JEE Main 2002Annual, Paper 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002If y=(x+1+x2)n,y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},y=(x+1+x2)n, then (1+x2)d2ydx2+xdydx\left( {1 + {x^2}} \right){{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}(1+x2)dx2d2y+xdxdy isAOption A. n2y{n^2}yn2yBOption B. −n2y-{n^2}y−n2yCOption C. −y-y−yDOption D. 2x2y2{x^2}y2x2yCheckSkip · show solution