Econ 300 - Summer 2009_Differentiation and Applications - SOLUTIONS

# Econ 300 - Summer 2009_Differentiation and Applications - SOLUTIONS

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Econ 300: Methods and Tools of Economic Analysis Summer 2009 Pablo M Federico DIFFERENTIATION AND APPLICATIONS: SOLUTIONS 1 Univariate Derivatives 1. y = 3 x 2 +4 x x 2 +1 y = f ( x ) g ( x ) f 0 ( x ) = 6 x + 4 g 0 ( x ) = 2 x y 0 = (6 x + 4)( x 2 + 1) (3 x 2 + 4 x )(2 x ) ( x 2 + 1) 2 y 0 = 4 x 2 + 6 x + 4 (1 + x 2 ) 2 Taking the second derivative,: y 00 = ( 8 x + 6)(1 + x 2 ) 2 ( 4 x 2 + 6 x + 4)(2( x 2 + 1) ± 2 x ) (1 + x 2 ) 4 Don±t worry about simplifying that. .. 2. y = x 2 ln( x + 1) y 0 = 2 x ln( x + 1) + x 2 x + 1 y 00 = 2 ln( x + 1) + 2 x x + 1 ± + 2 x ( x + 1) x 2 ( x + 1) 2 1

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3. y = x 3 e 3 x y 0 = 3 x 2 e 3 x 3 x 3 e 3 x y 00 = 6 xe 3 x 9 x 2 e 3 x 3(3 x 2 e 3 x 3 x 3 e 3 x ) 4. y = x 3 x y 0 = x (3 x ln 3) + 3 x ) = (ln(3) x + 1)3 x y 00 = (ln(3) x + 1)3 x ln 3 + ln(3)3 x 1.1 The Chain Rule Take the derivative of y with respect to t in each of the following cases: 1. y = x 2 2 xz 2 + z 2 x = t 2 z = 3 t dy dt = @y @x dx dt ± + @y @z dz dt ± = (2 x 2 z 2 ) dx dt (4 xz + 2 z ) dz dt = (2 x 2 z 2 )(2 t ) (4 xz + 2 z )(3) = (2 t 2 2(3 t ) 2 )(2 t ) (4 t 2 (3 t ) + 2(3 t ))(3) 2. y = 3 x + z + xz z = t + 7 x = 2 + t 2
dy dt = @y @x dx dt ± + @y @z dz dt ± = (3 + z ) dx dt + (1 + x ) dz dt = (3 + (7 t ))(1) + (1 + (2 + t ))( 1) 3. y = e 3 x 2 + 4 x ln z + tz x = t 1 = 2 z = t + 7 dy dt = @y @t + @y @x dx dt ± + @y @z dz dt ± = z + (6 xe 3 x 2 + 4 ln z ) dx dt + 4 x z + t ± dz dt = ( t + 7) + (6( t 1 = 2 ) e 3( t 1 = 2 ) 2 + 4 ln( t + 7)) t 1 = 2 2 + 4( t 1 = 2 ) ( t + 7) + t ± ± 1 2 Multivariate Functions minima and maxima is the key part of the material that we have covered in the last month.

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## This note was uploaded on 11/14/2010 for the course ECON Econ 326 taught by Professor Hayes during the Spring '08 term at University of Maryland Baltimore.

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Econ 300 - Summer 2009_Differentiation and Applications - SOLUTIONS

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