what is f'(x) when f(x) = (x^2-2)^2(4-2x)?

Question

is that a derivative

Answer 1

Assuming you mean this:
f(x) = [(x^2-2)^2]*[(4-2x)]

Product rule:
f(x) = p*q
f'(x) = p*q' + q*p'

So use the product rule and also apply the chain rule:
f'(x) = (x^2 - 2)^2 * (-2) + (4 - 2x)(2(x^2 -2) * 2x)
= -2(x^2 - 2)^2 + (4 - 2x)(4x(x^2 - 2))

Take out the common term:
= (x^2 - 2) [-2(x^2 - 2) + (4x)(4 - 2x)]
= (x^2 - 2) (-2x^2 + 4 + 16x - 8x)
= (x^2 - 2) (-2x^2 + 8x + 4)
= -2(x^2 - 2)(x^2 - 4x - 2)

Answer 2

Yep...sure is...but no one knows what your equation is without proper parenthesis use. Is (x^2-2) raised to the power of 2(4-2x) or is it just squared and multiplied by (4-2x)?

Come on people! Work with me!

Answer 3

It's not clear whether the power is just 2 or 2(4-2x). Lets us assume the power is 2.
u=(x^2-2)^2
v=(4-2x)
f'(x)=udv+vdu
=(x^2-2)^2(-2) + (4-2x)(2)(x^2-2)(2x)
=-2(x^2-2)+4x(4-2x)(x^2-2)

Answer 4

ya....ur looking for the derviative......hmmmm........

a(2x^a-2)^a-1(2ax^a-1) {2(4-2x)}+ (x^a2-2) -2

i think....... but i've done a question like this so....??? sorry

Answer 5

you need to stop swearing! to F's is 2 to many!

Answer 6

there are math blogs where people love to do problems, google it

Answer 7

wow if i knew that i would be.........smart..
give me a second while i go check my oven for that pooper scooper....
*distracted...somthing shiny*