Answer
$\frac{{dy}}{{dx}} = - 8x\left( {\ln 3} \right){\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^3}\left( {{3^{\cos \left( {{x^2}} \right)}}} \right)\sin \left( {{x^2}} \right)$
Work Step by Step
$$\eqalign{
& y = {\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^4} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)}^4}} \right] \cr
& \cr
& {\text{Use The Power Rule Combined with the Chain Rule}} \cr
& \frac{d}{{dx}}{\left[ {g\left( x \right)} \right]^n} = n{\left[ {g\left( x \right)} \right]^{n - 1}}g'\left( x \right) \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = 4{\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^{4 - 1}}\frac{d}{{dx}}\left[ {{3^{\cos \left( {{x^2}} \right)}} - 1} \right] \cr
& \frac{{dy}}{{dx}} = 4{\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^3}\left( {\frac{d}{{dx}}\left[ {{3^{\cos \left( {{x^2}} \right)}}} \right] - \frac{d}{{dx}}\left[ 1 \right]} \right) \cr
& \frac{{dy}}{{dx}} = 4{\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^3}\frac{d}{{dx}}\left[ {{3^{\cos \left( {{x^2}} \right)}}} \right] \cr
& \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {{a^u}} \right] = {a^u}\ln a\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = 4{\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^3}\left[ {{3^{\cos \left( {{x^2}} \right)}}} \right]\left( {\ln 3} \right)\frac{d}{{dx}}\left[ {\cos \left( {{x^2}} \right)} \right] \cr
& \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {\cos u} \right] = - \sin u\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = 4{\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^3}\left[ {{3^{\cos \left( {{x^2}} \right)}}} \right]\left( {\ln 3} \right)\left[ { - \sin \left( {{x^2}} \right)} \right]\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& \frac{{dy}}{{dx}} = 4{\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^3}\left[ {{3^{\cos \left( {{x^2}} \right)}}} \right]\left( {\ln 3} \right)\left[ { - \sin \left( {{x^2}} \right)} \right]\left( {2x} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = - 8x\left( {\ln 3} \right){\left( {{3^{\cos \left( {{x^2}} \right)}} - 1} \right)^3}\left( {{3^{\cos \left( {{x^2}} \right)}}} \right)\sin \left( {{x^2}} \right) \cr} $$
