Answer
$y'=\frac{t^4-8t^3+6t^2+9}{(t^2-4t+3)^2}$
Work Step by Step
$y=\frac{t^3+3t}{t^2-4t+3}$
Use the quotient rule.
$y'=\frac{(t^2-4t+3)(3t^2+3)-(t^3+3t)(2t-4)}{(t^2-4t+3)^2}$
Distribute.
$y'=\frac{3t^4+3t^2-12t^3-12t+9t^2+9-2t^4+4t^3-6t^2+12t}{(t^2-4t+3)^2}$
Combine like terms.
$y'=\frac{t^4-8t^3+6t^2+9}{(t^2-4t+3)^2}$
