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

1 Answer
NickTheTurtle
Oct 31, 2017

#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))#

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