90 4.1 Definition and Basic Properties of the Derivative ■ Example 4.1.1 (a) Let f : R→Rbe given by f (x) =x and let x0 ∈R. Then lim x→x0 f (x)−f (x0) x−x0 = lim x→x0 x−x0 x−x0 = lim x→x0 1=1. It follows that f is differentiable at x0 and f ′(x0) =1. Since x0 was an arbitrary point inR, we have f ′(x) =1 for all x ∈R. (b) Let f : R→Rbe given by f (x) =x2 and let x 0 ∈R. Then lim x→x0 f (x)−f (x0) x−x0 = lim x→x0 x2 −x2 0 x−x0 = lim x→x0 (x−x0)(x+x0) x−x0 = lim x→x0 (x+x0) =2x0. Thus, f is differentiable at every x ∈Rand f ′(x) =2x. (c) Let f : R→Rbe given by f (x) =|x| and let x0 =0. Then lim x→0+ f (x)−f (0) x−0 = lim x→0+ |x| x = lim x→0+ x x =1, and lim x→0− f (x)−f (0) x−0 = lim x→0− |x| x = lim x→0− −x x =−1. Therefore, limx→0 f (x)−f (0) x−0 does not exist and, hence, f is not differentiable at 0. Theorem 4.1.1 Let I be an open interval in Rand let f be defined on I. If f is differentiable at x0 ∈I, then f is continuous at this point. Proof: We have the following identity for x ∈I \ {x0}: f (x) =f (x)−f (x0)+f (x0) = f (x)−f (x0) x−x0 (x−x0)+f (x0). Thus, lim x→x0 f (x) = lim x→x0 f (x)−f (x0) x−x0 (x−x0)+f (x0) =f ′(x0)· 0+f (x0) =f (x0). Therefore, f is continuous at x0 by Theorem 3.3.1. □ Remark 4.1.1 The converse of Theorem 4.1.1 is not true. For instance, the absolute value function f (x) =|x| is continuous at 0, but it is not differentiable at this point (as shown in Example 4.1.1(c)). Theorem 4.1.2 Let I be an open interval in Rand let f,g: I →R. Suppose both f and g are differentiable at x0 ∈I. Then the following hold. (i) The function f +gis differentiable at x0 and (f +g)′(x0) =f ′(x0)+g′(x0). (ii) For anyc ∈R, the functionc f is differentiable at x0 and (c f )′(x0) =c f ′(x0).
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