix} \right)^\top \left( \begin{matrix} \partial_{x} f(x,y) \\ \partial_{y} f(x,y) \end{matrix} \right) \\ = f(x,y) + \partial_{x} f(x,y) \cdot u + \partial_{y} f(x,y) $$ Which is the correct solution ? Thanks for your help.

EDIT: I'm sorry. I forgot an important aspect, $u$ denotes a function depend from $x$.

Answer 1

They both are correct. Observe that if $z=x+u$ the the chain rule says that $\dfrac{\partial}{\partial x}=\dfrac{\partial z}{\partial x}\dfrac{\partial}{\partial z}=\dfrac{\partial}{\partial z}$ since $z$ depends only on $x$ ($u$ being a constant). So your first partials are the same (though one or the other may be more convenient).

Answer 2

In a Taylor expansion, we have a fixed base point at which derivatives are evaluated. Here this fixed point is $(x,y)$. So, any dependency of $u$ upon $x$ is moot, since $x$ itself is fixed.

To form 1st order Taylor expansion:

1. Evaluate the derivative at the base point.
2. Multiply them by displacement from the base point.

Result:

$$f(x+u,y+1)\approx f(x,y)+\partial_xf (x,y) u +\partial_y f(x,y)$$