University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 417: 34

Answer

$-\dfrac{\ln 2}{ \sqrt{1+4^\theta}}$

Work Step by Step

We are given: $y=csch^{-1} 2^{\theta}$ Recall the formula: $\dfrac{d (csch^{-1} x)}{dx}=\dfrac{-1}{|x| \sqrt{1 + x^2}}$ and $\dfrac{d m^x}{dx}=m^x \ln m$ We need to use the product rule to get the differentiation: Thus, $\dfrac{dy}{d \theta}=\dfrac{-1}{|2^{\theta}| \sqrt{1 + ((2)^{\theta})^2}}(2)^{\theta} \ln 2=-\dfrac{\ln 2}{ \sqrt{1+4^\theta}}$
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