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

Question

8710 views around the world

You can reuse this answer
Creative Commons License

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)#