Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 42

Answer

$y'=\dfrac{4\lambda(x+\lambda)^{3}(\lambda^{3}-x^{3})}{(x^{4}+\lambda^{4})^{2}}$

Work Step by Step

$y=\dfrac{(x+\lambda)^{4}}{x^{4}+\lambda^{4}}$ Start the differentiation process by using the quotient rule: $y'=\dfrac{(x^{4}+\lambda^{4})[(x+\lambda)^{4}]'-(x+\lambda)^{4}(x^{4}+\lambda^{4})'}{(x^{4}+\lambda^{4})^{2}}=...$ Use the chain rule to evaluate the derivatives indicated in the numerator: $...=\dfrac{(x^{4}+\lambda^{4})[4(x+\lambda)^{3}(1)]-(x+\lambda)^{4}(4x^{3})}{(x^{4}+\lambda^{4})^{2}}=...$ $...=\dfrac{4(x^{4}+\lambda^{4})(x+\lambda)^{3}-4x^{3}(x+\lambda)^{4}}{(x^{4}+\lambda^{4})^{2}}=...$ Take out common factor $4(x+\lambda)^{3}$ from the numerator and simplify: $...=\dfrac{4(x+\lambda)^{3}[x^{4}+\lambda^{4}-x^{3}(x+\lambda)]}{(x^{4}+\lambda^{4})^{2}}=...$ $...=\dfrac{4(x+\lambda)^{3}(x^{4}+\lambda^{4}-x^{4}-x^{3}\lambda)}{(x^{4}+\lambda^{4})^{2}}=\dfrac{4(x+\lambda)^{3}(\lambda^{4}-x^{3}\lambda)}{(x^{4}+\lambda^{4})^{2}}=...$ Take common factor $\lambda$ from the numerator: $...=\dfrac{4\lambda(x+\lambda)^{3}(\lambda^{3}-x^{3})}{(x^{4}+\lambda^{4})^{2}}$
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