THE TOTAL DERIVATIVE - Partial Differentiation - The Calculus Primer

The Calculus Primer (2011)

Part X. Partial Differentiation

Chapter 36. THE TOTAL DERIVATIVE

10—4.Tolal Differential. We have already learned how to find the derivative of a function of a function; thus, if y = f(u), and u = F(x), then

images

In a similar way, it may be shown that if z = f(x,y), and x = F1 (t) and y = F2(t) then

images

Equation [1] gives the total derivative of z with respect to t. In the same way, if u = f(x,y,z), and x, y, and z are functions of t, then

images

It should be understood that the meanings of images and images are definitely different. Thus the partial derivative images supposes that only the particular variable x varies; on the other hand, images is the limit of images, where Δu is the total increment in u brought about by changes in all the variables due to an increment in the independent variable x.

If we multiply equations [1] and [2], respectively, by dt, we obtain

images

The expression du is called the total differential of u, or the “complete differential.”

The expressions images dx, images dy, etc. are called partial differentials.

10—5.Implicit Functions. The expression f(x,y) = 0 represents an equation in x and y where all the terms have been transposed to one side of the equation. In other words, y is an implicit function of x (or x is an implicit function of y). Now, let us set z = f(x,y), and find the total differential of z; from [3], §10—4, we have:

images

But by hypothesis, since z = f(x,y) = 0, then for all values of x, z = 0; hence dz = 0. Therefore, if images ≠ 0:

images

or, dividing (2) through by dx and by images, we get:

images

EXAMPLE. If f(x,y) = x2 + y3 + xy = 0, find images by using partial derivatives.

Solution.

images

This result agrees, of course, with the value of images found by the previously learned method; thus, differentiating “directly”:

images

The relation above is quite general. Thus if

u = F(x,y,z) = 0,

then, by §10—4, equation [4]:

images

Analogous to the derivation in the earlier part of the present section, we remember that z is a function of x and y, so that

images

Substituting the value of dz from (2) in equation (1), and remembering that du = 0, we obtain:

images

or, by factoring:

images

Since z is a function of x and y, the variables x and y are independent variables, and so we may assign to the increments dx and dy such values as we please.

Let us first set dy = 0, ≠ 0; then, from (3), we get:

images

Next, we set dx = 0, dy ≠ 0; then, from (3) we get:

images