What is the direction of vector $\left[\begin{smallmatrix}d x \\ d y\end{smallmatrix}\right]$ Recently I came across a topic " total differential" which comes with a result
$d f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y$
As much I learned in multivariable calculus this can be simplified as
$\nabla f \cdot\left[\begin{array}{l}d x \\ d y\end{array}\right]$ which graphically means taking directional Derivative in diraction of $\left[\begin{array}{l}d x \\ d y\end{array}\right]$ But is it  Makes any sense ? someone plese explain
 A: Observe that the derivative of a function $f=f(x,y)$ is its gradient
$${\rm grad}f=\left[\begin{array}{cc}\dfrac{\partial f}{\partial x}&\dfrac{\partial f}{\partial y}\end{array}
\right].$$
But if $f(x,y)=x$ (the projection in the $x$-axis) then its gradient is
$\left[\begin{array}{cc}
\frac{\partial x}{\partial x}&\frac{\partial x}{\partial y}\end{array}
\right]=\left[\begin{array}{cc}1&0\end{array}
\right]$
and
similarly for
$f(x,y)=y$ (the projection in the $y$-axis), its gradient is $\left[\begin{array}{cc}\frac{\partial y}{\partial x}&\frac{\partial y}{\partial y}\end{array}
\right]=\left[\begin{array}{cc}0&1\end{array}
\right]$
Hence the relation
$df=\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial y}dy$
is the relation
$$\left[\begin{array}{cc}\dfrac{\partial f}{\partial x}&\dfrac{\partial f}{\partial y}\end{array}
\right]
=\frac{\partial f}{\partial x}\left[\begin{array}{cc}1&0\end{array}
\right]+\frac{\partial f}{\partial y}\left[\begin{array}{cc}0&1\end{array}
\right],$$
where anyone clearly sees that
$dx=\left[\begin{array}{cc}1&0\end{array}\right]$ and
$dy=\left[\begin{array}{cc}0&1\end{array}\right]$ are the matrix forms of the derivatives of the projections onto both  axis, that is, their gradients, respectively.
A: The direction vector for any nonzero vector $\mathbf v$ (in $\mathbb R^n$) is $\dfrac{\mathbf v}{\|\mathbf v\|}$. For instance, if $\mathrm dx$ and $\mathrm dy$ are small real numbers that aren't both $0$, then the direction vector for $\begin{bmatrix}\mathrm dx\\\mathrm dy\end{bmatrix}$ is $\begin{bmatrix}\mathrm dx/\sqrt{(\mathrm dx)^2+(\mathrm dy)^2}\\\mathrm dy/\sqrt{(\mathrm dx)^2+(\mathrm dy)^2}\end{bmatrix}$.
Now suppose that $f$ is differentiable at $(a,b)$ so that for small $\mathrm dx$ and $\mathrm dy$, $f(a+\mathrm dx,b+\mathrm dy)-f(a,b)$ is well-approximated by $\mathrm df=\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(a,b)}\mathrm dx+\left.\dfrac{\partial f}{\partial y}\right|_{(x,y)=(a,b)}\mathrm dy=\nabla f(a,b)\bullet\begin{bmatrix}\mathrm dx\\\mathrm dy\end{bmatrix}$.
Then $\mathrm df$ really is either "the directional derivative of $f$ in the direction of $\begin{bmatrix}\mathrm dx\\\mathrm dy\end{bmatrix}$", or "$\sqrt{(\mathrm dx)^2+(\mathrm dy)^2}$ times the directional derivative of $f$ in the direction of $\begin{bmatrix}\mathrm dx\\\mathrm dy\end{bmatrix}$", depending on how you define directional derivatives for non-unit vectors.
This sort of idea can come up in confirming calculations like the one in the question About the formula for the magnitude of the gradient of a scalar valued function, |∇T|.
A: The total differential expresses the linear part of the variation of the function when you vary the independent variables (excaclty like in 1D). In fact, there is no other possible way, as the most general  linear function of two variables has the form
$$a\,dx+b\,dy.$$
Then by definition, the coefficients are the partial derivatives.
