Use the quotient rule: g(t)=t^3-2t/t^2+3?

Answer 1

Here's how the quotient rule is set up:
f'(x)=g(x)f'(x)-g'(x)f(x)/(g(x))^2
f'(x)=((t^2+3)(3t^2-2))-((2t)(t^3-2t))/
(t^2+3)^2
=((3t^4+7t^2-6)-(2t^4-4t^2))/(t^2+3)^2
=(3t^4-2t^4+7t^2+4t^2-6)/(t^2+3)^2
=(t^4+11t^2-6)/(t^2+3)^2

I hope this helps!

Answer 2

g(t)=t³-2t/t²+3
t² = -3
t = \/-3
{"t" belongs to R | "t" must be different of \/-3} because the sqrt can't be negative radical.
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