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First things first: don't be a limit dropper!
Here is what I call "the most important definition in calculus" -- the limit definition of the derivative function, $f'(x)$: \[ f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \]
This is generalized into "partial" derivatives \[ f_x(x,y) = \frac{\partial f}{\partial x} = \lim_{h \to 0}\frac{f(x+h,y)-f(x,y)}{h} \] \[ f_y(x,y) = \frac{\partial f}{\partial y} = \lim_{h \to 0}\frac{f(x,y+h)-f(x,y)}{h} \]
Where we think of one variable as fixed, and the other as varying.
We consider each as a slope in a particular direction. But there are an infinite number of directions from which one can depart, of course. We are considering only two.
As we saw last time, partials are actually easy to compute, provided one has a formula. If one has only data, then it comes down to approximating with a finite difference.
We saw that higher partial derivatives are easy to calculate, as well, since each partial derivative is a multivariate function in its own right. So do it again!
(See notation on p. 930.)
We think of second derivatives in the univariate world as saying something about curvature, and the same is true in the bivariate case.
Let's revisit #2, and compute the derivatives, including higher derivatives.
Note that we'll be imposing the assumption of continuity of the second partials: we can't look at a formula and say "Ergo the second partials are continuous!")
Just like derivatives are closely linked to tangent lines, which kiss a curve (osculate); about the slopes of those linear approximations (tangent lines) to a univariate function; so are partial derivatives are about slopes of univariate functions obtained by slicing multivariate functions with planes. If the function is differentiable at a point, then there is a tangent plane, which kisses the surface.
An example from our author's TEC animations.
The slopes in those directions are given by $f_x(a,b)$ and $f_y(a,b)$.
If you take a one-unit step in either of those directions, the $z$-value would rise by either slope: \[ T(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b) \] If you take partial derivatives of $T$ at $(a,b)$, you will see that the partials (and the function value) agree with those of $f$ there.
To prevent problems, we introduce the following definition:
If $z=f(x,y)$, then $f$ is differentiable at $(a,b)$ if the increment $\Delta z$, which is \[ \Delta z = f(a+\Delta x,b+\Delta y) - f(a,b) \] be expressed in the form \[ \Delta z = f_x(a,b)(x-a) + f_y(a,b)(y-a) + \epsilon_1 \Delta x + \epsilon_2 \Delta y \] where $\epsilon_1$ and $\epsilon_2 \to 0$ as $(\Delta x, \Delta y) \to (0,0)$.
We can generally avoid using this definition, however, because of Theorem 8, p. 942:
Existence and continuity of partials $f_x$ and $f_y$ are sufficient to imply differentiability.
For a function $z=f(x,y)$, we define the differentials $dx$ and $dy$ to be independent variables; that is, they can be given any values. Then the differential $dz$, the total differential, is defined by \[ dz=f_x(x,y)dx + f_y(x,y)dy \] Differentials give us approximate changes to a function in the neighborhood of a point, generally when $dx$ and $dy$ are small, obtained by following along the tangent plane.
$\Delta{z}$ gives the actual change, so \[ \Delta{z}\approx dz=f_x(x,y)dx + f_y(x,y)dy \] The picture at the end of the section summarizes what we're up to nicely.