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