1. ## derivative cube root

hey guys, just wondering what the derivative of this is... thank u

$\displaystyle 8 \sqrt[3]{x}$

2. Hello !
Originally Posted by jvignacio
hey guys, just wondering what the derivative of this is... thank u

$\displaystyle 8 \sqrt[3]{x}$
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
$\displaystyle (x^n)'=nx^{n-1}$

3. Originally Posted by Moo
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
$\displaystyle (x^n)'=nx^{n-1}$
thank u! got it

4. Originally Posted by jvignacio
hey guys, just wondering what the derivative of this is... thank u

$\displaystyle 8 \sqrt[3]{x}$
$\displaystyle f(x)= 8 \sqrt[3]{x}$
$\displaystyle f(x)= 8 x^{\frac{1}{3}}$
$\displaystyle f'(x)= 8 (\frac{1}{3}) x^{\frac{-2}{3}}$
$\displaystyle f'(x)= \frac{8}{\sqrt[3]{x^2}}$