# How do you differentiate  f(x)=sqrt((7-2x^3) using the chain rule.?

$f ' \left(x\right) = \frac{- 3 {x}^{2}}{\sqrt{7 - 2 {x}^{3}}}$
$y = f \left(g \left(x\right)\right) \to \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) g ' \left(x\right)$
$\frac{d}{\mathrm{dx}} \left(\sqrt{7 - 2 {x}^{3}}\right) = \frac{1}{2 \sqrt{7 - 2 {x}^{3}}} \frac{d}{\mathrm{dx}} \left(7 - 2 {x}^{3}\right)$
$\therefore \frac{d}{\mathrm{dx}} \left(\sqrt{7 - 2 {x}^{3}}\right) = \frac{- 3 {x}^{2}}{\sqrt{7 - 2 {x}^{3}}}$