Answer
$y'=4(1-2x)\sin{(1-2x)}^2$.
Work Step by Step
$u=(1-2x)^2$; $\dfrac{du}{dx}=-4(1-2x)$.
$\dfrac{dy}{du}=-\sin{u}$.
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=4(1-2x)\sin{(1-2x)}^2$ .
Note: $\dfrac{du}{dx}$ is found using the Chain Rule as follows:
$\dfrac{du}{dx}=(-2)(2)(1-2x)^{2-1}=-4(1-2x).$
