Explanation Given: The function y = ( x − 3 ) x + 2 . Derivative rule: (1) Quotient Rule : d d x ( f g ) = g d d x ( f ) − f d d x ( g ) ( g ) 2 (2) Product Rule : d d x ( f g ) = f d d x ( g ) + g d d x ( f ) . (3) The Chain Rule : If y = f ( u ) is differentiable at u = g ( x ) and u = g ( x ) is differentiable at x , then the composite function y = f ( g ( x ) ) is differentiable at x , then derivative of y = f ( g ( x ) ) is d y d x = d y d u ⋅ d u d x Calculation: The first derivative of the function y = ( x − 3 ) x + 2 is computed as follows, d y d x = d d x ( ( x − 3 ) x + 2 ) = ( x − 3 ) d d x ( x + 2 ) + x + 2 d d x ( x − 3 ) [ by Product Rule ] = ( x − 3 ) ( 1 2 ( x + 2 ) 1 2 − 1 d d x ( x + 2 ) ) + x + 2 ( 1 x 1 − 1 − 0 ) [ by Chain Rule ] = ( x − 3 ) ( 1 2 ( x + 2 ) − 1 2 ( 1 x 1 − 1 + 0 ) ) + x + 2 ( 1 ) Simplify further, d y d x = ( x − 3 ) ( 1 2 ( x + 2 ) − 1 2 ) + x + 2 = 1 2 ( x − 3 ) ( x + 2 ) 1 2 + x + 2 = 1 2 ( x − 3 ) x + 2 + x + 2 Therefore, the first derivative of the function is d y d x = 1 2 ( x − 3 ) x + 2 + x + 2 . The second derivative of the function y = ( x − 3 ) x + 2 is computed as follows, d 2 y d x 2 = d d x ( d y d x ) = d d x ( 1 2 ( x − 3 ) x + 2 + x + 2 ) = d d x ( 1 2 ( x − 3 ) x + 2 ) + d d x ( x + 2 ) = 1 2 ( ( x + 2 ) d d x ( x − 3 ) − ( x − 3 ) d d x ( x + 2 ) ( x + 2 ) 2 ) + d d x ( x + 2 ) [ by Quotient Rule ] Simplify further, d 2 y d x 2 = 1 2 ( ( x + 2 ) ( 1 x 1 − 1 − 0 ) − ( x − 3 ) ( 1 2 ( x + 2 ) 1 2 − 1 d d x ( x + 2 ) ) ( x + 2 ) 2 ) + ( 1 2 ( x + 2 ) 1 2 − 1 d d x ( x + 2 ) ) = 1 2 ( ( x + 2 ) ( 1 ) − ( x − 3 ) ( 1 2 ( x + 2 ) − 1 2 ( 1 x 1 − 1 + 0 ) ) ( x + 2 ) 2 ) + ( 1 2 ( x + 2 ) − 1 2 ( 1 x 1 − 1 + 0 ) ) = 1 2 ( ( x + 2 ) − ( x − 3 ) 2 ( x + 2 ) 1 2 ( x + 2 ) ) + ( 1 2 ( x + 2 ) 1 2 ) = 1 2 ( ( 2 ( x + 2 ) 1 2 ( x + 2 ) 1 2 ) − ( x − 3 )

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 3, Problem 48RE

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To find: The higher order derivatives y′,y′′, and y′′′ .

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Chapter 3, Problem 47RE

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Calculus: Early Transcendentals (2nd Edition)

Ch. 3.1 - Use definition (1) (p. 127) for the slope of a...Ch. 3.1 - Explain why the slope of a secant line can be...Ch. 3.1 - Explain why the slope of the tangent line can be...Ch. 3.1 - Prob. 4E Ch. 3.1 - Given a function f and a point a in its domain,...Ch. 3.1 - Explain the relationships among the slope of a...Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Equations of tangent lines by definition (1) a....Ch. 3.1 - Equations of tangent lines by definition (1) a....

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The higher order derivatives y ′ , y ′ ′ , and y ′ ′ ′ .

Solution Summary: The author explains that the higher order derivatives are yprime,andsqrtx+2, and d2

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Chapter 3 Solutions