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# 14.3: Partial Differentiation

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When we first considered what the derivative of a vector function might mean, there was really not much difficulty in understanding either how such a thing might be computed or what it might measure. In the case of functions of two variables, things are a bit harder to understand. If we think of a function of two variables in terms of its graph, a surface, there is a more-or-less obvious derivative-like question we might ask, namely, how "steep'' is the surface. But it's not clear that this has a simple answer, nor how we might proceed. We will start with what seem to be very small steps toward the goal; surprisingly, it turns out that these simple ideas hold the keys to a more general understanding.

So far, using no new techniques, we have succeeded in measuring the slope of a surface in two quite special directions. For functions of one variable, the derivative is closely linked to the notion of tangent line. For surfaces, the analogous idea is the tangent plane---a plane that just touches a surface at a point, and has the same "steepness'' as the surface in all directions. Even though we haven't yet Figured out how to compute the slope in all directions, we have enough information to find tangent planes. Suppose we want the plane tangent to a surface at a particular point $$(a,b,c)$$. If we compute the two partial derivatives of the function for that point, we get enough information to determine two lines tangent to the surface, both through $$(a,b,c)$$ and both tangent to the surface in their respective directions. These two lines determine a plane, that is, there is exactly one plane containing the two lines: the tangent plane. Figure $$\PageIndex{3}$$ shows (part of) two tangent lines at a point, and the tangent plane containing them.

Figure $$\PageIndex{3}$$: Tangent vectors and tangent plane.

How can we discover an equation for this tangent plane? We know a point on the plane, $$(a,b,c)$$; we need a vector normal to the plane. If we can find two vectors, one parallel to each of the tangent lines we know how to find, then the cross product of these vectors will give the desired normal vector.

So it appears that to find a tangent plane, we need only find two quite simple ordinary derivatives, namely $$f_x$$ and $$f_y$$. This is true if the tangent plane exists. It is, unfortunately, not always the case that if $$f_x$$ and $$f_y$$ exist there is a tangent plane. Consider the function $$xy^2/(x^2+y^4)$$ pictured in Figure 14.2.1. This function has value 0 when $$x=0$$ or $$y=0$$, and we can "plug the hole'' by agreeing that $$f(0,0)=0$$. Now it's clear that $$f_x(0,0)=f_y(0,0)=0$$, because in the $$x$$ and $$y$$ directions the surface is simply a horizontal line. But it's also clear from the picture that this surface does not have anything that deserves to be called a "tangent plane'' at the origin, certainly not the $$x$$-\)y\) plane containing these two tangent lines.

When does a surface have a tangent plane at a particular point? What we really want from a tangent plane, as from a tangent line, is that the plane be a "good'' approximation of the surface near the point. Here is how we can make this precise:

Definition $$\PageIndex{4}$$

Let $$\Delta x=x-x_0$$, $$\Delta y=y-y_0$$, and $$\Delta z=z-z_0$$ where $$z_0=f(x_0,y_0)$$. The function $$z=f(x,y)$$ is differentiable at $$(x_0,y_0)$$ if

$\Delta z=f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y+\epsilon_1\Delta x + \epsilon_2\Delta y,$

and both $$\epsilon_1$$ and $$\epsilon_2$$ approach 0 as $$(x,y)$$ approaches $$(x_0,y_0)$$.

This definition takes a bit of absorbing. Let's rewrite the central equation a bit:

\eqalignno{ z&=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+f(x_0,y_0)+ \epsilon_1\Delta x + \epsilon_2\Delta y.& ($$\PageIndex{1}$$)\cr }

The first three terms on the right are the equation of the tangent plane, that is,

$f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)+f(x_0,y_0)$

is the $$z$$-value of the point on the plane above $$(x,y)$$. Equation $$\PageIndex{1}$$ says that the $$z$$-value of a point on the surface is equal to the $$z$$-value of a point on the plane plus a "little bit,'' namely $$\epsilon_1\Delta x + \epsilon_2\Delta y$$. As $$(x,y)$$ approaches $$(x_0,y_0)$$, both $$\Delta x$$ and $$\Delta y$$ approach 0, so this little bit $$\epsilon_1\Delta x + \epsilon_2\Delta y$$ also approaches 0, and the $$z$$-values on the surface and the plane get close to each other. But that by itself is not very interesting: since the surface and the plane both contain the point $$(x_0,y_0,z_0)$$, the $$z$$ values will approach $$z_0$$ and hence get close to each other whether the tangent plane is "tangent'' to the surface or not. The extra condition in the definition says that as $$(x,y)$$ approaches $$(x_0,y_0)$$, the $$\epsilon$$ values approach 0---this means that $$\epsilon_1\Delta x + \epsilon_2\Delta y$$ approaches 0 much, much faster, because $$\epsilon_1\Delta x$$ is much smaller than either $$\epsilon_1$$ or $$\Delta x$$. It is this extra condition that makes the plane a tangent plane.

We can see that the extra condition on $$\epsilon_1$$ and $$\epsilon_2$$ is just what is needed if we look at partial derivatives. Suppose we temporarily fix $$y=y_0$$, so $$\Delta y=0$$. Then the equation from the definition becomes

$\Delta z=f_x(x_0,y_0)\Delta x+\epsilon_1\Delta x$

or

${\Delta z\over\Delta x}=f_x(x_0,y_0)+\epsilon_1.$

Now taking the limit of the two sides as $$\Delta x$$ approaches 0, the left side turns into the partial derivative of $$z$$ with respect to $$x$$ at $$(x_0,y_0)$$, or in other words $$f_x(x_0,y_0)$$, and the right side does the same, because as $$(x,y)$$ approaches $$(x_0,y_0)$$, $$\epsilon_1$$ approaches 0. Essentially the same calculation works for $$f_y$$.

## Contributors

• Integrated by Justin Marshall.