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If $y \sqrt{x^{2}+1}=\log \left(\sqrt{x^{2}+1}-x\right)$, show that $\left(x^{2}+1\right) \frac{d y}{d x}+x y+1=0$


We are given with an equation $y \sqrt{x^{2}+1}=\log \left(\sqrt{x^{2}+1}-x\right)$, we have to prove that

$\left(x^{2}+1\right) \frac{d y}{d x}+x y+1=0$ by using the given equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,

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

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

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

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

$x y+\left(x^{2}+1\right) \frac{d y}{d x}+1=0$

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