Answer
$$\frac{{dy}}{{d\theta }} = - \sqrt 2 \sin \theta {\left( {\cos \theta } \right)^{\sqrt 2 - 1}}$$
Work Step by Step
$$\eqalign{
& y = {\left( {\cos \theta } \right)^{\sqrt 2 }} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\left( {\cos \theta } \right)}^{\sqrt 2 }}} \right] \cr
& {\text{Use the general power rule for differentiation }}\cr
& \frac{d}{{d\theta }}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{d\theta }}{\text{ }} \cr
& {\text{Let }}n = \sqrt 2 {\text{ and }}u = \cos \theta {\text{; Then}}{\text{,}} \cr
& \frac{{dy}}{{d\theta }} = \sqrt 2 {\left( {\cos \theta } \right)^{\sqrt 2 - 1}}\frac{d}{{d\theta }}\left[ {\cos \theta } \right] \cr
& {\text{solve the derivative}} \cr
& \frac{{dy}}{{d\theta }} = \sqrt 2 {\left( {\cos \theta } \right)^{\sqrt 2 - 1}}\left( { - \sin \theta } \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{d\theta }} = - \sqrt 2 \sin \theta {\left( {\cos \theta } \right)^{\sqrt 2 - 1}} \cr} $$
