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: 42

Answer

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

Work Step by Step

$$y=\sqrt{(x^2+1)(x-1)^2}$$ Logarithmic differentiation helps to simplify the differentiation of otherwise complex functions by using the algebraic properties of logarithm. In detail, 1) Take the natural logarithm of both sides and simplify: $$\ln y=\ln\sqrt{(x^2+1)(x-1)^2}=\ln\Big(\sqrt {x^2+1}\times\sqrt{(x-1)^2}\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$: $$\ln y=\ln\sqrt {x^2+1}+\ln\sqrt{(x-1)^2}$$ 2) Then we take the derivative of both sides with respect to $x$ and remember that $(\ln x)'=1/x$: $$\frac{1}{y}\times y'=\frac{1}{\sqrt {x^2+1}}(\sqrt {x^2+1})'+\frac{1}{\sqrt{(x-1)^2}}(\sqrt{(x-1)^2})'$$ $$\frac{y'}{y}=\frac{1}{\sqrt {x^2+1}}\times\frac{1}{2\sqrt{x^2+1}}(x^2+1)'+\frac{1}{\sqrt{(x-1)^2}}\times\frac{1}{2\sqrt{(x-1)^2}}((x-1)^2)'$$ $$\frac{y'}{y}=\frac{1}{2(x^2+1)}\times2x+\frac{1}{2(x-1)^2}\times2(x-1)(x-1)'$$ $$\frac{y'}{y}=\frac{x}{x^2+1}+\frac{2(x-1)}{2(x-1)^2}$$ $$\frac{y'}{y}=\frac{x}{x^2+1}+\frac{1}{x-1}=\frac{(x^2-x)+(x^2+1)}{(x^2+1)(x-1)}=\frac{2x^2-x+1}{(x^2+1)(x-1)}$$ 3) Solve for $y'$: $$y'=\frac{2x^2-x+1}{(x^2+1)(x-1)}\times y$$ 4) Finally, substitute for $y=\sqrt{(x^2+1)(x-1)^2}$ $$y'=\frac{(2x^2-x+1)\sqrt{(x^2+1)(x-1)^2}}{(x^2+1)(x-1)}$$
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