## Thomas' Calculus 13th Edition

$\frac{cos(\sqrt{1+\sqrt{t}})}{\sqrt{t+t\sqrt{t}}}$
y=4$sin(\sqrt{1+\sqrt{t}})$ Apply the chain rule: $\frac{dy}{dt}=4cos(\sqrt{1+\sqrt{t}}).\frac{d}{dt}(\sqrt{1+\sqrt{t}})$ =$4cos(\sqrt{1+\sqrt{t}}).\frac{1}{2\sqrt{\sqrt{1+\sqrt{t}}}}.\frac{d}{dt}(\sqrt{1+\sqrt{t}})$ =$\frac{cos(\sqrt{1+\sqrt{t}})}{\sqrt{t+t\sqrt{t}}}$