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

$\frac{sec^2q}{(1+tanq)^2}$

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