Explanation Given information: The given function is, f ( x ) = x 3 + x + 2 2 x 2 − 5 x + 4 . The value of x is 1.2 . Definition used: The first derivative of the function is the differentiation of the function, f ′ ( x ) = d d x f ( x ) (1) The second derivative of the function is the differentiation of the first derivative of the function, f ″ ( x ) = d d x f ′ ( x ) (2) The chain rule stated that the derivation of the function, f ( g ( x ) ) is f ′ ( g ( x ) ) ⋅ g ′ ( x ) . Formula used: The power rule is, d d x a x n = a ⋅ n ⋅ x n − 1 The quotient rule of the derivative is, d d x ( u v ) = v d u d x − u d v d x v 2 Calculation: Substitute x 3 + x + 2 2 x 2 − 5 x + 4 for f ( x ) in equation (1) to find the first derivative of the function, f ′ ( x ) = d d x [ x 3 + x + 2 2 x 2 − 5 x + 4 ] = ( 2 x 2 − 5 x + 4 ) d d x ( x 3 + x + 2 ) − ( x 3 + x + 2 ) d d x ( 2 x 2 − 5 x + 4 ) ( 2 x 2 − 5 x + 4 ) 2 ( Use qoutient rule ) = ( 2 x 2 − 5 x + 4 ) ( 3 x 2 + 1 ) − ( x 3 + x + 2 ) ( 4 x − 5 ) ( 2 x 2 − 5 x + 4 ) 2 ( Use power rule ) = ( 6 x 4 − 15 x 3 + 12 x 2 + 2 x 2 − 5 x + 4 ) − ( 4 x 4 + 4 x 2 + 8 x − 5 x 3 − 5 x − 10 ) ( 2 x 2 − 5 x + 4 ) 2 Further the simplification of the above equation is, f ′ ( x ) = 6 x 4 − 15 x 3 + 12 x 2 + 2 x 2 − 5 x + 4 − 4 x 4 − 4 x 2 − 8 x + 5 x 3 + 5 x + 10 ( 2 x 2 − 5 x + 4 ) 2 = 2 x 4 − 10 x 3 + 10 x 2 − 8 x + 14 ( 2 x 2 − 5 x + 4 ) 2 Substitute 2 x 4 − 10 x 3 + 10 x 2 − 8 x + 14 ( 2 x 2 − 5 x + 4 ) 2 for f ′ ( x ) in equation (2) to find the second derivative of the function, f ″ ( x ) = d d x [ 2 x 4 − 10 x 3 + 10 x 2 − 8 x + 14 ( 2 x 2 − 5 x + 4 ) 2 ] = ( 2

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 3 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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