Question

derive (x^4-3x^2)/(x^2-1)^2

Answer 1

\[\frac{ x^4-3x^2 }{ (x^2-1)^2 }\]

Answer 2

i appied quotient rule

Answer 3

i think it's correct, but would like to be sure

Answer 4

You mean "differentiate" or "find the derivative of"? Derive has a different meaning...

Answer 5

yes

Answer 6

\[\frac{ (4x^3-6x(x^2-1)^2) -(x^4-3x^2)(2(x^2-1))2x }{ (x^2-1)^4 }\]

Answer 7

i think that's it

Answer 8

\[\Large \frac{ d }{ dx} \frac{ x^4-3x^2 }{ (x^2-1)^2 } =\]\[\Large \frac{ (4x^3 -6x)(x^2-1)^2 - 2*2x(x^2-1)(x^4-3x^2) }{ (x^2-1)^4 }\]

Answer 9

alright, what does the 2*2x(x^2-1) become?

Answer 10

my professor wrote 2x^3-2x

Answer 11

i have something different..

Answer 12

It looks like he might've made a mistake, then.

Answer 13

Or he's factored it out somewhere else. Can't say w/o seeing the whole solution.

Answer 14

hmm, let me check

Answer 15

i'm guessing it's a mistake then, does it matter if it's the second derivative of something to find local extrema?

Answer 16

what i asked to differentiate was already diff. , so this is actually the second derivative

Answer 17

If he's setting it equal to zero, then yeah he might've factored out a 2 and divided it off (since you can factor a 2 out of the first set of brackets too)

Answer 18

hm? how..why?

Answer 19

oh i get it.., but i see the 2 still there in the first set

Answer 20

Factor a 2 out of the first brackets \[\Large \frac{2 (2x^3 -3x)(x^2-1)^2 - 2*2x(x^2-1)(x^4-3x^2) }{ (x^2-1)^4 } = 0\]
notice the common factor 2 on both brackets. It's equal to zero, so you can divide both sides by 2.

Answer 21

yea, he didn't do that on the left bracket, it stays 4x^3

Answer 22

i think he made a mistake

Answer 23

i'll just do it my way then

Answer 24

Yeah he prob made a mistake then