Confusion about the Total Derivative

Question

I just started multivariable calculus a little while ago and I'm confused about the concept of a total derivative of some function $z = z(x, y)$. I was taught that $dz = \frac{\partial z}{\partial x}\cdot dx + \frac{\partial z}{\partial y}\cdot dy$, and my understanding is that $dz$ represents an infinitesimal rate of change in $z$, and $\frac{\partial z}{\partial x}$ is the rate of change of $z$ w.r.t. $x$ while $y$ remains constant. So if $dz$ is the rate of change of $z$ and $z$ relies on two variables, shouldn't the rate of change be given by a vector $<\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}>$? Why is it simply the sum of $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$?

Answer 1

All the total derivative is saying is that we should have

$$f(x+\Delta x,y+\Delta y) \approx f(x,y)+\frac{\partial f}{\partial x} \Delta x +\frac{\partial f}{\partial y} \Delta y$$

We call the function $df\big|_{(x,y)}: \mathbb{R}^2 \to \mathbb{R}$ given by

$$ df\big|_{(x,y)}\left( \begin{bmatrix} \Delta x \\ \Delta y\end{bmatrix}\right) = \frac{\partial f}{\partial x} \Delta x +\frac{\partial f}{\partial y} \Delta y $$

The total derivative of $f$ at $(x,y)$. This is a linear map, whose matrix with respect to the standard basis is

$$ \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\end{bmatrix} $$

so sometimes we also call this matrix the total derivative (sometimes the "jacobian matrix of $f$").

In general the total derivative of a function from $\mathbb{R}^n \to \mathbb{R}^m$ will be a different linear function from $\mathbb{R}^n \to \mathbb{R}^m$ at each point of $\mathbb{R}^n$. The total derivative is recorded by the $m \times n$ matrix of partial derivatives.