Answer
$$\frac{{dy}}{{dt}} = - {8^{ - t}}\left( {\ln 8} \right)$$
Work Step by Step
$$\eqalign{
& y = {8^{ - t}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{8^{ - t}}} \right] \cr
& {\text{Use the general power rule for differentiation }}\cr
&\frac{d}{{dt}}\left[ {{a^u}} \right] = {a^u}\left( {\ln a} \right)\frac{{du}}{{dt}}{\text{ }} \cr
& {\text{for this exercise, let }}a = 8{\text{ and }}u = - t{\text{}}{\text{,}} \cr
& \frac{{dy}}{{dt}} = {8^{ - t}}\left( {\ln 8} \right)\frac{d}{{dt}}\left[ { - t} \right] \cr
& {\text{solve the derivative}} \cr
& \frac{{dy}}{{dt}} = {8^{ - t}}\left( {\ln 8} \right)\left( { - 1} \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{dt}} = - {8^{ - t}}\left( {\ln 8} \right) \cr} $$
