Chain Rule and Implicit Differentiation

# Chain Rule and Implicit Differentiation - 18.01 Calculus...

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Unformatted text preview: 18.01 Calculus Jason Starr Fall 2005 Lecture 4. September 15, 2005 Homework. No new problems. Practice Problems. Course Reader: 1F-1, 1F-6, 1F-7, 1F-8. 1. Product rule example. For u = 3 x + 1, what is u ( x )? Since u u = 3 x + 1, ( u u ) = (3 x + 1) = 3. By the product rule, ( u u ) = u u = 2 uu . Thus solving, u + u u ( x ) = 3 / (2 u ) = x 1 / 2 / 2 . 3(3 + 1) 2. The derivative of u n . From above, ( u 2 ) equals 2 uu . By a similar computation, ( u 3 ) equals 3 u 2 u . This suggests a pattern, d ( u n ) n 1 du = nu . dx dx 18.01 Calculus Jason Starr Fall 2005 This can be proved by induction on n . For n = 1 , 2 and 3, it was checked. Let n be a particular integer (for instance, 70119209472933054321). For that integer, suppose the result is known, d ( u n ) n 1 du = nu . dx dx The goal is to prove the result for n + 1, that is, n +1 ) d ( u du = ( n + 1) u n ....
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## This note was uploaded on 06/03/2008 for the course MATH B6A taught by Professor Moretti during the Spring '08 term at Bakersfield College.

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Chain Rule and Implicit Differentiation - 18.01 Calculus...

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