Let 4x = u

y = sin u

\(\frac{\text{dy}}{\text{du}}\) = cos u

\(\frac{\text{du}}{\text{dx}}\) = 4

\(\frac{\text{dy}}{\text{dx}}\) = \(\frac{\text{dy}}{\text{du}}\) x \(\frac{\text{du}}{\text{dx}}\) = 4 x cos u

but u = 4x

Therefore, \(\frac{\text{dy}}{\text{dx}}\) = 4cos4x