# Math Help - Quiotent rule

1. ## Quiotent rule

I understand all the steps in solving the problem except for the final answer.

Y= (-2t^2+1) / (3t^2-5)

I know that f (t) * g' (t) = (-12t^3+6t)

g (t) * f' (t) = (-12t^3+20t)

I know that the denominator is always gonna be the second part of the equation squared. But how in the heck do you get the 14?
Thanks for all the help Peace

2. Let $f(t) = -2t^{2} + 1$ and $g(t) = 3t^{2} - 5$.

$y' = \frac{f'(t)g(t) - f(t)g'(t)}{\left[g(t)\right]^{2}}$

You found $f'(t)g(t) = -12t^{3}+20t$ and $f(t)g'(t) = -12t^{3} + 6t$.

So:

$y' = \frac{-12t^{3} + 20t - \left(-12t^{3} + 6t\right)}{\left(3t^{2} - 5\right)^{2}}$

Distribute the negative sign and you should see that it works out to your answer. I'm guessing you forgot the parentheses when substituting everything in.