If $x = \mathop \smallint \limits_0^y \frac{1}{{\sqrt {1 + 4{t^2}} }}dt$, then $\frac{{{d^2}y}}{{d{x^2}}}$ is

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

x = \int \limits_0^y \frac{1}{{\sqrt {1 + 4{t^2}} }}dt

, then

\frac{{{d^2}y}}{{d{x^2}}}

View answer and explanation

Correct option: B

4y

Option B (

4y

) is correct because it is the result of differentiating the given integral equation twice and simplifying.

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