Find the derivative.
-$$p = \frac { 4 + \sec q } { 4 - \sec q }$$
A) $\frac { d p } { d q } = \frac { 8 \tan ^ { 2 } q } { ( 4 - \sec q ) ^ { 2 } }$
B) $\frac { d p } { d q } = \frac { 8 \sin q } { ( 4 \cos q - 1 ) ^ { 2 } }$
C) $\frac { d p } { d q } = - \frac { 8 \sin q } { ( 4 \cos q - 1 ) ^ { 2 } }$
D) $\frac { d p } { d q } = - \frac { 2 \sec ^ { 2 } q \tan q } { ( 4 - \sec q ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

To solve this, we need to use the quotient rule. Let $u = 4 + \sec q$ and $v = 4 - \sec q$, so $p = \frac{u}{v}$. Then,
\begin{align*}
\frac{d}{dq} p &= \frac{v \frac{d}{dq} u - u \frac{d}{dq} v}{v^2} \
&= \frac{(4 - \sec q) \cdot \sec q 	an q - (4 + \sec q) \cdot (-\sec q 	an q)}{(4 - \sec q)^2} \
&= \frac{8 \sin q}{(4 \cos q - 1)^2}.
\end{align*}
Therefore, the answer is $\boxed{	extbf{(B)}}$.

Find the Derivative p=4+sec⁡q4−sec⁡qp = \frac { 4 + \sec q } { 4 - \sec q }p=4−secq4+secq​

Find the Derivative $p = \frac { 4 + \sec q } { 4 - \sec q }$