Chap3_Sec6

Chap3_Sec6 - f ( x ) = l n x , then f ( x ) = 1/ x . Thus,...

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1 (log ) ln a d x dx x a = DERIVATIVES OF LOG FUNCTIONS (Section 3.6) Prove s Let y = log a x. Then, a y = x . s Differentiating this equation implicitly with respect to x , we get: s So, (ln ) 1 y dy a a dx = 1 1 ln ln y dy dx a a x a = =
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Differentiate y = l n( x 3 + 1). s To use the Chain Rule, we let u = x 3 + 1. s Then y = l n u . Find DERIVATIVES OF LOG FUNCTIONS (Section 3.6) Example 1 Example 2 ln(sin ) d x dx
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Differentiate Differentiate f ( x ) = log 10 (2 + sin x ). ( ) ln f x x = DERIVATIVES OF LOG FUNCTIONS (Section 3.6) Example 3 Example 4
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1 st way : Find 2 nd way : Simplify the given function using the laws of logarithms, then differentiate: 1 ln 2 d x dx x + - DERIVATIVES OF LOG FUNCTIONS Example 5 (3.6)
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Find f ’ ( x ) if f ( x ) = l n | x |. s Since s Differentiate ln if 0 ( ) ln( ) if 0 x x f x x x > = - < DERIVATIVES OF LOG FUNCTIONS (Section 3.6) Example 7 3 / 4 2 5 1 (3 2) x x y x + = + Example 6
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Differentiate . s 1 st way : Use logarithmic differentiation, we have: s 2 nd way : Write LOGARITHMIC DIFFERENTIATION Example 8 (3.6) x y x = ln ( ) x x x x e =
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We have shown that, if
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Unformatted text preview: f ( x ) = l n x , then f ( x ) = 1/ x . Thus, f (1) = 1 . Now, we use this fact to express the number e as a limit. From the definition of a derivative as a limit, we have: THE NUMBER e AS A LIMIT (Section 3.6) 1 (1 ) (1) (1 ) (1) '(1) lim lim ln(1 ) ln1 1 lim lim ln(1 ) lim ln(1 ) h x x x x x f h f f x f f h x x x x x x +-+-= = +-= = + = + As f (1) = 1 , we have If we put n = 1/ x , then n as x + . So, an alternative expression for e is: 1 lim ln(1 ) 1 x x x + = 1/ 1 lim ln(1 ) 1 ln(1 ) 1 lim lim(1 ) x x x x x x x x e e e e x + + = = = = + THE NUMBER e AS A LIMIT (Section 3.6) 1 lim(1 ) x x e x = + 1 lim 1 n n e n = +...
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Chap3_Sec6 - f ( x ) = l n x , then f ( x ) = 1/ x . Thus,...

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