University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 40

Answer

$$y'=-\frac{5(3x+2)}{2(x+1)(x+2)}$$

Work Step by Step

$$y=\ln\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}$$ The derivative of $y$: $$y'=\frac{1}{\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}}\Big(\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}\Big)'=\frac{1}{\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}}\Big(\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}\Big)'$$ We have $(\sqrt u)'=(u^{1/2})'=\frac{1}{2}u^{-1/2}=\frac{1}{2\sqrt u}$ Therefore, $$y'=\frac{1}{\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}}\times\frac{1}{2\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}}\times\Big(\frac{(x+1)^5}{(x+2)^{20}}\Big)'$$ $$y'=\frac{1}{2\frac{(x+1)^5}{(x+2)^{20}}}\times\Big(\frac{(x+1)^5}{(x+2)^{20}}\Big)'=\frac{(x+2)^{20}}{2(x+1)^5}\times\Big(\frac{(x+1)^5}{(x+2)^{20}}\Big)'$$ We now examine $\Big(\frac{(x+1)^5}{(x+2)^{20}}\Big)'$, using the Quotient Rules: $$\Big(\frac{(x+1)^5}{(x+2)^{20}}\Big)'=\Big[\Big(\frac{x+1}{(x+2)^4}\Big)^5\Big]'=5\Big(\frac{x+1}{(x+2)^4}\Big)^4\Big(\frac{x+1}{(x+2)^4}\Big)'$$ $$=\frac{5(x+1)^4}{(x+2)^{16}}\times\frac{(x+2)^4-4(x+2)^3(x+1)}{(x+2)^8}$$ $$=\frac{5(x+1)^4}{(x+2)^{16}}\times\frac{(x+2)^3(x+2-4(x+1))}{(x+2)^8}$$ $$=\frac{5(x+1)^4}{(x+2)^{16}}\times\frac{x+2-4x-4}{(x+2)^5}$$ $$=\frac{5(x+1)^4}{(x+2)^{16}}\times\frac{-3x-2}{(x+2)^5}=-\frac{5(x+1)^4(3x+2)}{(x+2)^{21}}$$ Apply the result back to the calculation of $y'$: $$y'=\frac{(x+2)^{20}}{2(x+1)^5}\times\Big(-\frac{5(x+1)^4(3x+2)}{(x+2)^{21}}\Big)$$ $$y'=-\frac{5(3x+2)}{2(x+1)(x+2)}$$
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