My Math Forum Chain RULE and higher derivatives!! Anyone?

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 February 22nd, 2010, 12:26 AM #1 Newbie   Joined: Feb 2010 Posts: 1 Thanks: 0 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
 February 22nd, 2010, 02:27 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond Re: Chain RULE and higher derivatives!! Anyone? Does this article help? 1. $y\,=\,(2u\,-\,3)^{2/3},\,u\,=\,4x^3\,+\,1$ \begin{align*}y'\,=#39\\ &=\,\frac{2}{3}(2(4x^3\,+\,1)\,-\,3)^{-1/3}(2(4x^3\,+\,1)'\\ &=\,\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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