Question

Find the derivative. (Chain Rule)

f(y) = 4 / (4-y^2)^4

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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