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Unformatted text preview: Lecture 15 Derivatives and rates of change Some differentiation rules (cont’d) The Chain Rule (Relevant section from Stewart, Sixth Edition: Section 3.4) The “Chain Rule,” with which most, if not all, of you are familiar deals with the differentiation of composite functions. If you have a function of a function of x , then how does this composite function change with respect to x . Mathematically, what is d dx ( f ◦ g )( x ) = d dx f ( g ( x )) ? (1) I think that all of you would be able to perform the following differentiation, d dx ( x 2 + 5) 3 / 2 . (2) The answer is 3 2 ( x 2 + 5) 1 / 2 (2 x ) = 3 x ( x 2 + 5) 1 / 2 . (3) And the same goes for the following differentiation, d dx (sin 2 x ) 7 . (4) The answer is 7(sin 2 x ) 6 (2cos 2 x ) = 14(sin 6 2 x )(cos 2 x ) . (5) Going back to the first example, you all know somehow that you have to multiply the expression by the exponent 3 2 , then reduce the exponent of the term x 2 + 5 by one, and then multiply by the derivative of the inside. In terms of composite functions, we can express this function in the form f ( g ( x )), where f ( x ) = x 3 / 2 and g ( x ) = x 2 + 5 . (6) But the occurrence of x as arguments in both functions may be a bit confusing. For this reason, it is standard practice to do the following: We let u = g ( x ) = x 2 + 5 (7) 100 and then define y = f ( u ) where f ( u ) = u 3 / 2 . (8) The “Chain Rule” may then be written as dy dx = df du du dx . (9) To check, the terms on the RHS are df du = 3 2 u 1 / 2 and du dx = 2 x. (10) Then dy dx = df du du dx = 3 2 u 1 / 2 (2 x ) , (11) which agrees with our earlier result when we substitute u = x 2 + 5. Likewise, in the second example, we have y = f ( u ) , u = g ( x ) , (12) where f ( u ) = u 7 , g ( x ) = sin 2 x. (13) Then dy dx = df du du dx = 7 u 6 (2cos 2 x ) , (14) in agreement with our earlier result when we substitute u = sin 2 x . In summary, the Chain Rule may be written as dy dx = df du du dx , where u = g ( x ) . (15) You have probably seen the Chain Rule written in the following, equivalent way, d dx f ( g ( x )) = f ′ ( g ( x )) g ′ ( x ) . (16) These formulas are stated with the assumption that f and g are differentiable functions. We now prove the Chain Rule. 101 Proof of Chain Rule: We’ll have to consider the – Guess what? – Newton quotient for the composite function f ◦ g , i.e., d dx f ( g ( x )) = lim h → f ( g ( x + h )) − f ( g ( x )) h . (17) We assume that f is differentiable at g ( x ) and g ( x ) is differentiable at x . With an eye to the desired result in Eq. (16), us write the Newton quotient as follows, f ( g ( x + h )) − f ( g ( x )) h = f ( g ( x + h )) − f ( g ( x )) g ( x + h ) − g ( x ) g ( x + h ) − g ( x ) h . (18) The second term on the RHS looks like it will become g ′ ( x ) in the limit h → 0, without any problem....
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This note was uploaded on 09/22/2011 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.
 Fall '08
 SPEZIALE
 Calculus, Chain Rule, Derivative, The Chain Rule

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