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: 25

Answer

The Derivative is: $v'=-\frac{1}{x^2}+2x^{-\frac{3}{2}}$

Work Step by Step

$v=\frac{1+x-4\sqrt{x}}{x}$ Using Quotient Rule: $y'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$ $v'=\frac{(0+(1)x^{1-1}-(4)\frac{1}{2\sqrt{x}})(x)-(1+x-4\sqrt{x})((1)x^{1-1})}{(x)^2}$ $v'=\frac{(1-\frac{2}{\sqrt{x}})(x)-(1+x-4\sqrt{x})}{x^2}$ $v'=\frac{\frac{(\sqrt{x}-2)(x)}{\sqrt{x}}-1-x+4\sqrt{x}}{x^2}$ $v'=\frac{\frac{x\sqrt{x}-2x-\sqrt{x}-x\sqrt{x}+4x}{\sqrt{x}}}{x^2}$ $v'=\frac{\frac{2x-\sqrt{x}}{\sqrt{x}}}{x^2}$ $v'=\frac{2x-\sqrt{x}}{\sqrt{x}}\cdot \frac{1}{x^2}$ $v'=\frac{\sqrt{x}(2\sqrt{x}-1)}{x^2\sqrt{x}}$ $v'=\frac{2\sqrt{x}-1}{x^2}$ $v'=-\frac{1}{x^2}+2x^{-\frac{3}{2}}$
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