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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