May 1st, 2008, 03:57 PM  #1 
Newbie Joined: May 2008 Posts: 1 Thanks: 0  Chain Rule
I'm stuck on this one problem. y= ((t^3 +1)/(t^3 1)^1/4 I keep coming up to the answer 1/4((t^3 +1)/(t^3 1)^3/4 * ((6t^2)/((t^3 1)^2) Am, I on the right track at least? I'm a little lost on chain rule, product rule combinations. 
May 2nd, 2008, 02:18 PM  #2 
Member Joined: Apr 2008 Posts: 41 Thanks: 0 
On this one you just have to take the numerator and take its derivative. Then multiply that by the denominator. Now you subtract that by the derivative of the denominator and then multiply that by the numerator. And under all of this you square the denominator.

May 3rd, 2008, 09:40 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
If you have trouble remembering the rule for division (I did), just remember that X/Y = XY^{1}, so you can just do (X/Y)' = (XY^{1})' = X'Y^{1} + X(Y^{1})' = X'Y^{1} + X(Y^{2}Y') = (XY+XY')/Y^2 
May 3rd, 2008, 11:33 AM  #4 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 
It may help to do a variable substitution: ((t^3 +1)/(t^3 1)^1/4 = u/v where u = t^3+1 and v = (t^3 1)^1/4 So the derivative is (uv'  vu')/(v^2) Figure out u', figure out v' (using the chain rule), and then substitute back in. 

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