1. ## derivative cube root

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

$
8 \sqrt[3]{x}
$

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

$
8 \sqrt[3]{x}
$
Remember that $\sqrt[n]{x}$ can be rewritten $x^{\tfrac 1n}$

So here, $\sqrt[3]{x}=\dots$

Then remember some basic rules of differentiation :
$(a \cdot f(x))'=a \cdot f'(x)$, a a constant
$(x^n)'=nx^{n-1}$

3. Originally Posted by Moo
Hello !

Remember that $\sqrt[n]{x}$ can be rewritten $x^{\tfrac 1n}$

So here, $\sqrt[3]{x}=\dots$

Then remember some basic rules of differentiation :
$(a \cdot f(x))'=a \cdot f'(x)$, a a constant
$(x^n)'=nx^{n-1}$
thank u! got it

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

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