Answer
$f^{\prime}(x)=-4 \cos (-4x-5)$
Work Step by Step
We have: $f(x)=\sin (-4x-5)$
We need to differentiate both sides with respect to $x$.
$f^{\prime}(x)=\cos (-4x-5) \dfrac{d}{dx} (-4x-5) \\=\cos (-4x-5) \times (-4)$
Therefore, $f^{\prime}(x)=-4 \cos (-4x-5)$
