University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Practice Exercises - Page 204: 136



Work Step by Step

$$y=\frac{2u2^u}{\sqrt{u^2+1}}$$ Take the natural logarithm of both sides: $$\ln y=\ln(2u2^u)-\ln\sqrt{u^2+1}$$ $$\ln y =\ln2+\ln u+\ln2^u-\ln(u^2+1)^{1/2}$$ $$\ln y =\ln2+\ln u+\ln2^u-\frac{1}{2}\ln(u^2+1)$$ Now take the derivative of both sides, and remember that $(\ln u)'=1/u$: $$\frac{1}{y}\times y'=0+\frac{1}{u}+\frac{(2^u)'}{2^u}-\frac{1}{2}\times\frac{(u^2+1)'}{u^2+1}$$ $$\frac{y'}{y}=\frac{1}{u}+\frac{2^u\ln2}{2^u}-\frac{2u}{2(u^2+1)}$$ $$\frac{y'}{y}=\frac{1}{u}+\ln2-\frac{u}{u^2+1}$$ Find $y'$: $$y'=\Big(\frac{1}{u}+\ln2-\frac{u}{u^2+1}\Big)y$$ $$y'=\Big(\frac{1}{u}+\ln2-\frac{u}{u^2+1}\Big)\frac{2u2^u}{\sqrt{u^2+1}}$$
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