Question

How do you differentiate `y=sqrt(t/(t^2+4))`?

Answer 1

`dy/dt=(4-t^2)/(2sqrt(t)sqrt((4+t^2)^3))`

Explanation:

`y=sqrt(t/(t^2+4))`

If `s=t/(t^2+4)`, we can use the quotient rule `d/dx(u/v)=((du)/dxv-u(dv)/dx)/v^2` with `u=t` and `v=t^2+4` to find that,

`(ds)/dt=(t^2+4-2t^2)/(t^2+4)^2=(4-t^2)/(4+t^2)^2`.

Now, `y=sqrt(s)`, then, according to the chain rule, `dy/dt=d/dt(sqrt(s))=d/(ds)sqrt(s)*(ds)/dt=1/(2sqrt(s))(4-t^2)/(4+t^2)^2`.

Substitute `s` back in terms of `t`:
`=1/(2sqrt(t/(t^2+4)))(4-t^2)/(4+t^2)^2`
`=(4-t^2)/(2sqrt(t)sqrt((4+t^2)^3))`