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

$$\frac{dw}{ds}=\cos(e^{\sqrt{3/2}})e^{\sqrt{3/2}}\frac{3\sqrt2}{4}$$
$$w=\sin(e^{\sqrt r})\hspace{2cm}r=3\sin(s+\pi/6)$$ 1) First, we find $dw/ds$ as normal, using $w=\sin(e^{\sqrt r})$: $$\frac{dw}{ds}=\cos(e^{\sqrt r})\frac{d(e^{\sqrt r})}{ds}=\cos(e^{\sqrt r})e^{\sqrt r}\frac{d(\sqrt r)}{ds}$$ $$=\cos(e^{\sqrt r})e^{\sqrt r}\frac{1}{2\sqrt r}\frac{dr}{ds}$$ 2) Next, we find $dr/ds$, using $r=3\sin(s+\pi/6)$: $$\frac{dr}{ds}=3\cos(s+\pi/6)\frac{d(s+\pi/6)}{ds}$$ $$=3\cos(s+\pi/6)(1+0)=3\cos(s+\pi/6)$$ At $s=0$, we have: $$\frac{dr}{ds}=3\cos(0+\pi/6)=3\cos(\pi/6)=\frac{3\sqrt3}{2}$$ $$r=3\sin(0+\pi/6)=3\sin(\pi/6)=\frac{3}{2}$$ 3) Replace these results back to $dw/ds$ to find $dw/ds$ at $s=0$: $$\frac{dw}{ds}=\cos(e^{\sqrt{3/2}})e^{\sqrt{3/2}}\frac{1}{2\sqrt{3/2}}\frac{3\sqrt3}{2}$$ $$\frac{dw}{ds}=\cos(e^{\sqrt{3/2}})e^{\sqrt{3/2}}\frac{3\sqrt3}{\frac{4\sqrt3}{\sqrt2}}=\cos(e^{\sqrt{3/2}})e^{\sqrt{3/2}}\frac{3}{\frac{4}{\sqrt2}}$$ $$\frac{dw}{ds}=\cos(e^{\sqrt{3/2}})e^{\sqrt{3/2}}\frac{3\sqrt2}{4}$$