Explanation Result used: Quotient Rule : “If f ( x ) = u ( x ) v ( x ) , then f ′ ( x ) = v ( x ) u ( x ) − u ( x ) v ′ ( x ) [ v ( x ) ] 2 ”. Calculation: Obtain the derivative of y = ln | 2 x − 1 | x + 3 . Differentiate y with respect to x . y ′ ( x ) = d d x ( y ( x ) ) = d d x ( ln | 2 x − 1 | x + 3 ) Apply Quotient rule as shown in equation (1), y ′ ( x ) = ( x + 3 ) ( ln | 2 x − 1 | ) ′ − ( ln | 2 x − 1 | ) ( x + 3 ) ′ ( x + 3 )

Calculus with Applications Plus MyLab Math with Pearson eText -- Access Card Package (11th Edition) (Lial, Greenwell & Ritchey, The Applied Calculus & Finite Math Series)

11th Edition

ISBN: 9780133886832

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 4, Problem 40RE

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To find: The derivative of y=ln|2x−1|x+3 using the rule of derivatives.

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Chapter 4, Problem 39RE

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Calculus with Applications Plus MyLab Math with Pearson eText -- Access Card Package (11th Edition) (Lial, Greenwell & Ritchey, The Applied Calculus & Finite Math Series)

Ch. 4.1 - YOUR TURN 1 find f′(t). Ch. 4.1 - YOUR TURN 2 find dy/dx. Ch. 4.1 - Prob. 3YT Ch. 4.1 - Prob. 4YT Ch. 4.1 - Prob. 5YT Ch. 4.1 - Prob. 1WE Ch. 4.1 - Prob. 2WE Ch. 4.1 - Find the derivative of each function defined as...Ch. 4.1 - Find the derivative of each function defined as...Ch. 4.1 - Find the derivative of each function defined as...

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The derivative of y = ln | 2 x − 1 | x + 3 using the rule of derivatives.

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To find: The derivative of y=ln|2x−1|x+3 using the rule of derivatives.

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