Handouts7 - Calculus of Composite Functions

# Handouts7 - Calculus of Composite Functions - MATH0201...

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MATH0201 BASIC CALCULUS MATH0201 BASIC CALCULUS Calculus of Composite Functions Dr. WONG Chi Wing Department of Mathematics, HKU MATH0201 BASIC CALCULUS Composite Functions Chain Rule Calculus of Natural Logarithmic Function General Power Rules Exponential Function and Logarithmic Function with general base Antidifferentiation by Substitution Reference § 1.1, 2.4, 3.3, 3.5, 4.3–4, 5.2 of the textbook. MATH0201 BASIC CALCULUS Composite Functions Deﬁnition 2 (Composition of Functions (p.7)) Let f ( x ) and g ( u ) be two functions so that the range of f ( x ) is a subset of dom ( g ) . The function h ( x ) = g ( f ( x )) ; x 2 dom ( f ) is called the composition of g with f or simply g composite with f . ( x 2 + x + 1 ) n is the composition of g ( u ) = u n with f ( x ) = x 2 + x + 1. MATH0201 BASIC CALCULUS Composite Functions Example 3 (Exercise 1.1.51) Consider h ( x ) = ( x + 1 ) 3 + e x + 1 + 3. The form of the given function is h ( x ) = ± 3 + e ± + 3 where ± contains the expression ( x + 1 ) . Therefore, h ( x ) = g ( f ( x )) where f ( x ) = x + 1 and g ( u ) = u 3 + e u + 3 : Read also Example 1.1.9.

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MATH0201 BASIC CALCULUS Composite Functions Example 4 1. (Power Function composite with Polynomials) ( x 2 + x + 1 ) n , 3 p x 3 ± 1. 2. (Exponential Function composite with Polynomials) e x 2 + x , 3 x 3 ± x + 1 . 3. (Polynomials composite with Exponential Function) e 2 x + 2 e x + 3. 4. (Logarithmic Function composite with Polynomials) ln ( 1 + x ) , log 2 ( x 2 ± 3 ) . 5. (Trigonometric Function composite with Polynomials) sin ( 3 ± x ) , cos ( ± x 2 + e ) , tan ( 1 ± x + x 3 ) . MATH0201 BASIC CALCULUS Chain Rule Theorem 5 (Chain Rule (p.143)) If y = g ( u ) and u = f ( x ) are differentiable, then d dx g ( f ( x )) = g 0 ( f ( x )) ² f 0 ( x ) : We may write also dy dx = dy du ² du dx . ± dy du = dy du ² ² ² u = f ( x ) ³ Observe that g ( f ( x + h )) ± g ( f ( x )) h = g ( f ( x + h )) ± g ( f ( x )) f ( x + h ) ± f ( x ) ² f ( x + h ) ± f ( x ) h lim h ! 0 g ( f ( x + h )) ± g ( f ( x )) h = lim h ! 0 g ( f ( x + h )) ± g ( f ( x )) f ( x + h ) ± f ( x ) ² lim h ! 0 f ( x + h ) ± f ( x ) h = lim k ! 0 g ( y + k ) ± g ( y ) k ² lim h ! 0 f ( x + h ) ± f ( x ) h = g 0 ( f ( x )) ² f 0 ( x ) MATH0201 BASIC CALCULUS Chain Rule Example 6 Find the derivative of ( x 2 + x + 1 ) 3 . Solution. Let y = g ( u ) = u 3 and u = f ( x ) = x 2 + x + 1. By the chain rule, ´ ( x 2 + x + 1 ) 3 µ 0 = dy du ² ² ² ² u = f ( x ) ² du dx = 3 ( x 2 + x + 1 ) 2 ( 2 x + 1 ) : Read Examples 2.4.1–4. MATH0201 BASIC CALCULUS
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## This note was uploaded on 04/22/2011 for the course MATH 201 taught by Professor C.wong during the Spring '11 term at HKU.

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Handouts7 - Calculus of Composite Functions - MATH0201...

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