Explanation Given information: The given function is, f ( x ) = x 1 / 2 + 2 x 3 / 2 + 1 2 x 1 / 2 + 3 . The value of x is 0.5 . 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 1 / 2 + 2 x 3 / 2 + 1 2 x 1 / 2 + 3 for f ( x ) in equation (1) to find the first derivative of the function, f ′ ( x ) = d d x [ x 1 / 2 + 2 x 3 / 2 + 1 2 x 1 / 2 + 3 ] = ( 2 x 1 / 2 + 3 ) d d x ( x 1 / 2 + 2 x 3 / 2 + 1 ) − ( x 1 / 2 + 2 x 3 / 2 + 1 ) d d x ( 2 x 1 / 2 + 3 ) ( 2 x 1 / 2 + 3 ) 2 ( Use qoutient rule ) = ( 2 x 1 / 2 + 3 ) ( 1 2 x 1 / 2 + 2 ⋅ 3 2 x 1 / 2 ) − ( x 1 / 2 + 2 x 3 / 2 + 1 ) ( 2 ⋅ 1 2 x 1 / 2 ) ( 2 x 1 / 2 + 3 ) 2 ( Use power rule ) = ( 2 x 1 / 2 + 3 ) ( x − 1 / 2 2 + 3 x 1 / 2 ) − ( x 1 / 2 + 2 x 3 / 2 + 1 ) ( x − 1 / 2 ) ( 2 x 1 / 2 + 3 ) 2 Further the simplification of the above equation is, f ′ ( x ) = ( 2 x 1 / 2 + 3 ) ( x − 1 / 2 2 + 3 x 1 / 2 ) − ( x 1 / 2 + 2 x 3 / 2 + 1 ) ( x − 1 / 2 ) ( 2 x 1 / 2 + 3 ) 2 = 6 x + 3 x − 1 / 2 2 + 9 x 1 / 2 − 2 x − x − 1 / 2 ( 2 x 1 / 2 + 3 ) 2 = 4 x + 9 x 1 / 2 +

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 3.5, Problem 7TE

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The value of second derivative of the function at given value of x.

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Chapter 3 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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The value of second derivative of the function at given value of x .

Solution Summary: The author explains the value of second derivative of the function at x=0.5 is 0.1973.

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Chapter 3 Solutions