Verified answer

If I interpretted your question correctly, they basically want you to find the point at which the derivative is 0.

so do this:

ds/dt=d/dt((3t)*((t^2+11)^-1))

notice there are 2 functions to s so write this:

s(t)=x(t)y(t)

where x(t)=3t

and y(t)=(t^2+11)^-1

now the product rule is:

ds/dt=d/dt(xy)=xy'+x'y

evaluate dx/dt:

dx/dt=3

evaluate dy/dt using the chain rule (df(g(t))/dt=df/dg*dg/dt):

dy/dt=(-1)((t^2+11)^-2)(2t)

plug it back into the product rule:

ds/dt=(3t)*((-1)((t^2+11)^-2)(2t))+(3)*((t^2+11)^-1)

now to find where the particle is at rest, solve for when ds/dt=0

to do this, set ds/dt to 0 so that you have,

0=(3t)*((-1)((t^2+11)^-2)(2t))+(3)*((t^2+11)^-1)

(simplify the immediate above if you wish)

now t is the only variable in the equation so solve for t. if you have trouble solving for t, you should go simplify and rearrange as needed. if it's STILL too much trouble, then go use a program called Mathematica.