PT 3 solutions

PT 3 solutions - PRACTICE TEST 3 SOLUTIONS dy [by implicit...

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PRACTICE TEST 3 SOLUTIONS 1. Find dy dx [by implicit differentiation]: 3 3 2 3 x y xy + = ( 29 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 2 3 ; 3 6 3 3 ; 3 6 2 dy dy dy dy y x x y x y x y y y xy y x dx dx dx dx y xy y xy - - + = + - = - = = - - 2. Use implicit differentiation to find all points on the graph of ( 29 4 2 1 y y x x + = - at which the tangent line is vertical. This means where dy/dx is not defined. ( 29 4 2 2 3 3 3 2 1 ; 4 2 2 1; 4 2 2 1; 4 2 dy dy dy dy x y y x x y y x y y x dx dx dx dx y y - + = - + = - + = - = + The denominator is 0 [hence dy/dx undefined] when y = 0 [ ( 29 3 2 4 2 2 2 1 y y y y + = + ] so, substituting y = 0 in the given equation, we get ( 29 2 1 0 x x x x - = - = So the required points are (0, 0) and (1, 0) 3. Use logarithmic differentiation to find dy dx : 3 2 1 y x x = + [Look familiar?] ( 29 ( 29 ( 29 3 2 2 3 2 2 2 1 ln ln 1 ln ln 1 3 1 1 2 1 2 ; 1 3 1 3 1 y x x x x dy x dy x x x y dx x dx x x x = + = + + = + = + + + + 4. Find dy dx : a)
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This note was uploaded on 12/26/2011 for the course MAC 2311 taught by Professor Staff during the Fall '08 term at FIU.

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PT 3 solutions - PRACTICE TEST 3 SOLUTIONS dy [by implicit...

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