If y - sin - 1x-1 - x - -x-1 - x2- then dydx -

The correct option is D None of these y= sin^ 1(x√(1 x^2)+√(x)√(1 x^2)) dydx=1√(1 (x√(1 x^2)+√(x(1 x^2)^2)))[√(1 x^2)+x( 2x)2√(1 x^2)+(1 x^2)+x( 2x)2√( ...

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Byju's Answer

Standard XII

Mathematics

General Solution of a Differential Equation

If y = sin ...

Question

If $y = {sin}^{- 1} (x \sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^{2})}$ , then $\frac{d y}{d x} =$

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

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

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

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None of these

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Solution

The correct option is D None of these
$y = s i n^{- 1} (x \sqrt{1 - x^{2}} + \sqrt{x} \sqrt{1 - x^{2}})$
$\frac{d y}{d x} = \frac{1}{\sqrt{1 - (x \sqrt{1 - x^{2}} + \sqrt{x (1 - x^{2})^{2}})}} [\frac{\sqrt{1 - x^{2}} + x (- 2 x)}{2 \sqrt{1 - x^{2}}} + \frac{(1 - x^{2}) + x (- 2 x)}{2 \sqrt{x (1 - x^{2})}}]$
$= \frac{1}{\sqrt{1 - (x \sqrt{1 - x^{2}} + \sqrt{x (1 - x^{2})})^{2}}} [\frac{2 \sqrt{x (1 - x^{2})^{2}} - 2 x^{2} \sqrt{x} + (1 - x^{2}) - 2 x^{2}}{2 \sqrt{(1 - x^{2}) x}}]$
Wrong question.

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If y=sin−1(x√1−x+√x√1−x2), then dydx=

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If $y = {sin}^{- 1} (x \sqrt{1 - x} + \sqrt{x} \sqrt{1 - x^{2})}$ , then $\frac{d y}{d x} =$