# Solve this

Question:

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

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

Solution:

as $x=\cos ^{-1} \frac{1}{\sqrt{1+t^{2}}}$

Differentiating it with respect to $t$ using chain rule,

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

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

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

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

$\frac{\mathrm{dx}}{\mathrm{dt}}=-\frac{1}{1+\mathrm{t}^{2}} \ldots \ldots(1)$

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

Differentiating it with respect to $t$ using chain rule,

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

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dt}}=-\frac{1}{1+\mathrm{t}^{2}} \ldots \ldots(2)$

dividing equation (2) by (1),

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=-\frac{1}{1+\mathrm{t}^{2}} \times-\frac{1+\mathrm{t}^{2}}{1}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=1$