Introduction to Mathematical Analysis I - Second Edition

102 4.1 DEFINITION AND BASIC PROPERTIES OF THE DERIVATIVE Example 4.1.1 (a) Let f : R → R be given by f ( x ) = x and let a ∈ R . Then lim x → a f ( x ) − f ( a ) x − a = lim x → a x − a x − a = lim x → a 1 = 1 . It follows that f is differentiable at a and f 0 ( a ) = 1. (b) Let f : R → R be given by f ( x ) = x 2 and let a ∈ R . Then lim x → a f ( x ) − f ( a ) x − a = lim x → a x 2 − a 2 x − a = lim x → a ( x − a )( x + a ) x − a = lim x → a ( x + a ) = 2 a . Thus, f is differentiable at every a ∈ R and f 0 ( a ) = 2 a . (c) Let f : R → R be given by f ( x ) = | x | and let a = 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, lim x → 0 f ( x ) − f ( 0 ) x − 0 does not exist and, hence, f is not differentiable at 0. Theorem 4.1.1 Let G be an open subset of R and let f be defined on G . If f is differentiable at a ∈ G , then f is continuous at this point. Proof: We have the following identity for x ∈ G \ { a } : f ( x ) = f ( x ) − f ( a )+ f ( a ) = f ( x ) − f ( a ) x − a ( x − a )+ f ( a ) . Thus, lim x → a f ( x ) = lim x → a f ( x ) − f ( a ) x − a ( x − a )+ f ( a ) = f 0 ( a ) · 0 + f ( a ) = f ( a ) . Therefore, f is continuous at a by Theorem 3.3.2 . Remark 4.1.2 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 the example above). Theorem 4.1.3 Let G be an open subset of R and let f , g : G → R . Suppose both f and g are differentiable at a ∈ G . Then the following hold. (a) The function f + g is differentiable at a and ( f + g ) 0 ( a ) = f 0 ( a )+ g 0 ( a ) . (b) For a constant c , the function c f is differentiable at a and ( c f ) 0 ( a ) = c f 0 ( a ) .