I just want to check another answer to see I am on the right track. I do not think I done this one correctly.
Totally wrong I'm afraid (I have no idea what you did). This requires the product rule and the chain rule.
Let $\displaystyle f(x)=x^{2},g(x)=(x-2)^{4},h(x)=f(x)g(x)$
The product rule: $\displaystyle h'(x)=f(x)g'(x)+f'(x)g(x)$
Applying the chain rule here:
$\displaystyle h'(x)=x^{2}(\frac{d}{dx}(x-2)^{4})+(\frac{d}{dx}x^{2})(x-2)^{4}$
You will need to apply the chain rule to $\displaystyle (x-2)^{4}$, which ends up being $\displaystyle 4(x-2)^{3}(\frac{d}{dx}(x-2))$ which is simply $\displaystyle 4(x-2)^{3}$. Getting back to the whole derivative we will have:
$\displaystyle h'(x)=4x^{2}(x-2)^{3}+2x(x-2)^{4}$
You have a function such that $\displaystyle h(x)=f(x)g(x)$, where $\displaystyle f(x)=x^{2}$ and $\displaystyle g(x)=(x-2)^{4}$. If you have not learned the product rule yet, then your only other option is to expand $\displaystyle (x-2)^{4}$ and multiply every term by $\displaystyle x^{2}$, and then differentiate.