If log 1 + x 2 - x = y 1 + x 2 , then 1 + x 2 d y d x + x y =

Given,

$\log \left(\sqrt{1+x^2}-x\right)=y\left(\sqrt{1+x^2}\right)$

Differentiating, both sides w.r.t $x$,

$\frac{d}{dx}\left[\log \left(\sqrt{1+x^2}-x\right)\right]=\frac{d}{dx}\left[y\left(\sqrt{1+x^2}\right)\right]$

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

$\Rightarrow \frac{1}{\left(\sqrt{1+x^2}-x\right)}\left[\frac{x}{\sqrt{1+x^2}}-1\right]=\sqrt{1+x^2}\frac{dy}{dx}+\frac{xy}{\sqrt{1+x^2}}$
Multiplying by $\sqrt{1+x^2}$ both the sides,

$\Rightarrow \sqrt{1+x^2}\times \frac{1}{\left(\sqrt{1+x^2}-x\right)}\left[\frac{x}{\sqrt{1+x^2}}-1\right]=\sqrt{1+x^2}\frac{dy}{dx}+xy$

$\sqrt{1+x^2}\frac{dy}{dx}+xy=\sqrt{1+x^2}\times \frac{1}{\left(\sqrt{1+x^2}-x\right)}\times \frac{\left(\sqrt{1+x^2}-x\right)}{\sqrt{1+x^2}}=-1$