Question

f'(x) given f(x)=(2-2x^2)/(2+2x^2)

Answer 1

pulled out a 2, used the quotient rule and now stuck factoring further

Answer 2

what do I do with -4x/(x^2+1)^2

Answer 3

i dont think u can do anything else with it..

Answer 4

why?

Answer 5

wolfram manages to break it down further

Answer 6

polpak! I owe you answer still!

Answer 7

are you sure the quotient rule was correct?

Answer 8

Hrm?

Answer 9

yes, has to be quotient rule

Answer 10

\[\frac{d}{dx}[{(2-2x^2)\over (2+2x^2)}]\]\[ = \frac{d}{dx}[{(2-2x^2)(2+2x^2)^{-1}}]\]\[=(2-2x^2)[-1(2+2x^2)^{-2}(4x)] + [-4x](2+2x^2)^{-1}\]\[= {-4x(2-2x^2 + (2+2x^2)) \over (2+2x)^2}\]\[= {-16x \over (2+2x)^2} \]\[= {-16x \over [2(1+x^2)]^2}\]\[={-16x \over 4(x^2 +1)^2} = {-4x \over (x^2 +1)^2}\]

Answer 11

that is exactly what i got

Answer 12

my issue is that the program wolfram breaks it down further

Answer 13

No it doesn't.

Answer 14

yea.. it can't have..

Answer 15

i had a capital X in the denom! haha! thanks for your help polpak - you rock out!