Differentiation - Power rule Find derivative of this function
$$g(t)=-3t(6t^4-1)^{4/3}$$
I have tried it till the answer:
$ -3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1)  $
$-3t x \frac{4}{3} (6t^4-1)^\frac{1}{3} ( 6X4 t^3 -0 ) $
$ -4t (6t^4 -1)^\frac{4}{3} (24t^3) $ 
$-96t(6t^4-1)^\frac{1}{3} $ 
However , I checked and saw that the answer is
$$-3(6t^4-1)^{1/3}(38t^4-1)$$
I just started learning differentiation, and I don't understand my mistake on why I can't achieve the answer. Thanks!

As there are two terms. You need to use product rule.
Product rule -
$(UV)' = UV' + U'V$
So it should be -
$g(t) = -3t(6t^4-1)^\frac{4}{3}$
$-3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) - 3.\frac{d}{dt} t.(6t^4-1)^\frac{4}{3}$
Take common terms and simplify.
