JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},

then

\left( {1 + {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}}

View answer and explanation

Correct option: A

{n^2}y

Option A (

{n^2}y

) is correct because after differentiating twice and rearranging, the expression equals

{n^2}y

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