Answer
$$\frac{dp}{dq}=\frac{\sec^2q}{(1+\tan q)^2}$$
Work Step by Step
$$p=\frac{\tan q}{1+\tan q}$$
$$\frac{dp}{dq}=\frac{d}{dx}\frac{\tan q}{1+\tan q}=\frac{(\tan q)'(1+\tan q)-(\tan q)(1+\tan q)'}{(1+\tan q)^2}$$ $$=\frac{\sec^2q(1+\tan q)-\tan q(0+\sec^2q)}{(1+\tan q)^2}$$ $$=\frac{\sec^2q(1+\tan q)-\tan q\sec^2q}{(1+\tan q)^2}$$ $$=\frac{\sec^2q(1+\tan q-\tan q)}{(1+\tan q)^2}$$ $$=\frac{\sec^2q\times1}{(1+\tan q)^2}$$ $$=\frac{\sec^2q}{(1+\tan q)^2}$$
