1549901369-Elements_of_Real_Analysis__Denlinger_

6.2 Rules for Differentiation 309

Theorem 6.2.3 (The Chain Rule) Suppose f is differentiable at an inte-
rior point Xo of its domain, and g is differentiable at f(x 0 ), an interior point
of its domain. Then the composite function go f is differentiable at x 0 , and
(go f)'(xo) = (g' o f)(xo) · f'(xo) = g'(f(xo)) · f'(xo).

f g

.~.~.

x~y

goj

Figure 6.2

Proof.^4 Suppose f is differentiable at an interior point x 0 of its domain,
and g is differentiable at f(x 0 ), an interior point of its domain. Define the
function h : V(g) ----; JR by

{

g(u) -^9 (f(xo)) if u "/:-f(xo);

h(u) = u - f(xo)

g' (f(xo)) if u = f(xo).

Then h is continuous at f ( xo), since

lim h(u) = lim g(u) -^9 (f(xo)) (u "/:-f(xo) as u----; f(xo))

u->f(xo) u->f(xo) U - f(xo)

= g' (f(x 0 )) by definition of derivative

= h (f(x 0 )) by definition of h. (1)

Now Vu E V(g), even if u = f(xo), the definition of h(u) yields

g(u) - g (f(x o)) = h(u) (u - f(xo)). (2)

Thus, for all x in some deleted neighborhood of xo,

(go f)(x) - (go f)(xo) g (f(x)) - g (f(xo))

x-xo x -xo

h (f(x)) (f(x) - f(xo))

x -xo

= h (f(x)) f(x) - f(xo).

x - xo

4. This proof may seem unnecessarily complicated. To see the fallacy of a simpler approach,
   see Exercise 8.