308 Chapter 6 • Differentiable Tunctions
Now, f is differentiable at x 0. Hence, by Theorem 6.1.8, f is continuous at
x 0. That is, lim f(x) = f(x 0 ). Substituting this into the last line above, we
X--+Xo
have
(f g)'(xo) = lim (fg)(x) - (fg)(xo) = f(xo)g'(xo) + g(xo)f'(xo).
x->xo x - xo
(e) Since g(x 0 ) =I 0 and g is differentiable (hence continuous) at xo, g(x) =I
0 on some neighborhood of xo, so
lim (f) (x) - (f) (xo)
X--+Xo X -Xo X--+Xo x-xo= lim [ - 1 _ f(x)g(xo) -g(x)f(xo)]
x->xo x - xo g(x)g(xo)
= lim [ 1 ] lim [ f(x)g(xo) -g(x)f(xo)]
x->xo g(x)g(xo) x->xo x - Xo= 1 lim [ f(x)g(xo) - f(xo)g(xo) + f(xo)g(xo) - f(xo)g(x)]
g(xo)g(xo) x->xo x -xo x - Xo
(Remember, differentiability =}continuity, so lim g(x) = g(x 0 ).)
X--+Xo_ - - (^1) - 1. Im [ ( g xo )f(x) - f(xo) - f( xo )g(x) - g(xo)]
g^2 (xo) x->xo X - Xo X - Xo
= --^1 [ gxo ( ) 1 Im. f(x) - f(xo) - f( Xo ) 1. Im---g(x) -g(xo)- ]
g2(x0) X->Xo x -Xo X->Xo x - Xo
= ~(l ) [g(xo)f'(xo) - f(xo)g'(xo)]
g Xo
g(xo)f'(xo) - f(xo)g'(xo)
g^2 (xo) •
The next theorem, the chain rule, is easy to state and almost as easy to
believe on an intuitive basis. Its proof, however, requires some finesse (to avoid
dividing by zero at a crucial point in the proof). The chain rule is known to
elementary calculus students in its familiar form: if y is a differentiable function
of u, and u is a differentiable function of x, then
dy dy du
dx du. dx ·
For our purposes, however, we must restate this as a theorem with more
precise hypotheses and a more precise conclusion.