# calculus

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Okay, so I have this problem...

this was the original: (t^2-1?t^2+1)^3
and then I got the derivative which looked like this after I finished all the other steps:
3(t^2-1/t^2+1)^2d(t^2+1)(2t)-(t^2-1)(2t)/ (t^2+1)^2

So now, the answer is 3(t^2-1/t^2+1)^2d 4t/(t^2+1)^2

But I am not sure how to from the part I did to the answer...can someone please explain?

• calculus -

whew - try some proofreading. It appears you want the derivative of

((t^2-1)/(t^2+1))^3

and you applied the chain rule to get

3(t^2-1/t^2+1)^2 * ((t^2+1)(2t))-(t^2-1)(2t))/(t^2+1)^2

So far so good, with the calculus. Now start using that half-forgotten algebra:

(t^2+1)(2t))-(t^2-1)(2t) = 2t^3 + 2t - 2t^3 + 2t = 4t

and go directly from there to the given answer (where "d" apparently means *)

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