Question

Find $\frac{dy}{dx}$, when

$x={\cos }^{-1}\frac{1}{\sqrt{1+t^2}}\mathrm{and}y={\sin }^{-1}\frac{t}{\sqrt{1+t^2}},t\in R$

Answer 1

$\mathrm{We}\mathrm{have},x={\cos }^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$

$$\begin{gathered} \Rightarrow \frac{dx}{dt}=\frac{-1}{\sqrt{1-{\left({\displaystyle \frac{1}{\sqrt{1+t^2}}}\right)}^2}}\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right) \\ \Rightarrow \frac{dx}{dt}=\frac{-1}{\sqrt{1-{\displaystyle \frac{1}{\left(1+t^2\right)}}}}\left\{\frac{-1}{2{\left(1+t^2\right)}^{{\displaystyle \frac{3}{2}}}}\right\}\frac{d}{dt}\left(1+t^2\right) \\ \Rightarrow \frac{dx}{dt}=\frac{{\left(1+t^2\right)}^{{\displaystyle \frac{1}{2}}}}{\sqrt{1+t^2-1}}\times \frac{1}{2{\left(1+t^2\right)}^{{\displaystyle \frac{3}{2}}}}\left(2t\right) \\ \Rightarrow \frac{dx}{dt}=\frac{t}{\sqrt{t^2}\times \left(1+t^2\right)} \\ \Rightarrow \frac{dx}{dt}=\frac{1}{1+t^2}...\left(\mathrm{i}\right) \\ \mathrm{Now},y={\sin }^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dy}{dt}=\frac{1}{\sqrt{1-{\left({\displaystyle \frac{1}{\sqrt{1+t^2}}}\right)}^2}}\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right) \\ \Rightarrow \frac{dy}{dt}=\frac{1}{\sqrt{1-{\displaystyle \frac{1}{\left(1+t^2\right)}}}}\left\{\frac{-1}{2{\left(1+t^2\right)}^{{\displaystyle \frac{3}{2}}}}\right\}\frac{d}{dt}\left(1+t^2\right) \\ \Rightarrow \frac{dx}{dt}=\frac{{\left(1+t^2\right)}^{{\displaystyle \frac{1}{2}}}}{\sqrt{1+t^2-1}}\times \frac{-1}{2{\left(1+t^2\right)}^{{\displaystyle \frac{3}{2}}}}\left(2t\right) \\ \Rightarrow \frac{dx}{dt}=\frac{-1}{2\sqrt{t^2}\times \left(1+t^2\right)}\left(2t\right) \\ \Rightarrow \frac{dx}{dt}=\frac{-1}{1+t^2}...\left(\mathrm{ii}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}\left(\mathrm{i}\right), \\ \frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{1}{\left(1+t^2\right)}\times \frac{\left(1+t^2\right)}{-1} \\ \Rightarrow \frac{dy}{dx}=-1 \end{gathered}$$