Answer
$$y'=-\frac{1}{2v\sqrt{v-1}}$$
Work Step by Step
$$y=\sin^{-1}\Big(\frac{1}{\sqrt{v}}\Big)\hspace{1cm}v \gt1$$
According to the Chain Rule, we have $$y'=\frac{d}{dv}\sin^{-1}\Big(\frac{1}{\sqrt{v}}\Big)$$
$$y'=\frac{d}{d\Big(\frac{1}{\sqrt v}\Big)}\sin^{-1}\Big(\frac{1}{\sqrt{v}}\Big)\times\frac{d}{dv}\Big(\frac{1}{\sqrt v}\Big)$$
$$y'=\frac{1}{\sqrt{1-\Big(\frac{1}{\sqrt v}\Big)^2}}\times\frac{(1)'\sqrt v-1(\sqrt v)'}{v}=\frac{1}{\sqrt{1-\frac{1}{v}}}\times\frac{0\times\sqrt v-\frac{1}{2\sqrt v}}{v}$$
$$y'=\frac{1}{\sqrt{1-\frac{1}{v}}}\times\Big(-\frac{\frac{1}{2\sqrt v}}{v}\Big)=-\frac{1}{\sqrt{1-\frac{1}{v}}}\times\frac{1}{2v\sqrt v}$$
$$y'=-\frac{1}{2v\sqrt{v\Big(1-\frac{1}{v}\Big)}}=-\frac{1}{2v\sqrt{v-1}}$$
