Answer
$k^{\prime}(x)=-2 \sin x \cos x$
Work Step by Step
We have: $k(x)=\cos^2 x$
We need to differentiate both sides with respect to $x$.
$k^{\prime}(x)=\dfrac{d}{dx} [\cos^2 x]\\=2 \cos x \times \dfrac{d}{dx} (\cos x) \\=2 \cos x \times (-\sin x)$
Therefore, $k^{\prime}(x)=-2 \sin x \cos x$
