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DIFFERENTIABILITY AND CONTINUITY f (a + h) f (a) In the formula lim for the derivative f (a), we can let x = a + h and rewrite f (a) as h 0 h f (x) f (a) . If f is differentiable at a, then lim x a x a
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lim f (x) = lim [f (x) f (a) + f (a)]
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= lim [f (x) f (a)] + lim f (a)
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f (x) f (a) (x a) + f (a) = lim x a x a f (x) f (a) = lim lim (x a) + f (a) x a x a x a = f (a) 0 + f (a) = f (a) Thus, f is continuous at a. This proves: Theorem 13.1: If f is differentiable at a, then f is continuous at a. EXAMPLE Differentiability is a stronger condition than continuity. In other words, the converse of Theorem 13.1 is not true. To see this, consider the absolute-value function f (x) = |x| (see Fig. 13-1). f is obviously continuous at x = 0; but it is not differentiable at x = 0. In fact,
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f (0 + h) f (0) h 0 = lim = lim 1 = 1 + h h h 0 h 0+ f (0 + h) f (0) h 0 lim = lim = lim 1 = 1 h h h 0 h 0 h 0
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and so the two-sided limit needed to de ne f (0) does not exist. (The sharp corner in the graph is a tip-off. Where there is no unique tangent line, there can be no derivative.)
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FURTHER RULES FOR DERIVATIVES
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Theorem 13.1 will enable us to justify Rule 3, the product rule, of 12 and to establish two additional rules.
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MORE ON THE DERIVATIVE
[CHAP. 13
Fig. 13-1 If f and g are differentiable at x and if g(x) = 0, then Dx For a proof, see Problem 13.2. EXAMPLES
(a) Dx x+1 x2 2 = (x 2 2) Dx (x + 1) (x + 1) Dx (x 2 2) (x 2 2)2 (x 2 2)(1) (x + 1)(2x) x 2 2 2x 2 2x = 2 2)2 (x (x 2 2)2 x 2 2x 2 x 2 + 2x + 2 = 2 2 2)2 (x (x 2)2
RULE 6 (Quotient Rule).
f (x) g(x)
g(x) Dx f (x) f (x) Dx g(x) [g(x)]2
= = 1 x2
x 2 Dx (1) 1 Dx (x 2 ) x 2 (0) 1(2x) 2x 2 = = 4 = 3 (x 2 )2 x4 x x
The quotient rule allows us to extend Rule 4 of 12: RULE 7. Dx (x k ) = kx k 1 for any integer k (positive, zero, or negative). For a proof, see Problem 13.3. EXAMPLES
(a) (b) Dx Dx 1 x 1 x2 1 1 = Dx (x 1 ) = ( 1)x 2 = ( 1) 2 = 2 x x 2 = Dx (x 2 ) = 2x 3 = 3 x
Solved Problems
13.1 Prove Rule 3, the product rule: If f and g are differentiable at x, then Dx (f (x) g(x)) = f (x) Dx g(x) + g(x) Dx f (x)
CHAP. 13]
MORE ON THE DERIVATIVE
By simple algebra, f (x + h)g(x + h) f (x)g(x) = f (x + h)[g(x + h) g(x)] + g(x)[f (x + h) f (x)] Hence, Dx (f (x) g(x)) = lim = lim f (x + h)g(x + h) f (x)g(x) h f (x + h)[g(x + h) g(x)] + g(x)[f (x + h) f (x) h f (x + h)[g(x + h) g(x)] g(x)[f (x + h) f (x)] + lim h h h 0 g(x + h) g(x) f (x + h) f (x) + lim g(x) lim h h h 0 h 0 h 0
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