Question

derivative of
f(x)= 56x/(x^2;4)^2

Answer 1

Use the quotient rule.

Answer 2

\[
\left(\frac{f}{g}\right)' = \frac{f'g-fg'}{g^2}
\]

Answer 3

yea i did and i keeping getting wrong anser

Answer 4

answer

Answer 5

what is \(f\) and \(g\)?

Answer 6

56x/(x^2+4)^2

Answer 7

You could also use product rule if you feel more comfortable using that. Remember: \(\frac{ 1 }{ (a+b)^n } = (a+b)^{-n}\)

Answer 8

I know, so what is \(f'\) and \(g'\)?

Answer 9

56 and 2x

Answer 10

Nope, your \(g'\) is wrong.

Answer 11

You either have to expand it, or use the chain rule.

Answer 12

ohhhhhh

Answer 13

3x^3+32x

Answer 14

I need to see your steps

Answer 15

(x^2+4)^2
(x^2+4)(x^2+4)
x^4+4x^2+4x^2+16
x^4+8x^2+16
3x^2+16x

Answer 16

you messed up on \((x^4)'\)

Answer 17

4x^3

Answer 18

Alright, now you can use the quotient rule.

Answer 19

(56)(x^2+4)^2-(56)(4x^3+16)/((x^2+4)^2)^2

Answer 20

you want 56x on the second term

Answer 21

ops

Answer 22

then simplify a bit and you're done.

Answer 23

simplyging is the problem !