Calculus: Analytic Geometry and Calculus, with Vectors

11.2 Increments, chain rule, and gradients 573

The extrema (if any) occur when x, y, X are such that

aw

ax =2x - 2A = 0

Ow

ay =2y+X=0

and

y=x2-4.

These relations imply that X = 1, y =- , x = ± N/3.5, and x2 + y2 = 3.75.

20 Formulas more or less like

(1) w= fuf(x,Y)dY

often appear in pure and applied mathematics. It is supposed that u and v are

functions of x and that, for each x in some interval, the integral in the rightmem-

ber exists and is a number w. More advanced courses set forth conditions under

which dw/dx can be obtained from the chain formula

(2) dw=aw dv aw du + aw

dx av dx T. dx ax

When appropriate conditions are satisfied, applications of the fundamental

theorem of the calculus give

(3) aeff(x,y) dy= f(x,v)

(n

auf f(x,y) dy = - au f U f(x,y) dy = -f(x,u)-

When (see the last of Problems 11.19) we can differentiate under the integral

sign, we get

(4) ax =

fua( X

f(x,y) dy.

Substituting in (2) then gives the formula

r

(5) dxfu f(x,y) dy = f(x,v) ax - f(x,u)dx+ fur

aJ Y) dy.

Verify that (5) is correct when

(a) u = x, v = 2x,f(x,y) = x + y

(b) u = x, v = 2x, f(x,y) =x2+y2

(c) u=x,v=a+x,f(x,y) =1/y

(d) u = x2, v = x3, f(x,y) _ (x + y)e-v

(e) u=x,v=x2,f(x,y) =logy

(^21) It can be observed that our proof of the chain rule for functions of more
than one variable is more straightforward than our proof of the chain rule (Theo-
rem 3.65) for functions'of one variable. Can this phenomenon be explained?
.4nr.: Yes. When we proved Theorem 3.65, we did not know about the mean-
value theorem and, moreover, the mean-value theorem was inapplicable because
we did not have the hypothesis that the derivatives exist over intervals and are