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## anonymous 4 years ago Find the derivative. (Chain Rule) f(y) = 4 / (4-y^2)^4

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1. anonymous

set u = 4-y^2 du = -2y f(u) = 4/u^4 = 4u^-4 f'(u) = -16u^-5 = -16/u^5 f'(y) = f'(u)*du f'(y) = [-16/(4-y^2)^5]*(-2y) f'(y) = 32y/(4-y^2)^5

2. anonymous

$f(y)=4(4-y^2)^{-4}$ $f'(y)=-16(4-y^2)^{-5}*(-2y)=\frac{32y}{(4-y^2)^5}$

3. anonymous

The back of the textbook with answers says that its 24y/(4-y^2)^5 ...

4. anonymous

@ Kevin rick is correct,it must be a misprint in your textbook.

5. anonymous

I think you guys are correct. I sure hope that it's a misprint. :) Thank you!

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