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  4. If y= sin -1(x √1-x+√x √1-x2) (d y/d x)=(1/2 √x(1-x))+p, then p=

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Q. If $y=\sin ^{-1}\left(x \sqrt{1-x}+\sqrt{x} \sqrt{1-x^2}\right) \& \frac{d y}{d x}=\frac{1}{2 \sqrt{x(1-x)}}+p$, then $p=$

Continuity and Differentiability

Solution:

$ x=\sin \theta ; x=\sin ^2 \phi$
$y=\sin ^{-1}(\sin \theta \cos \phi+\sin \phi \cos \theta)=\sin ^{-1}\left((\sin (\theta+\phi))=\theta+\phi=\sin ^{-1} x+\sin ^{-1} \sqrt{x}\right.$
$Dy =\frac{1}{\sqrt{1- x ^2}}+\frac{1}{2 \sqrt{ x (1- x )}}$ assuming $x ^2+ x \leq 1$ i.e. $0 \leq x <\frac{\sqrt{5}-1}{2}$ ]
Note: $\sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right) $
$=\sin ^{-1} x+\sin ^{-1} y \text { if } x^2+y^2 \leq 1=\pi-\left(\sin ^{-1} x+\sin ^{-1} y\right) \text { if } x^2+y^2 \geq 1$