## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dy}{dx} = 12x^{3}$
$\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}$ Now, $\frac{dy}{du} = \frac{d(6u-9)}{du} = 6$ And, $\frac{du}{dx} = \frac{d (\frac{1}{2})x^{4}}{dx} = 2x^{3}$ Now, $\frac{dy}{dx} = 12x^{3}$