Calculus: Analytic Geometry and Calculus, with Vectors

11.3 Formulas involving partial derivatives 583

(6) Implicit function theorem Let G be a given function of two variables
x and y such that G(xo,yo) = 0 and, moreover, the function G and its two partial
derivatives G. and G are continuous and G((x,y) 0 over some rectangular region
R1 having its center at the point (xo,yo). Then, for some positive number h, there
is one and only one function f, defined and differentiable over the interval xo - h <
x < xo + h, for which f(xo) = yo and

(7) G(x,f(x)) =0 (xo - h < x<xo+h).

Moreover, when xo - h < x < xo + h and f(x) = y, f'(x) and dv can be obtained

by differentiating (7) or

(8)

to obtain

(9)

or

(10)

and solving to obtain

G(x,y) = 0

G.T(x, f(x)) + Gv(x, f(x))f'(x) = 0

aG aG dyax+aydx_0_

f, (x) G=(x, f(x))Gu(x, f(x))

or

(12)

aG

dy ax

dx aG

ay

Before proving the theorem, we observe that it provides very solid information

about the graph of the equation G(x,y) = 0; the graph must contain a curve C

which contains the point Po(xo,yo), and this curve C has a tangent at the point

(xo,yo) which has slope f'(xo). As sometimes happens in other cases, our proof

of the theorem reveals some facts that are not stated in the conclusion of the

theorem. To prove the theorem, we suppose that 0; in case Gr,(xo,yo)

< 0, the proof is similar. Then the hypothesis that 0 over R1 and the

intermediate-value theorem imply that G5,(x,y) > 0 over R1. Choose numbers

yl and y2 such that yl < yo < Y2 and R1 contains the line segment consisting of

points (xo,y) for which y1 < y < y2, Since Gl,(xo,y) > 0 when y, S y :-!9 y2, it

follows that G(xo,y) is an increasing function of y over the interval yl - y < y2,

But G(xo,yo) = 0, and therefore

(13) G(xo,yl) < 0 < G(xo,y2)

Since G is continuous, we can choose a positive number h such that the rectangu-