Answer
$$\frac{\partial }{{\partial p}}\left[ {{e^{ - 7p/q}}} \right] = - \frac{7}{q}{e^{ - 7p/q}}{\text{ and }}\frac{\partial }{{\partial q}}\left[ {{e^{ - 7p/q}}} \right] = \frac{{7p{e^{ - 7p/q}}}}{{{q^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{\partial }{{\partial p}}\left[ {{e^{ - 7p/q}}} \right] \cr
& {\text{Differentiating with respect to }}p,{\text{ treat }}q{\text{ as a constant}} \cr
& = {e^{ - 7p/q}}\frac{\partial }{{\partial p}}\left[ { - \frac{{7p}}{q}} \right] \cr
& = {e^{ - 7p/q}}\left( { - \frac{7}{q}} \right) \cr
& = - \frac{7}{q}{e^{ - 7p/q}} \cr
& \cr
& \frac{\partial }{{\partial q}}\left[ {{e^{ - 7p/q}}} \right] \cr
& {\text{Differentiating with respect to }}q,{\text{ treat }}p{\text{ as a constant}} \cr
& = {e^{ - 7p/q}}\frac{\partial }{{\partial q}}\left[ { - \frac{{7p}}{q}} \right] \cr
& = - 7p{e^{ - 7p/q}}\frac{\partial }{{\partial q}}\left[ {\frac{1}{q}} \right] \cr
& = - 7p{e^{ - 7p/q}}\left( { - \frac{1}{{{q^2}}}} \right) \cr
& = \frac{{7p{e^{ - 7p/q}}}}{{{q^2}}} \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& \frac{\partial }{{\partial p}}\left[ {{e^{ - 7p/q}}} \right] = - \frac{7}{q}{e^{ - 7p/q}}{\text{ and }}\frac{\partial }{{\partial q}}\left[ {{e^{ - 7p/q}}} \right] = \frac{{7p{e^{ - 7p/q}}}}{{{q^2}}} \cr} $$
