find f '(x)
f(x)=4/(3xto2nd-2x+1)to3rd
Note that $\displaystyle \frac{4}{(3x^2-2x+1)^3}=4(3x^2-2x+1)^{-3}$
Do you know how to apply the chain rule?
$\displaystyle \frac{\,d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)$
Let $\displaystyle f(x)=4x^{-3}$ and let $\displaystyle g(x)=3x^2-2x+1$
Can you take it from here?
--Chris
$\displaystyle \frac{\,d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)$
Let $\displaystyle f(x)=4x^{-3}$ and let $\displaystyle g(x)=3x^2-2x+1$
Then we see that $\displaystyle f'(x)=-12x^{-4}$ and $\displaystyle g'(x)=6x-2$
So when we apply the chain rule, we see that $\displaystyle \frac{\,d}{\,dx}\left[4(3x^2-2x+1)^{-3}\right]=-12(3x^2-2x+1)^{-4}\cdot (6x-2)$
Can you take it from here and simplify this?
--Chris