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

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

Mathematics
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Dx(t/b) = (bt'-b't)/b^2
or Dx(t/b) = Dx(t b^-1) = t' b^-1 + -b^-2 t
thank you sir!

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Other answers:

or ma'am
:) yw
2x(x^3 +1)- (x^2)3x^2 --------------------- ; simplify as needed (x^3 +1)^2
[(x^2)/(x^3 +1)]'=(-x^4+1)/(x^3+1)^2
(U/V)'=9U'V-UV')/V^2
uvs are fine and all; but I perfor t and b to distiguish top and bottom easier
like the product rule: r'l + rl' , not that its hard to keep track of those lol
ok but i meant Dx(t/b) = (bt'-b't)/b^2 must b b't-bt'/b^2
nope; bt' is first.... always has been you bring the bottom up to derive the top
just like yours ;)
let me see smth pls :)
(U/V)'=U'V-UV')/V^2 t/b = t'b - tb' / b^2 t/b = bt' - b't / b^2
its all the same ;)
it is? O.o
i mean in this case u subtract them...not like...U*V which would b ok if u switched places...
nobody switched places; they all 3 are identical
u'v = t'b = bt'
oops :$ :$ ...i got it.....i thought u had written smth else...sorry ^_^
uv' = tb' = b't
I got it :)

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