Question

$y=ta{n}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right),x^2<1$, then find $\frac{dy}{dx}$.

Answer 1

$y=ta{n}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right),x^2<1$
$\frac{dy}{dx}={\frac{1}{1+\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)}}^2.\frac{d}{dx}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$
$\frac{dy}{dx}=\frac{(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^2}{(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^2+(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^2}.\frac{d}{dx}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$
$\frac{d}{dx}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{(\sqrt{1+x^2}-\sqrt{1-x^2}).(\frac{2x}{2\sqrt{1+x^2}}-\frac{2x}{2\sqrt{1-x^2}})-(\sqrt{1+x^2}-\sqrt{1-x^2}).(\frac{2x}{2\sqrt{1+x^2}}+\frac{2x}{2\sqrt{1-x^2}})}{(\sqrt{1+x^2}-\sqrt{1-x}{)}^2}$
Here Numerator $=x\left[\right(\sqrt{1+x^2}-\sqrt{1-x^2})(\frac{1}{\sqrt{1+x^2}}-\frac{1}{\sqrt{1+x^2}})-(\sqrt{1+x^2}-\sqrt{1-x^2}\left)(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}})\right]$
$=x[(1-\sqrt{\frac{1+x^2}{1-x^2}}-\sqrt{\frac{1-x^2}{1+x^2}}+1)-(1+\sqrt{\frac{1+x^2}{1-x^2}}+\sqrt{\frac{1-x^2}{1+x^2}}+1)]$
$=x[1-\sqrt{\frac{1+x^2}{1-x^2}}-\sqrt{\frac{1-x^2}{1+x^2}}+1-1-\sqrt{\frac{1+x^2}{1-x^2}}+\sqrt{\frac{1-x^2}{1+x^2}}-1]$
Numerator $=-2x=x(\sqrt{\frac{1+x^2}{1-x^2}}+\sqrt{\frac{1-x^2}{1+x^2}})$
Now, $\frac{dy}{dx}=\frac{(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^2}{(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^2+(\sqrt{1+x^2}+\sqrt{1-x^2}{)}^2}\times \frac{-2x(\sqrt{\frac{1+x^2}{1-x^2}}+\sqrt{\frac{1+x^2}{1+x^2}})}{(\sqrt{1+x^2}-\sqrt{1-x^2}{)}^2}=$
$\frac{dy}{dx}=\frac{\sqrt{(1+x^2)(1-x^2)}}{\sqrt{(1+x^2)(1-x^2)}+\sqrt{(1+x^2)(1-x^2)}}=\frac{\frac{4x}{\sqrt{(1+x^2)(1-x^2)}}}{(1+x^2)(1-x^2)-2\sqrt{(1+x^2)(1-x^2)}+(1+x^2)(1-x^2)+2\sqrt{(1+x^2)(1-x^2)}}\frac{dy}{dx}=\frac{-4x}{4\sqrt{(1+x^2)(1-x^2)}}\therefore \frac{dy}{dx}=\frac{-x}{\sqrt{(1+x^2)(1-x^2)}}$