I just want to check another answer to see I am on the right track. I do not think I done this one correctly.

http://i244.photobucket.com/albums/g...g?t=1237847751

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- Mar 23rd 2009, 02:37 PMrobascUsing the chain rule
I just want to check another answer to see I am on the right track. I do not think I done this one correctly.

http://i244.photobucket.com/albums/g...g?t=1237847751 - Mar 23rd 2009, 02:46 PMPinkk
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:

$\displaystyle h'(x)=4x^{2}(x-2)^{3}+2x(x-2)^{4}$ - Mar 23rd 2009, 06:05 PMrobasc
So, how did you determine that you had to use the product rule?

My book states nothing about learning to use the product rule to solve this problem?

I am very confused? - Mar 23rd 2009, 06:24 PMPinkk
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.