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May 1st, 2008, 03:57 PM   #1
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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.
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May 2nd, 2008, 02:18 PM   #2
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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.
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May 3rd, 2008, 09:40 AM   #3
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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
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May 3rd, 2008, 11:33 AM   #4
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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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