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: 44



Work Step by Step

$$y=\sqrt{\frac{1}{t(t+1)}}=\sqrt{\frac{1}{t^2+t}}$$ 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: $$\ln y=\ln\sqrt{\frac{1}{t^2+t}}=\ln\Big(\frac{1}{\sqrt{t^2+t}}\Big)$$ Recall that $\ln(A/ B)=\ln A-\ln B$: $$\ln y=\ln1-\ln\sqrt{t^2+t}=0-\ln\sqrt{t^2+t}$$ $$\ln y =-\ln\sqrt{t^2+t}$$ 2) Then we take the derivative of both sides with respect to $x$ and remember that $(\ln x)'=1/x$: $$\frac{1}{y}\times y'=-\frac{1}{\sqrt{t^2+t}}(\sqrt{t^2+t})'$$ $$\frac{y'}{y}=-\frac{1}{\sqrt{t^2+t}}\times\frac{1}{2\sqrt{t^2+t}}(t^2+t)'$$ $$\frac{y'}{y}=-\frac{(t^2+t)'}{2(t^2+t)}$$ $$\frac{y'}{y}=-\frac{2t+1}{2(t^2+t)}$$ 3) Solve for $y'$: $$y'=-\frac{2t+1}{2(t^2+t)}\times y$$ 4) Finally, substitute for $y=\sqrt{\frac{1}{t^2+t}}$ $$y'=-\frac{2t+1}{2(t^2+t)}\times\sqrt{\frac{1}{t^2+t}}=-\frac{2t+1}{2(t^2+t)\sqrt{t^2+t}}$$
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