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

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Answer 1 · 2017-07-02T16:46:16

$y'=(-2t^5-8t^3+14t)/((t^4-3t^2+1)^2)$

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

$y=f(t)/g(t)->y'=(f'(t)*g(t)-f(t)*g'(t))/g^2(t)$

Let's apply it to:

$y=(t^2+2)/(t^4-3t^2+1)$

Then

$y'=(2t*(t^4-3t^2+1)-(t^2+2)(4t^3-6t))/((t^4-3t^2+1)^2)$

Let's expand the numerator:

$y'=(2t^5cancel(-6t^3)+2t-4t^5cancel(+6t^3)-8t^3+12t)/((t^4-3t^2+1)^2)$

and sum the like terms:

$y'=(-2t^5-8t^3+14t)/((t^4-3t^2+1)^2)$

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