Categories > differentiation > standard-differentiation > If $y\sqrt {{x^2} + 1} = \log \left\{ {\sqrt {{x^2} + 1} - x} \right\}$, then$({x^2} + 1)\frac{{dy}}{{dx}} + xy + 1 = $

differentiation

Question: If $y\sqrt {{x^2} + 1} = \log \left\{ {\sqrt {{x^2} + 1} - x} \right\}$, then$({x^2} + 1)\frac{{dy}}{{dx}} + xy + 1 = $


1) 0
2) 1
3) 2
4) None of these
Solution: Explanation: Explanation
standard-differentiation

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