How do you find the derivative for sqrt(2x^3 - 3x- 4)?

Apr 29, 2018

$\frac{6 {x}^{2} - 3}{2 \sqrt{2 {x}^{3} - 3 x - 4}}$

Explanation:

By using the chain rule

IF F(x)=f(g(x))

THEN F'(x)=f'(g(x))*g'(x)

$\frac{d}{\mathrm{dx}} \sqrt{2 {x}^{3} - 3 x - 4} = \frac{1}{2 \sqrt{2 {x}^{3} - 3 x - 4}} \times \left(6 {x}^{2} - 3\right)$

=$\frac{6 {x}^{2} - 3}{2 \sqrt{2 {x}^{3} - 3 x - 4}}$