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Unformatted text preview: 18.01 Calculus Jason Starr Fall 2005 Lecture 5. September 16, 2005 Homework. Problem Set 2 Part I: (a)–(e); Part II: Problem 2. Practice Problems. Course Reader: 1I1, 1I4, 1I5 1. Example of implicit differentiation . Let y = f ( x ) be the unique function satisfying the equation, 1 1 + = 2 . x y What is slope of the tangent line to the graph of y = f ( x ) at the point ( x, y ) = (1 , 1)? Implicitly differentiate each side of the equation to get, d 1 d 1 d (2) + = = . dx x dx y dx Of course (1 /x ) = ( x − 1 ) = − x − 2 . And by the rule d ( u n ) /dx = nu n − 1 ( du/dx ), the derivative of 1 /y is − y − 2 ( dy/dx ). Thus, − y − 2 dy − x − 2 = . dx Plugging in x equals 1 and y equals 1 gives, − 1 − 1 y (1) = , whose solution is, y (1) = In fact, using that 1 /y equals 2 − 1 /x , this can be solved for every x , − 1 . dy 1 1 1 = . dx = ( x − 2 ) / ( y − 2 ) = x 2 · (2 − 1 /x ) 2 (2 x − 1) 2 2. Rules for exponentials and logarithms. Let a be a positive real number. The basic rules of exponentials are as follows. c Rule 1. If a b equals B and a equals C , then a b + c equals B · C , i.e., b + c b c a = a a . · Rule 2. If a b equals B and B d equals D , then a bd equals D , i.e., b ) d bd ( a = a . If a b equals B , the logarithm with base a of B is defined to be b . This is written log a ( B ) = b . The function B → log a ( B ) is defined for all positive real numbers B . Using this definition, the rules of exponentiation become...
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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.
 Spring '08
 Moretti
 Derivative, Implicit Differentiation

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