If $f\left( x \right) = \mathop \smallint \limits_0^{{x^2}} \,\,\sqrt {1 + \,\,{t^2}} \,dt$, then $f'\,\,\left( x \right)$ equals

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

f\left( x \right) = \int \limits_0^{{x^2}} \,\,\sqrt {1 + \,\,{t^2}} \,dt

, then

f'\,\,\left( x \right)

equals

View answer and explanation

Correct option: C

2x\,\sqrt {1\,\, + \,\,{x^4}}

Option C (

2x\,\sqrt {1\,\, + \,\,{{{x^4}}}}

) is correct because it is the result of applying the Leibniz rule for differentiating under the integral sign.

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