Find the derivative.
-$$p = \frac { \sec q + \csc q } { \csc q }$$
A) $\frac { d p } { d q } = - \csc q \cot q$
B) $\frac { \mathrm { dp } } { \mathrm { dq } } = \sec ^ { 2 } \mathrm { q }$
C) $\frac { d p } { d q } = \sec q \tan q$
D) $\frac { \mathrm { dp } } { \mathrm { dq } } = \sec ^ { 2 } \mathrm { q } + 1$

Question

Quizplus · Accepted Answer

The Answer of Find the derivative.
-$$p = \frac { \sec q + \csc q } { \csc q }$$
A) $\frac { d p } { d q } = - \csc q \cot q$
B) $\frac { \mathrm { dp } } { \mathrm { dq } } = \sec ^ { 2 } \mathrm { q }$
C) $\frac { d p } { d q } = \sec q \tan q$
D) $\frac { \mathrm { dp } } { \mathrm { dq } } = \sec ^ { 2 } \mathrm { q } + 1$

Find the Derivative p=sec⁡q+csc⁡qcsc⁡qp = \frac { \sec q + \csc q } { \csc q }p=cscqsecq+cscq​

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