Find dy dx , when x= 1-t^2 1+t^2 and y= 2t 1+t^2

Question

Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{1-\text{t}^2}{1+\text{t}^2}\text{ and y}=\frac{2\text{t}}{1+\text{t}^2}$

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Answer

We have, $\text{y}=\frac{2\text{t}}{1+\text{t}^{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(2\text{t})-2\text{t}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
[using quotient rule]
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(2)-2\text{t}(2\text{t})}{(1+\text{t}^{2})^{2}}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{2+2\text{t}^{2}-4\text{t}^{2}}{(1+\text{t}^{2})}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\bigg[\frac{2-2\text{t}^{2}}{(1+\text{t}^{2})^{2}}\bigg]...(\text{i})$
and, $\text{x}=\frac{1-\text{t}^{2}}{1+\text{t}^{2}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})\frac{\text{d}}{\text{dt}}(1-\text{t}^{2})-(1-\text{t}^{2})\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})}{(1+\text{t}^{2})^{2}}\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\bigg[\frac{(1+\text{t}^{2})(-2\text{t})-(1-\text{t}^{2})(2\text{t})}{(1+\text{t}^{2})^{2}}\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\bigg[\frac{-4\text{t}}{(1+\text{t}^{2})^{2}}\bigg].....(\text{ii})$
Dividing equation (i) by (ii), We get,
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2(1-\text{t}^{2})}{(1+\text{t}^{2})^{2}}\times\frac{(1+\text{t}^{2})^{2}}{-4\text{t}} $
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{t}^{2})}{-4\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{t}^{2}-1}{2\text{t}}$

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