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February 22nd, 2010, 12:26 AM   #1
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Chain RULE and higher derivatives!! Anyone?

Can somebody please help me with my take home quiz? I've been absent in school for a week because I have a sick and I asked my classmates if they had a quiz, so I rush to my professor and asked if she will give me a take home quiz. so I went to one of my classmate's house to borrow some notes, but sadly I can't understand a thing.

I will be glad forever if someone will explain and answer this for me.

1. y=(2u-3) ^2/3, u=4x^3+1
2. y= cube u+2, u=4x-2
3. y=cube root x, find: y^5
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February 22nd, 2010, 02:27 PM   #2
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Re: Chain RULE and higher derivatives!! Anyone?

Does this article help?

1.

\\
&=\,\frac{2}{3}(2(4x^3\,+\,1)\,-\,3)^{-1/3}(2(4x^3\,+\,1)&#39\\
&=\,\frac{2}{3}(8x^3\,-\,1)^{-1/3}(2(12x^2))\\
&=\,\frac{2}{3}(8x^3\,-\,1)^{-1/3}(24x^2)\\
&=\,\frac{16x^2}{(8x^3\,-\,1)^{1/3}\\
\end{align*}" />
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