differentiate. y = (t^2 + 2)/(t^4 - 5t^2 + 3) y =

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = t^{2}+2$, $v=t^{4}-5t^{2}+3$.

Step2: Find $u'$ and $v'$

Differentiate $u$ with respect to $t$: $u'=\frac{d}{dt}(t^{2}+2)=2t$. Differentiate $v$ with respect to $t$: $v'=\frac{d}{dt}(t^{4}-5t^{2}+3)=4t^{3}-10t$.

Step3: Apply quotient - rule

$y'=\frac{(2t)(t^{4}-5t^{2}+3)-(t^{2}+2)(4t^{3}-10t)}{(t^{4}-5t^{2}+3)^{2}}$. Expand the numerator: [ \begin{align*} &(2t)(t^{4}-5t^{2}+3)-(t^{2}+2)(4t^{3}-10t)\ =&2t^{5}-10t^{3}+6t-(4t^{5}-10t^{3}+8t^{3}-20t)\ =&2t^{5}-10t^{3}+6t - 4t^{5}+10t^{3}-8t^{3}+20t\ =&(2t^{5}-4t^{5})+(-10t^{3}+10t^{3}-8t^{3})+(6t + 20t)\ =&-2t^{5}-8t^{3}+26t \end{align*} ] So, $y'=\frac{-2t^{5}-8t^{3}+26t}{(t^{4}-5t^{2}+3)^{2}}$.

Answer:

$\frac{-2t^{5}-8t^{3}+26t}{(t^{4}-5t^{2}+3)^{2}}$