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

Question

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

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Answer

We have, $\text{x}=\cos^{-1}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{-1}{\sqrt{1-\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)^{2}}}\frac{\text{d}}{\text{dt}}\Big(\frac{1}{\sqrt{1+\text{t}}^{2}}\Big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{-1}{\sqrt{1-\frac{1}{(1+\text{t}^{2})}}}\left\{\frac{-1}{2(1+\text{t}^{2})^\frac{3}{2}}\right\}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{(1+\text{t}^{2})^{\frac{1}{2}}}{\sqrt{1+\text{t}^{2}-1}}\times\frac{1}{2(1+\text{t}^{2})^\frac{3}{2}}(2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{t}}{\sqrt{\text{t}^{2}}\times(1+\text{t}^{2})}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{1+\text{t}^{2}}...(\text{i})$
Now, $\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{1}{\sqrt{1-\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)^{2}}}\frac{\text{d}}{\text{dt}}\Big(\frac{1}{\sqrt{1+\text{t}^{2}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{1}{\sqrt{1-\frac{1}{(1+\text{t}^{2})}}}\left\{\frac{-1}{2(1+\text{t}^{2})\frac{3}{2}}\right\}\frac{\text{d}}{\text{dt}}(1+\text{t}^{2})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{(1+\text{t}^{2})^{\frac{1}{2}}}{\sqrt{1+\text{t}^{2}-1}}\times\frac{-1}{2(1+\text{t}^{2})^{\frac{3}{2}}}(2\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-1}{2\sqrt{\text{t}^{2}}\times(1+\text{t}^{2})}(2\text{t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-1}{1+\text{t}^{2}}....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{1}{(1+\text{t}^{2})}\times\frac{(1+\text{t}^{2})}{-1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$

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