3 years ago

Question 8 of 10

Mathematics

1 answer:

Alex787 [66]3 years ago

4 0

The answer is c hope I helped!

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Step-by-step explanation:

$f(x)=\frac{u}{v} ,u~ and~ v~ functions~ of~ x\\f'(x)=\frac{v ~u'-u~v' }{v^2} \\f(x)=\frac{x^4}{x^2-4} \\f'(x)=\frac{(x^2-4)*4x^3-x^4(2x)}{(x^2-4)^2} \\=\frac{2x^3(2x^2-8-x^2) }{(x^2-4)^2} \\=\frac{2x^3(x^2-8)}{(x^2-4)^2}$

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