Question

What is the derivative of (x^2)/(x^3 +1) and why?

Answer 1

Dx(t/b) = (bt'-b't)/b^2

Answer 2

or

Dx(t/b) = Dx(t b^-1) = t' b^-1 + -b^-2 t

Answer 3

thank you sir!

Answer 4

or ma'am

Answer 5

:) yw

Answer 6

2x(x^3 +1)- (x^2)3x^2
--------------------- ; simplify as needed
(x^3 +1)^2

Answer 7

[(x^2)/(x^3 +1)]'=(-x^4+1)/(x^3+1)^2

Answer 8

(U/V)'=9U'V-UV')/V^2

Answer 9

uvs are fine and all; but I perfor t and b to distiguish top and bottom easier

Answer 10

like the product rule: r'l + rl' , not that its hard to keep track of those lol

Answer 11

ok but i meant Dx(t/b) = (bt'-b't)/b^2 must b b't-bt'/b^2

Answer 12

nope; bt' is first.... always has been

you bring the bottom up to derive the top

Answer 13

just like yours ;)

Answer 14

let me see smth pls :)

Answer 15

(U/V)'=U'V-UV')/V^2
t/b = t'b - tb' / b^2
t/b = bt' - b't / b^2

Answer 16

its all the same ;)

Answer 17

it is? O.o

Answer 18

i mean in this case u subtract them...not like...U*V which would b ok if u switched places...

Answer 19

nobody switched places; they all 3 are identical

Answer 20

u'v = t'b = bt'

Answer 21

oops :$ :$ ...i got it.....i thought u had written smth else...sorry ^_^

Answer 22

uv' = tb' = b't

Answer 23

I got it :)