## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dt}=-\frac{1}{4}(3t+(2+(1-t)^{1/2})^{1/2})^{-1/2}\Big(3+\frac{1}{2}(2+(1-t)^{1/2})^{-1/2}\Big)(1-t)^{-1/2}$$
$$\frac{dy}{dt}=\frac{d}{dt}\sqrt{3t+\sqrt{2+\sqrt{1-t}}}=\frac{d}{dt}(3t+(2+(1-t)^{1/2})^{1/2})^{1/2}$$ Don't freak out! We will start deriving from the most outer part to the inner one, following the Chain Rule: $$\frac{dy}{dt}=\frac{1}{2}(3t+(2+(1-t)^{1/2})^{1/2})^{-1/2}\frac{d}{dt}(3t+(2+(1-t)^{1/2})^{1/2})$$ $$\frac{dy}{dt}=\frac{1}{2}(3t+(2+(1-t)^{1/2})^{1/2})^{-1/2}\Big(3+\frac{1}{2}(2+(1-t)^{1/2})^{-1/2}\Big)\frac{d}{dt}(2+(1-t)^{1/2})$$ $$\frac{dy}{dt}=\frac{1}{2}(3t+(2+(1-t)^{1/2})^{1/2})^{-1/2}\Big(3+\frac{1}{2}(2+(1-t)^{1/2})^{-1/2}\Big)\Big(\frac{1}{2}(1-t)^{-1/2}\frac{d}{dt}(1-t)$$ $$\frac{dy}{dt}=-\frac{1}{4}(3t+(2+(1-t)^{1/2})^{1/2})^{-1/2}\Big(3+\frac{1}{2}(2+(1-t)^{1/2})^{-1/2}\Big)(1-t)^{-1/2}$$