JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

y=\sin ^{-1} 2 x \cdot \sqrt{1-x^2}+\sec ^{-1} \frac{1}{\sqrt{\left(1-x^2\right)}}

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: C

\frac{3}{\sqrt{1-x^2}}

Option C (

\frac{3}{\sqrt{1-x^2}}

) is correct because after substituting

x = \sin \theta

and simplifying the expression, the derivative is found to be

\frac{3}{\sqrt{1-x^2}}

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