Explanation Given: x y + 4 x + 2 y = 1 …….(1) Calculation: If y 1 and y 2 are functions of x then d d x ( y 1 × y 2 ) = y 1 × d d x y 2 + y 2 × d d x y 1 Now differentiating (1) with respect to x , we get d d x x y + 4 d d x x + 2 d d x y = d d x 1 or x d d x y + y d d x x + 4 d d x x + 2 d d x y = d d x 1 or x d y d x + y + 4 + 2 d y d x = 0 ( x + 2 ) d y d x + y + 4 = 0 ……...

Loose-leaf Version For Calculus: Early Transcendentals

4th Edition

ISBN: 9781319055912

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Question

Chapter 3.8, Problem 2E

To determine

To show:

That if we differentiate both sides of $x y + 4 x + 2 y = 1$ , the result is $(x + 2) \frac{d y}{d x} + y + 4 = 0$ and then solve for $\frac{d y}{d x}$ and evaluate for at the given point.

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Chapter 3.8, Problem 1E

Chapter 3.8, Problem 3E

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Chapter 3 Solutions

Loose-leaf Version For Calculus: Early Transcendentals

Ch. 3.1 - Prob. 1PQ Ch. 3.1 - Prob. 2PQ Ch. 3.1 - Prob. 3PQ Ch. 3.1 - Prob. 4PQ Ch. 3.1 - Prob. 5PQ Ch. 3.1 - Prob. 6PQ Ch. 3.1 - Prob. 7PQ Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E

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That if we differentiate both sides of x y + 4 x + 2 y = 1 , the result is ( x + 2 ) d y d x + y + 4 = 0 and then solve for d y d x and evaluate for at the given point.

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To show: That if we differentiate both sides of xy+4x+2y=1, the result is (x+2)dydx+y+4=0 and then solve for dydxand evaluate for at the given point.

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That if we differentiate both sides of $x y + 4 x + 2 y = 1$ , the result is $(x + 2) \frac{d y}{d x} + y + 4 = 0$ and then solve for $\frac{d y}{d x}$ and evaluate for at the given point.