Explanation Given: The function is y = x 1 3 e x − 1 and the value of a and d x is 1 and 0.1 respectively. Concept used: Linear approximation: For a differentiable function at x = a the linearization is expressed as follows: f ( x ) ≈ f ( a ) + ( x − a ) f ′ ( a ) For a small change in x that is Δ x , the value of Δ y is equal to d y . Calculation: The given function is expressed as follows: y = x 1 3 e x − 1 The given function is in the form of y = f ( x ) . For a small change in x that is Δ x , the expression for Δ y is as follows: Δ y = f ( x + Δ x ) − f ( x ) Divide the above equation by Δ x as shown below: Δ y Δ x = f ( x + Δ x ) − f ( x ) Δ x Apply limit in the above equation as shown below: lim Δ x → 0 Δ y Δ x = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x d y d x = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x To further simplify the above equation, substitute 1 for x in the above equation as shown below: d y d x = lim Δ x → 0 f ( 1 + Δ x ) − f ( 1 ) Δ x d y d x = f ′ ( 1 ) d y = f ′ ( 1 ) d x Label the above equation as equation (1)

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

4th Edition

ISBN: 9781319057664

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.1, Problem 32E

To determine

To find:

The value of $Δy$ for the function $y=x13ex−1$ such that the value of $a$ and $dx$ is $1$ and $0.1$ respectively.

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Chapter 4.1, Problem 31E

Chapter 4.1, Problem 33E

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4 In the integral dxf1dy (7)², make the change of variables x = ½(r− s), y = ½(r + s), and evaluate the integral. Hint: Find the limits on r and s by sketching the area of integration in the (x, y) plane along with the r and s axes, and then show that the same area can be covered by s from 0 to r and r from 0 to 1.

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can you solve this question using the right triangle method and explain the steps used along the way

Chapter 4 Solutions

CALCULUS EARLY TRANSCENDENTALS W/ACHIEVE

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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The value of Δ y for the function y = x 1 3 e x − 1 such that the value of a and d x is 1 and 0.1 respectively.

Solution Summary: The author calculates the value of Delta y for a differentiable function.

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The value of $Δy$ for the function $y=x13ex−1$ such that the value of $a$ and $dx$ is $1$ and $0.1$ respectively.

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Chapter 4 Solutions