Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 6

Answer

y' = -$ \frac{1}{x(lnx)^{2}}$

Work Step by Step

y = $\frac{1}{lnx}$ Recall for this problem: Quotient Rule Differentiate: y' = ($\frac{1}{lnx}$)' y' = $\frac{lnx(1)' - 1(lnx)'}{(lnx)^{2}}$ y' = $\frac{0 - \frac{1}{x}}{(lnx)^{2}}$ y' = $\frac{-\frac{1}{x}}{(lnx)^{2}}$ y' = -$ \frac{1}{x(lnx)^{2}}$
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