Introduction to Mathematical Analysis I - Second Edition

103 (c) The function f g is differentiable at a and ( f g ) 0 ( a ) = f 0 ( a ) g ( a )+ f ( a ) g 0 ( a ) . (d) Suppose additionally that g ( a ) 6 = 0. Then the function f g is differentiable at a and f g 0 ( a ) = f 0 ( a ) g ( a ) − f ( a ) g 0 ( a ) ( g ( a )) 2 . Proof: The proofs of (a) and (b) are straightforward and we leave them as exercises. Let us prove (c) . For every x ∈ G \ { a } , we can write ( f g )( x ) − ( f g )( a ) x − a = f ( x ) g ( x ) − f ( a ) g ( x )+ f ( a ) g ( x ) − f ( a ) g ( a ) x − a = ( f ( x ) − f ( a )) g ( x ) x − a + f ( a )( g ( x ) − g ( a )) x − a . By Theorem 4.1.1 , the function g is continuous at a and, hence, lim x → a g ( x ) = g ( a ) . (4.1) Thus, lim x → a ( f g )( x ) − ( f g )( a ) x − a = f 0 ( a ) g ( a )+ f ( a ) g 0 ( a ) . This implies (c) . Next we show (d) . Since g ( a ) 6 = 0, by ( 4.1 ) , there exists an open interval I containing a such that g ( x ) 6 = 0 for all x ∈ I . Let h = f g . Then h is defined on I . Moreover, h ( x ) − h ( a ) x − a = f ( x ) g ( x ) − f ( a ) g ( x ) + f ( a ) g ( x ) − f ( a ) g ( a ) x − a = 1 g ( x ) ( f ( x ) − f ( a ))+ f ( a ) g ( x ) g ( a ) ( g ( a ) − g ( x )) x − a = 1 g ( x ) g ( a ) g ( a ) f ( x ) − f ( a ) x − a − f ( a ) g ( x ) − g ( a ) x − a . Taking the limit as x → a , we obtain (d) . The proof is now complete. Example 4.1.2 Let f : R → R be given by f ( x ) = x 2 and let a ∈ R . Using Example 4.1. 1(a) and Theorem 4.1.3( c) we can provide an alternative derivation of a formula for f 0 ( a ) . Indeed, let g : R → R be given by g ( x ) = x . Then f = g · g so f 0 ( a ) = ( gg ) 0 ( a ) = g 0 ( a ) g ( a )+ g ( a ) g 0 ( a ) = 2 g 0 ( a ) g ( a ) = 2 a . Proceeding by induction, we can obtain the derivative of g : R → R given by g ( x ) = x n for n ∈ N as g 0 ( a ) = nx n − 1 . Furthermore, using this and Theorem 4.1.3( a)( b) we obtain the familiar formula for the derivative of a polynomial p ( x ) = a n x n + · · · + a 1 x + a 0 as p 0 ( x ) = na n x n − 1 + · · · + 2 a 2 x + a 1 .