Calc03_9 - ln8 ln5 e x ln7 tan x ln x-2 3 x-15 5 4 ln 4 x 4...

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Unformatted text preview: ln8 ln5 e x ln7 tan x ln x-2 3 x-15 5 4 ln 4 x 4 x = 19 3 2.68 x = 18 - 5 ( 29 5 1.5 x = 3 2 - 3 -2.71-1 1 2 3-3-2-1 1 2 3 x Look at the graph of x y e = The slope at x=0 appears to be 1. If we assume this to be true, then: lim 1 h h e e h + - = definition of derivative Now we attempt to find a general formula for the derivative of using the definition. x y e = ( 29 lim x h x x h d e e e dx h + - = lim x h x h e e e h - = 1 lim h x h e e h - = 1 lim h x h e e h - = 1 x e = x e = This is the slope at x=0, which we have assumed to be 1. ( 29 x x d e e dx = x e is its own derivative! If we incorporate the chain rule: u u d du e e dx dx = We can now use this formula to find the derivative of x a ( 29 x d a dx ( 29 ln x a d e dx ( and are inverse functions.) x e ln x ( 29 ln x a d e dx ( 29 ln ln x a d e x a dx (chain rule) (...
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This note was uploaded on 02/16/2012 for the course CALCULUS 0064 taught by Professor Waldron during the Fall '10 term at Broward College.

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Calc03_9 - ln8 ln5 e x ln7 tan x ln x-2 3 x-15 5 4 ln 4 x 4...

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