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 ……...

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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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

CALCULUS:EARLY TRANS.-SAPLINGPLUS

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

Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 39E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 41E Ch. 3.1 - Prob. 42E Ch. 3.1 - Prob. 43E Ch. 3.1 - Prob. 44E Ch. 3.1 - Prob. 45E Ch. 3.1 - Prob. 46E Ch. 3.1 - Prob. 47E Ch. 3.1 - Prob. 48E Ch. 3.1 - Prob. 49E Ch. 3.1 - Prob. 50E Ch. 3.1 - Prob. 51E Ch. 3.1 - Prob. 52E Ch. 3.1 - Prob. 53E Ch. 3.1 - 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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.