Explanation Given: The data set is x 1 3 7 y 8 4 1.5 Formula used: The regression line with values m and b is given by, y = m x + b The chain rule of differentiation states that, d d x ( f ( x ) ) n = n ( f ( x ) ) n − 1 f ′ ( x ) Calculation: Consider the data set, x 1 3 7 y 8 4 1.5 For an exponential function of the form y = a ⋅ b x logarithm is taken on both sides, ln y = ln ( a ⋅ b x ) Use the property, ln ( m ⋅ n ) = ln m + ln n , ln y = ln a + ln b x Apply the property, ln ( m n ) = n ln m , ln y = ln a + x ln b Consider ln y to be y 0 , ln a to be a 0 and ln b to be b 0 , thus, the equation becomes, y 0 = b 0 x + a 0 The table for the new data set is x 1 3 7 y 8 4 1.5 y 0 ln 8 ln 4 ln 1.5 The data points are ( 1 , l n 8 ) , ( 3 , l n 4 ) and ( 7 , ln 1.5 ) . According to the least-squares assumption, the line which best fits is the line for which the sum of the squares of the y -deviations is minimum. For that, the y -deviations are y 1 − ln 8 , y 2 − ln 4 , y 3 − ln 1.5 The sum of the squares of y -deviations has to be minimised. Thus, consider the sum of the squares is S , S = ( y 1 − ln 8 ) 2 + ( y 2 − ln 4 ) 2 + ( y 3 − ln 1.5 ) 2 …… (1) The points ( 1 , y 1 ) , ( 3 , y 2 ) and ( 7 , y 3 ) must be solutions of the equation y = m x + b , thus, y 1 = m ( 1 ) + b y 2 = m ( 3 ) + b y 3 = m ( 7 ) + b Substitute the values of y 1 , y 2 and y 3 in the equation (1), S = ( m + b − ln 8 ) 2 + ( 3 m + b − ln 4 ) 2 + ( 7 m + b − ln 1.5 ) 2 Differentiate the both sides with respect to b , d S d b = d ( ( m + b − ln 8 ) 2 + ( 3 m + b − ln 4 ) 2 + ( 7 m + b − ln 1.5 ) 2 ) d b Apply the chain rule of differentiation, d S d b = d d b ( m + b − ln 8 ) 2 + d d b ( 3 m + b − ln 4 ) 2 + d d b ( 7 m + b − ln 1.5 ) 2 = 2 ( m + b − ln 8 ) ( 0 + b 1 − 1 − 0 ) + 2 ( 3 m + b − ln 4 ) ( 0 + b 1 − 1 − 0 ) + 2 ( 7 m + b − ln 1.5 ) ( 0 + b 1 − 1 − 0 ) = 2 m + 2 b − 2 ln 8 + 6 m + 2 b − 2 ln 4 + 14 m + 2 b − 2 ln 1.5 = 22 m + 6 b − 7.7424 Differentiate the both sides with respect to m , d S d m = d ( ( m + b − ln 8 ) 2 + ( 3 m + b − ln 4 ) 2 + ( 7 m + b − ln 1.5 ) 2 ) d m Apply the chain rule of differentiation, d S d m = d d m ( m + b − ln 8 ) 2 + d d m ( 3 m + b − ln 4 ) 2 + d d m ( 7 m + b − ln 1.5 ) 2 = 2 ( m + b − ln 8 ) ( m 1 − 1 + 0 − 0 ) + 2 ( 3 m + b − ln 4 ) ( 3 m 1 − 1 + 0 − 0 ) + 2 ( 7 m + b − ln 1

Calculus and Its Applications Plus MyLab Math with Pearson eText -- Access Card Package (11th Edition) (Bittinger, Ellenbogen & Surgent, The Calculus and Its Applications Series)

11th Edition

ISBN: 9780133795561

Author: Marvin L. Bittinger, David J. Ellenbogen, Scott J. Surgent

Publisher: PEARSON

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Chapter 6.4, Problem 7E

To determine

To calculate: The exponential regression curve for the data set represented by the table,

x 1 3 7

y 8 4 1.5

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Chapter 6.4, Problem 6E

Chapter 6.4, Problem 8E

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Calculus and Its Applications Plus MyLab Math with Pearson eText -- Access Card Package (11th Edition) (Bittinger, Ellenbogen & Surgent, The Calculus and Its Applications Series)

Ch. 6.1 - Forf(x,y)=x23xy,find(0,2),f(2,3),andf(10,5).Ch. 6.1 - 2. . Ch. 6.1 - 3. . Ch. 6.1 - 4. . Ch. 6.1 - Forf(x,y)=Inx+y3,findf(e,2),f(e2,4),andf(e3,5).Ch. 6.1 - 6. . Ch. 6.1 - Forf(x,y,z)=x2y2+z2,findf(1,2,3)andf(2,1,3).Ch. 6.1 - 8. . Ch. 6.1 - In Exercises 9-14, determine the domain of each...Ch. 6.1 - In Exercises 9-14, determine the domain of each...

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The exponential regression curve for the data set represented by the table, x 1 3 7 y 8 4 1.5

Solution Summary: The author explains the exponential regression curve for the given data set, x 1 3 7 y 8 4 1.5. The chain rule of differentiation states that, mathrmln(m