hey guys, just wondering what the derivative of this is... thank u
$\displaystyle
8 \sqrt[3]{x}
$
Hello !
Remember that $\displaystyle \sqrt[n]{x}$ can be rewritten $\displaystyle x^{\tfrac 1n}$
So here, $\displaystyle \sqrt[3]{x}=\dots$
Then remember some basic rules of differentiation :
$\displaystyle (a \cdot f(x))'=a \cdot f'(x)$, a a constant
$\displaystyle (x^n)'=nx^{n-1}$