Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.3 Differentiation Rules - Exercises 3.3: 24

Answer

The Derivative is: $ y'=\frac{5x-1}{4x\sqrt{x}} $

Work Step by Step

$ f(x)=\frac{5x+1}{2\sqrt{x}} $ Using Quotient rule to find the Derivative: $ y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)} $ $ y'=\frac{1}{2}\frac{(5x^{1-1}+0)(\sqrt{x})-(5x+1)(\frac{1}{2}x^{\frac{1}{2}-1})}{(\sqrt{x})^2} $ $ y'=\frac{1}{2}\frac{(5)(\sqrt{x})-(5x+1)(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2} $ $ y'=\frac{1}{2}\frac{5\sqrt{x}-\frac{1}{2\sqrt{x}}(5x+1)}{(\sqrt{x})^2} $ $ y'=\frac{1}{2}\frac{\frac{10x-5x-1}{2\sqrt{x}}}{x} $ $ y'=\frac{\frac{5x-1}{2\sqrt{x}}}{2x} $ $ y'=\frac{5x-1}{2x\cdot 2\sqrt{x}} $ $ y'=\frac{5x-1}{4x\sqrt{x}} $
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