Answer
$ f'( x ) =\frac{-12 (6x-2)}{(3x^2-2x+1)^4}$
Work Step by Step
Given that $f(x)=\frac{4}{(3x^2-2x+1)^3}$
We can first write it as $f(x)=4(3x^2-2x+1)^{-3}$ so that it is easier to deal with.
Next we write this as $f(x)=4(g(x))^{−3}$, where $f(x)$ is the outside function and $g(x)=3x^2-2x+1$ is the inside function.
To find the answer we need to differentiate the outside function $f(x)$ and times it by the derivative of the inside function $g(x)$.
$f(x)=4(3x^2-2x+1)^{-3}$
$ f (x)=4(g (x))^{-3}$
$f'(x)=-12(g ( x ))^{-3-1}×(g' ( x )) $
$f'(x)=-12(g ( x ))^{-4}×(6x^{2-1}-2) $
$f'(x)=-12(3x^2-2x+1)^{-4}×(6x^1-2) $
$f'(x)=-12(6x-2)(3x^2-2x+1)^{-4}$
$ f'( x ) =\frac{-12 (6x-2)}{(3x^2-2x+1)^4}$
