If I given $z=f(x,y)$, and I am asked to find the total derivative, and gradient of the functions the answers are slightly similar is there a particular reason why?
\begin{align}dz&=f_xdx+f_ydy\\ \nabla z&=f_x \hat{i}+f_y\hat{j}\end{align}
I can imagine that this works for a four space function as well: \begin{align}dw&=f_xdx+f_ydy+f_zdz \\ \nabla w&= f_x\hat{i}+f_y\hat{j}+f_z\hat{z}\end{align}
Could anybody explain how this relationship is possible, and better yet the connection, because I attempted a problem where I had to find the gradient, and I had the function. 1st I took the total derivative then I set the $\hat{i}$, and $\hat{j}$ terms, and the answer matched the textbook solution.
Recall the directional derivative, $$ \lim_{h \rightarrow 0}\dfrac{f(w+hw') - f(w)}{h} = \nabla f(w) \cdot w' $$ where $\nabla = (\partial/\partial w_1, ..., \partial/\partial w_n)$ is the gradient.
If you replace $w'$ with some arbitrarily small step along the principal axes, $(dx,dy,dz)$, you get the differential $dz$, $\nabla f(z) \cdot dz$. If you substituted $w'$ as the matrix $[\array{i &j &k}]$, you get the notation that I personally associate with physics and very boring calculus problems.
But in the end, both notations are essentially just directional derivatives. In the direction $w'$, how is the function changing, when $w'$ is just the principal axes? Of course, $w'$ could be other things, and that's why the directional derivative is a very useful concept, in optimization (necessary conditions, variational inequalities) and function spaces (Frechet, Gateaux derivatives).