# How do you differentiate sqrt(2x^3 - 3x- 4)?

Dec 30, 2015

$\frac{3 \left(2 {x}^{2} - 1\right)}{2 \sqrt{2 {x}^{3} - 3 x - 4}}$

#### Explanation:

This can be rewritten as ${\left(2 {x}^{3} - 3 x - 4\right)}^{\frac{1}{2}}$.

According to the chain rule, $\frac{d}{\mathrm{dx}} \left({u}^{\frac{1}{2}}\right) = \frac{1}{2} {u}^{- \frac{1}{2}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Thus,

$\frac{d}{\mathrm{dx}} {\left(2 {x}^{3} - 3 x - 4\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(2 {x}^{3} - 3 x - 4\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(2 {x}^{3} - 3 x - 4\right)$

$\implies \frac{1}{2 \sqrt{2 {x}^{3} - 3 x - 4}} \cdot \left(6 {x}^{2} - 3\right)$

$\implies \frac{3 \left(2 {x}^{2} - 1\right)}{2 \sqrt{2 {x}^{3} - 3 x - 4}}$