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

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Answer 1 · 2017-10-31T13:20:47

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

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

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

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