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Unformatted text preview: Math 2250 Exam #2 Solutions 1. Find the derivatives of the following functions. (a) f ( x ) = ln(tan 1 x ) Answer: Using the Chain Rule, f ( x ) = 1 tan 1 x d dx (tan 1 x ) Remembering that the derivative of tan 1 x is 1 1+ x 2 , the above becomes f ( x ) = 1 tan 1 x 1 1 + x 2 . (b) g ( x ) = sin x cos 2 x Answer: The hard way to solve this is using the Quotient Rule: g ( x ) = cos 2 x · d dx (sin x ) sin x · d dx (cos 2 x ) (cos 2 x ) 2 = cos 2 x (cos x ) sin x ( 2cos x · ( sin x )) cos 4 x = cos 3 x + 2sin 2 x cos x cos 4 x = 1 cos x + 2 sin 2 x cos 3 x . The easy way is to notice that g ( x ) = sin x cos 2 x = 1 cos x sin x cos x = sec x tan x. Therefore, by the Product Rule, g ( x ) = (sec x tan x )tan x + sec x (sec 2 x ) = sec x (tan 2 x + sec 2 x ) . To see that the two answers we got are the same, remember that tan 2 x +1 = sec 2 x , so the second expression for g ( x ) becomes sec x (tan 2 x + tan 2 x + 1) = sec x (1 + 2tan 2 x ) = 1 cos x + 2 sin 2 x cos 3 x . 2. Let f ( x ) = 3 x ln x . What is f (0)? Answer: Unless you remember the derivative of 3 x and use the Chain Rule, the best way to solve this is to use logarithmic differentiation. Taking the natural log of both sides, we have ln( f ( x )) = ln(3 x ln x ) = x ln x ln3 ....
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This note was uploaded on 01/18/2012 for the course MATH 2250 taught by Professor Chestkofsky during the Fall '08 term at UGA.
 Fall '08
 CHESTKOFSKY
 Calculus, Chain Rule, Derivative, The Chain Rule

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