Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 141: 30



Work Step by Step

Given that $p=\frac{tanq}{1+tanq}$ Differentiating both sides, we obtain: $p^{\prime}=\frac{(1+tanq).tanq^{\prime}-tanq.(1+tan q)^{\prime}}{(1+tanq)^2}$ $\implies p^{\prime}=\frac{(1+tanq).sec^2q-tanq.sec^2q}{(1+tanq)^2}$ $p^{\prime}=\frac{sec^2q+sec^2q\space tanq-tanq\space sec^2q}{(1+tanq)^2}$ $p^{\prime}=\frac{sec^2q}{(1+tanq)^2}$
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