Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 374: 42



Work Step by Step

Rewrite $ y $ as follows $$ y=\tan^{-1}u, \quad u:=\frac{1+t}{1-t}.$$ Now, uisng the chain rule, the derivative $ y'$ is given by $$ y'=\frac{1}{1+u^2}\frac{du}{dt}$$ where $\frac{du}{dt}$, by the qoutient rule, is given by $$\frac{du}{dt}=\frac{(1-t)+(1+t)}{(1-t)^2}=\frac{2}{(1-t)^2}.$$ Hence, we get $$ y'=\frac{2}{(1-t)^2+(1+t)^2} .$$ This can be simplified to: $$y'=\dfrac{2}{1^2-2t+t^2+1+2t+t^2}\\ y'=\dfrac{2}{2+2t^2}\\ y'=\dfrac{1}{1+t^2}$$
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