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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=(2u3) ^2/3, u=4x^3+1 2. y= cube u+2, u=4x2 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: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  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)'\\ &=\,\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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