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: 43


$$y'=\frac{\sqrt t}{2t(t+1)\sqrt{t+1}}$$

Work Step by Step

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