Answer
$H'\left( t \right) = - \sin 2t$
Work Step by Step
$$\eqalign{
& H\left( t \right) = {\cos ^2}t \cr
& {\text{Rewrite the function }} \cr
& H\left( t \right) = \left( {\cos t} \right)\left( {\cos t} \right) \cr
& {\text{Differentiating}} \cr
& H'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {\cos t} \right)\left( {\cos t} \right)} \right] \cr
& {\text{Use the product rule}} \cr
& H'\left( t \right) = \cos t\frac{d}{{dt}}\left[ {\cos t} \right] + \cos t\frac{d}{{dt}}\left[ {\cos t} \right] \cr
& H'\left( t \right) = 2\cos t\frac{d}{{dt}}\left[ {\cos t} \right] \cr
& {\text{Use the Derivative formula }}\frac{d}{{dx}}\left[ {\cos x} \right] = - \sin x{\text{ }} \cr
& H'\left( t \right) = 2\cos t\left( { - \sin t} \right) \cr
& H'\left( t \right) = - 2\sin t\cos t \cr
&H'\left( t \right) = -\sin 2t} $$
