#### Answer

$\frac{1-7x^{3}}{2√x}$

#### Work Step by Step

Applying product rule,
$\frac{d}{dx}[√x(1-x^{3})]$= $[\frac{d(√x)}{dx} $•$(1-x^{3})]$ + [√x •$\frac{d}{dx}(1-x^{3})$]
= $\frac{1}{2}x^{\frac{1}{2}-1}(1-x^3) + (\frac{d(1)}{dx}-\frac{d(x^{3})}{dx})√x$
= $\frac{1-x^{3}}{2√x} + (0-3x^{2})√x$
=$\frac{1-x^{3}}{2√x} - 3x^{5/2}$
Simplifying, we get
f'(x)= $\frac{1-7x^{3}}{2√x}$