How do you differentiate y=(t^2+2)/(t^4-3t^2+1)?

Jul 2, 2017

$y ' = \frac{- 2 {t}^{5} - 8 {t}^{3} + 14 t}{{\left({t}^{4} - 3 {t}^{2} + 1\right)}^{2}}$

Explanation:

We know the following rule for the derivative of the ratio:

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

Let's apply it to:

$y = \frac{{t}^{2} + 2}{{t}^{4} - 3 {t}^{2} + 1}$

Then

$y ' = \frac{2 t \cdot \left({t}^{4} - 3 {t}^{2} + 1\right) - \left({t}^{2} + 2\right) \left(4 {t}^{3} - 6 t\right)}{{\left({t}^{4} - 3 {t}^{2} + 1\right)}^{2}}$

Let's expand the numerator:

$y ' = \frac{2 {t}^{5} \cancel{- 6 {t}^{3}} + 2 t - 4 {t}^{5} \cancel{+ 6 {t}^{3}} - 8 {t}^{3} + 12 t}{{\left({t}^{4} - 3 {t}^{2} + 1\right)}^{2}}$

and sum the like terms:

$y ' = \frac{- 2 {t}^{5} - 8 {t}^{3} + 14 t}{{\left({t}^{4} - 3 {t}^{2} + 1\right)}^{2}}$