Find dy dx ,if y = sin^ -1 [x1 - x-x 1 - x^ 2 ]. - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}},\text{if y = sin}^{-1}[\text{x}\sqrt{1 - x}-\sqrt{x}\sqrt{1 - x^{2}}].$

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Answer

$\text{y}=\sin^{-1}\text[{x}\sqrt{1-x}-\sqrt{x}\sqrt{1 - x^{2}}]............(i)$
Let $x = \sin$
$\alpha\text{ and }\sqrt{x}=\sin\theta$
$\therefore (i)$ Becomes $y = \sin^{-1 }$
$[\sin\alpha\cos\theta-\cos\alpha\sin\theta]$.
$=\sin^{-1}[\sin(\alpha-\theta)]=\alpha-\theta$
$=\sin^{-1}\text{x}-\sin^{-1}\sqrt{x}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1}-x^2}-\frac{1}{2\sqrt{x}\sqrt{1-x}}$

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