Asked by gagaformath
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?
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?
Answers
Answered by
Steve
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 *)
((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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