JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because it is the derivative of the simplified form of the given function after using trigonometric substitution.

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