If $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ then $$\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}$$ is
JEE · Math · previous-year question
- A.$${n^2}y$$
- B.$$-{n^2}y$$
- C.$$-y$$
- D.$$2{x^2}y$$
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