Introduction to Mathematical Analysis I - Second Edition

104 4.1 DEFINITION AND BASIC PROPERTIES OF THE DERIVATIVE The following lemma is very convenient for studying the differentiability of the composition of functions. Lemma 4.1.4 Let G be an open subset of R and let f : G → R . Suppose f is differentiable at a . Then there exists a function u : G → R satisfying f ( x ) − f ( a ) = [ f 0 ( a )+ u ( x )]( x − a ) for all x ∈ G and lim x → a u ( x ) = 0. Proof: Define u ( x ) =   f ( x ) − f ( a ) x − a − f 0 ( a ) , x ∈ G \ { a } 0 , x = a . Since f is differentiable at a , we have lim x → a u ( x ) = lim x → a f ( x ) − f ( a ) x − a − f 0 ( a ) = f 0 ( a ) − f 0 ( a ) = 0 . Therefore, the function u satisfies the conditions of the lemma. Theorem 4.1.5 — Chain rule. Let f : G 1 → R and let g : G 2 → R , where G 1 and G 2 are two open subsets of R with f ( G 1 ) ⊂ G 2 . Suppose f is differentiable at a and g is differentiable at f ( a ) . Then the function g ◦ f is differentiable at a and ( g ◦ f ) 0 ( a ) = g 0 ( f ( a )) f 0 ( a ) . Proof: Since f is differentiable at a , by Lemma 4.1.4 , there exists a function u defined on G 1 with f ( x ) − f ( a ) = [ f 0 ( a )+ u ( x )]( x − a ) for all x ∈ G 1 , and lim x → a u ( x ) = 0. Similarly, since g is differentiable at f ( a ) , there exists a function v defined on G 2 with g ( t ) − g ( f ( a )) = [ g 0 ( f ( a ))+ v ( t )][ t − f ( a )] for all t ∈ G 2 , (4.2) and lim t → f ( a ) v ( t ) = 0. Applying ( 4.2 ) for t = f ( x ) , we have g ( f ( x )) − g ( f ( a )) = [ g 0 ( f ( a ))+ v ( f ( x ))][ f ( x ) − f ( a )] . Thus, g ( f ( x )) − g ( f ( a )) = [ g 0 ( f ( a ))+ v ( f ( x ))][ f 0 ( a )+ u ( x )]( x − a ) for all x ∈ G 1 . This implies g ( f ( x )) − g ( f ( a )) x − a = [ g 0 ( f ( a ))+ v ( f ( x ))][ f 0 ( a )+ u ( x )] for all x ∈ G 1 \ { a } . By the continuity of f at a and the property of v , we have lim x → a v ( f ( x )) = 0 and, hence, lim x → a g ( f ( x )) − g ( f ( a )) x − a = g 0 ( f ( a )) f 0 ( a ) . The proof is now complete.