Tutorial 4 Solns - Math 116 Lab 4 Solutions Fall 2011 1 d...

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Math 116 - Lab 4 - Solutions - Fall 2011. 1. d dx ( x ) = d dx ( ( y - x ) 4 ) 1 = 4( y - x ) 3 ( y 0 - 1). Therefore y 0 = 1 4( y - x ) 3 + 1 . 2. Implicit differentiation gives: d dx 3 e cos( x ) + 7 e sin( y ) = d dx (10) 3 e cos( x ) ( - sin( x )) + 7 e sin( y ) (cos( y )) y 0 = 0 which implies y 0 = 3 e cos( x ) sin( x ) 7 e sin( y ) cos( y ) . Hence the slope of the tangent line at ( π 2 , 0 ) is 3 7 and the equation of the tangent line is y = 3 7 x + arcsin ln 10 - 3 e 7 . 3. Implicit differentiation gives, d dx ( y 3 ) = d dx ( ( y - x ) 2 ) 3 y 2 y 0 = 2( y - x )( y 0 - 1) y 0 = 2( x - y ) 3 y 2 + 2( x - y ) . A vertical tangent line occurs when y 0 is undefined so solving 3 y 2 + 2( x - y ) = 0 for x gives x = - y (3 y - 2) 2 . Substituting into the original equation gives y = 0 or y = 4 9 . Hence there is a vertical tangent line at the points (0 , 0) and ( 4 27 , 4 9 ) . 4. a) We have, ln( y ) = ln x cot( x ) = cot( x ) ln( x ) . Therefore, 1
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y 0 y = cot( x ) 1 x - csc 2 ( x ) ln( x ) y 0 = x cot( x ) cot( x ) x - csc 2 ( x ) ln( x ) . b) We have, ln( y ) = ln x csc( x ) e 8 x ! = csc( x ) ln( x ) - 8 x. Therefore, y 0 y = csc( x ) 1 x - csc( x ) cot( x ) ln( x ) - 8 y 0 = x csc( x ) e 8 x csc( x ) x - csc( x ) cot( x ) ln( x ) - 8 . 5. f 00 ( x ) = 64 (4 x +3) 2 + 32(5 - 8 x ) (4 x +3) 3 . 6. a) Since f 0 ( x ) = 4 csc(3 - 2 x ) cot(3 - 2 x ) we get f 00 ( x ) = 8 csc(3 - 2 x ) ( cot 2 (3 - 2 x ) + csc 2 (3 - 2 x ) ) which implies f 00 (1 . 5) is undefined. b) neither 7. a) f (54) ( x ) = 0. b) f (4) ( x ) = 4 e x + xe x . 8. Taking the derivative of both sides with respect to x gives, sec 2 ( x ) + cos( y ) y 0 = 0 . Differentiating again gives, 2 tan( x ) sec 2 ( x ) - sin( y ) ( y 0 ) 2 + cos( y
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