Calculus II Notes 7.4

# Calculus II Notes 7.4 - Calculus II-Stewart Dr. Berg Spring...

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Calculus II- Stewart Dr. Berg Spring 2010 Page 1 7.4 7.4 Logarithmic Functions Derivatives Theorem d dx ln x ( ) = 1 x and d dx ln g ( x ) ( ) = 1 g ( x ) g ( x ) . Example A d dx ln tan x ( ) ( ) = 1 tan x sec 2 x . Example B d dx ln x 3 2 x ( ) = 1 x 3 2 x 3 x 2 2 ( ) . Antiderivatives Theorem 1 x dx = ln x + C and 1 g ( x ) g ( x ) dx = ln g ( x ) + C . Notice that we can now integrate any power function f ( x ) = x n , even if n = 1 . Example C Evaluate x 2 x 3 1 dx . Solution : Let u = x 3 1 . Then du = 3 x 2 dx so that x 2 x 3 1 dx = 1 3 1 u du = 1 3 ln u + C = 1 3 ln x 3 1 + C . Example D Evaluate x + 1 x 2 + 2 x + 2 dx 1 1 . Question : Is this function eligible for application of the FTC?

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Calculus II- Stewart Dr. Berg Spring 2010 Page 2 7.4 Solution : Let u = x 2 + 2 x + 2 . Then du = 2 x + 2 ( ) dx , and u ( 1) = 1 , and u (1) = 5 , so that x + 1 x 2 + 2 x + 2 dx 1 1 = 1 2 1 u 1 5 du = 1 2 ln
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## This note was uploaded on 02/28/2010 for the course M 56495 taught by Professor Berg during the Spring '10 term at University of Texas at Austin.

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Calculus II Notes 7.4 - Calculus II-Stewart Dr. Berg Spring...

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