find f' of the equation?

Question

of f(x) = (x^2-2)^2(4-2x)
i dont get it

Answer 1

use product rule,

2(x^2 - 2)(2x) (4 - 2x) + (x^2 - 2)(-2)
4x(x^2 - 2)(4 - 2x) - 2(x^2 - 2)

= 2(x^2 - 2) [ x(4 - 2x) - 1]
= 2(x^2 - 2) [ 4x - 2x^2 - 1]

Answer 2

Method 1
f (x) = (x² - 2)² (4 - 2x)
Using product rule and function of a function:-
f ` (x) = 2 (x² - 2) (2x) (4 - 2x) - 2 (x² - 2)²
f `(x) = 2 (x² - 2) [ (2x) (4 - 2x) - (x² - 2) ]
f `(x) = 2 (x² - 2)] [ 8x - 4x² - x² + 2 ]
f `(x)= 2 (x² - 2) (2 + 8x - 5x²)

Method 2
Multiply out at start:-
f(x) = (x^4 - 4x² + 4) (4 - 2x)
f (x) = 4x^4 - 2x^5 - 16x² + 8x³ + 16 - 8x
f ` (x) = 16x³ - 10x^4 - 32x + 24x² - 8

Method 1 answer (when expanded) = Method 2 answer

Answer 3

you have to use the product rule

let assume,
(x^2-2)^2 = u
thus u = x^4 + 4 + 4 = x^4 + 8

differentiating u we get,
u' = 4x^3

let also assume (4 - 2x) = v
thus differentiating v we get,
v' = -2

using the product rule: uv' + vu'

similarly,

f(x)' = (x^4 + 8)(-2) + (4 - 2x)(4x^3)
= -2x^4 - 16 + 16x^3 - 8x^4
= -10x^4 + 16x^3 - 16 ................ ANS


hope this helps!! :-)

Answer 4

let u = (x²-2)² and v = (4-2x)
1. find u'
u' = 2(x²-2)(2x)
2. find v'
v' = -2

the formula for the product rule is
dy/dx = u'v + uv'
therefore,
f'(x) = 2(x²-2)(2x)(4-2x) + (x²-2)²(-2)
= 4x(x²-2)(4-2x) - 2(x²-2)²
= 2(x²-2)[2x(4-2x) - (x²-2)]
= 2(x²-2)(8x - 4x² - x² + 2)
= 2(x²-2)(8x - 5x² + 2)

Answer 5

f(x)= (x^2-2)^2(4-2x)
=(x^4+4-4x^2)(4-2x)
=(4x^4+16-16x^2-2x^5-8x+8x^3)
= -2x^5+4x^4+8x^3-16x^2-8x+16

f'=df(x)/dx
= -2*5x^4+4*4x^3+8*3x^2-16*2x-8+0
= -10x^4 + 16x^3 + 24x^2 - 32x - 8

Answer 6

using product rule: uv'-vu'

-> (x^2-2)^2(-2)-[(4-2x)2(x^2)(2x)]
-> f'(x)= -2(x^2-2)^2[1+x(4-2x)^2]

Answer 7

hmm i don't get it either. usually there would be a number or something in the f(x)...for example f(4)=(4^2-2)^2(4-8)

Answer 8

what would be the value of x?

example: x=1

then substitute the value of x..

f(1)=(1^2-2)^2[4-2(1)]

f=(1^0)^6

f=1^6

f=1

that's it!

Answer 9

you factor out an X

Answer 10

dayum let me know when you figured that out lol