Question

Differentiate the following function w.r.t.$x$
$\sqrt[3]{3(2x^2-7x-4{)}^5}$

Answer 1

$\frac{d}{dx}\left[{(2x^2-7x-4)}^{\frac{5}{3}}\right]$

$=\frac{5}{3}{(2x^2-7x-4)}^{\frac{5}{3}-1}\cdot \frac{d}{dx}[2x^2-7x-4]$

$=\frac{5(2\cdot \frac{d}{dx}\left[x^2\right]-7\cdot \frac{d}{dx}\left[x\right]+\frac{d}{dx}[-4]){(2x^2-7x-4)}^{\frac{2}{3}}}{3}$

$=\frac{5(2\cdot 2x-7\cdot 1+0){(2x^2-7x-4)}^{\frac{2}{3}}}{3}$

$=\frac{5(4x-7){(2x^2-7x-4)}^{\frac{2}{3}}}{3}$

simplifying

$\frac{(20x-35){(2x^2-7x-4)}^{\frac{2}{3}}}{3}$