How do you differentiate  y =(sqrtx-3)^3  using the chain rule?

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2 \sqrt{x}} \times {\left(\sqrt{x} - 3\right)}^{2}$

Explanation:

$y = {\left(\sqrt{x} - 3\right)}^{3}$

Let
$t = \sqrt{x} - 3$

Let
$u = \sqrt{x}$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$
$v = 3$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 0$
$t = u - v$

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} - \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}} - 0$

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$

$y = {t}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {t}^{2} \frac{\mathrm{dt}}{\mathrm{dx}}$

Substituting for t and dt/dx

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \times {\left(\sqrt{x} - 3\right)}^{2} \times \frac{1}{2 \sqrt{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2 \sqrt{x}} \times {\left(\sqrt{x} - 3\right)}^{2}$