# Solve this

Question:

Find $\frac{d y}{d x}$, when

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

Solution:

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

$\Rightarrow \frac{d x}{d t}=\frac{-1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}} \frac{d}{d t}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$

$\Rightarrow \frac{d x}{d t}=\frac{-1}{\sqrt{1-\frac{1}{\left(1+t^{2}\right)}}}\left\{\frac{-1}{2\left(1+t^{2}\right)^{\frac{3}{2}}}\right\} \frac{d}{d t}\left(1+t^{2}\right)$

$\Rightarrow \frac{d x}{d t}=\frac{\left(1+t^{2}\right)^{\frac{1}{2}}}{\sqrt{1+t^{2}-1}} \times \frac{1}{2\left(1+t^{2}\right)^{\frac{3}{2}}}(2 t)$

$\Rightarrow \frac{d x}{d t}=\frac{t}{\sqrt{t^{2} \times\left(1+t^{2}\right)}}$

$\Rightarrow \frac{d x}{d t}=\frac{1}{1+t^{2}}$               .....(1)

Now, $y=\sin ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$

$\Rightarrow \frac{d y}{d t}=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}} \frac{d}{d t}\left(\frac{1}{\sqrt{1+t^{2}}}\right)$

$\Rightarrow \frac{d y}{d t}=\frac{1}{\sqrt{1-\frac{1}{\left(1+t^{2}\right)}}}\left\{\frac{-1}{2\left(1+t^{2}\right)^{\frac{3}{2}}}\right\} \frac{d}{d t}\left(1+t^{2}\right)$

$\Rightarrow \frac{d x}{d t}=\frac{\left(1+t^{2}\right)^{\frac{1}{2}}}{\sqrt{1+t^{2}-1}} \times \frac{-1}{2\left(1+t^{2}\right)^{\frac{3}{2}}}(2 t)$

$\Rightarrow \frac{d x}{d t}=\frac{-1}{2 \sqrt{t^{2}} \times\left(1+t^{2}\right)}(2 t)$

$\Rightarrow \frac{d x}{d t}=\frac{-1}{1+t^{2}}$            .....(2)

Dividing equation (ii) by (i),

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{1}{\left(1+t^{2}\right)} \times \frac{\left(1+t^{2}\right)}{-1}$

$\Rightarrow \frac{d y}{d x}=-1$