## Thomas' Calculus 13th Edition

$(\frac{t+2}{2(t+1)^{3/2}})cos(\frac{t}{\sqrt{t+1}})$
$q=sin(\frac{t}{\sqrt{t+1}})$ Apply the chain rule: $\frac{dq}{dt} =cos(\frac{t}{\sqrt{t+1}}).\frac{d}{dt}(\frac{t}{\sqrt{t+1}})$ =cos $(\frac{t}{\sqrt{t+1}}).\frac{\sqrt{t+1} (1)-t\frac{d}{dt}\sqrt{t+1}}{(\sqrt{t+1})^2}$ =cos $(\frac{1}{\sqrt{t+1}}).\frac{\sqrt{t+1}.\frac{1}{2\sqrt{t+1}}}{t+1}$ =cos $(\frac{t}{\sqrt{t+1}})(\frac{2(t+1)-t}{2(t+1)^{3/2}})$ =$(\frac{t+2}{2(t+1)^{3/2}})cos(\frac{t}{\sqrt{t+1}})$