Question

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

$x=\frac{1-t^2}{1+t^2}\mathrm{and}y=\frac{2t}{1+t^2}$

Answer 1

$\mathrm{We}\mathrm{have},y=\frac{2t}{1+t^2}$

$$\begin{gathered} \Rightarrow \frac{dy}{dt}=\left[\frac{\left(1+t^2\right){\displaystyle \frac{d}{dt}}\left(2t\right)-2t{\displaystyle \frac{d}{dt}}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\right]\left[\mathrm{using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dy}{dt}=\left[\frac{\left(1+t^2\right)\left(2\right)-2t\left(2t\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dy}{dt}=\left[\frac{2+2t^2-4t^2}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dy}{dt}=\left[\frac{2-2t^2}{{\left(1+t^2\right)}^2}\right]...\left(\mathrm{i}\right) \\ \mathrm{and}, \\ x=\frac{1-t^2}{1+t^2} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{dx}{dt}=\left[\frac{\left(1+t^2\right){\displaystyle \frac{d}{dt}}\left(1-t^2\right)-\left(1-t^2\right){\displaystyle \frac{d}{dt}}\left(1+t^2\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{\left(1+t^2\right)\left(-2t\right)-\left(1-t^2\right)\left(2t\right)}{{\left(1+t^2\right)}^2}\right] \\ \Rightarrow \frac{dx}{dt}=\left[\frac{-4t}{{\left(1+t^2\right)}^2}\right]...\left(\mathrm{ii}\right) \\ \mathrm{Dividing}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{by}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}, \\ \frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{2\left(1-t^2\right)}{{\left(1+t^2\right)}^2}\times \frac{{\left(1+t^2\right)}^2}{-4t} \\ \Rightarrow \frac{dy}{dx}=\frac{2\left(1-t^2\right)}{-4t} \\ \Rightarrow \frac{dy}{dx}=\frac{t^2-1}{2t} \end{gathered}$$