Answer
$2[(6.4x-1)^{2}+(5.4x-2)^{3}]\cdot[12.8(6.4x-1)+16.2(5.4x-2)^{2}]$
Work Step by Step
f(x) is a composite function, $f(x)=[u(x)]^{2}$
$u(x)$ is a sum of two composite functions,
$u(x)=[v_{1}(x)]^{2}+[v_{2}(x)]^{3}$
Apply the Generalized Power Rule to $[v_{1}(x)]^{2}$ and $[v_{2}(x)]^{3}$,
$\displaystyle \frac{d}{dx}[v_{1}^{2}]=nv_{1}^{n-1}\frac{dv}{dx}=2(6.4x-1)^{1}\cdot 6.4=12.8(6.4x-1)$
$\displaystyle \frac{d}{dx}[v_{2}^{2}]=nv_{1}^{n-1}\frac{dv}{dx}=3(5.4x-2)^{2}\cdot 5.4=16.2(5.4x-2)^{2}$
So,
$\displaystyle \frac{du}{dx}=12.8(6.4x-1)+10.8(5.4x-2)^2$.
Apply the Generalized Power Rule to $f(x)=[u(x)]^{2}$
$f^{\prime}(x)= \displaystyle \frac{d}{dx}[u^{n}]=nu^{n-1}\frac{du}{dx}$
$=2[(6.4x-1)^{2}+(5.4x-2)^{3}]\cdot[12.8(6.4x-1)+16.2(5.4x-2)^{2}]$
