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  4. If y= sin -1[√x-a x-√a-a x] then (d y/d x)=

Q. If $y=\sin ^{-1}[\sqrt{x-a x}-\sqrt{a-a x}]$ then $\frac{d y}{d x}=$

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Solution:

$y =\sin ^{-1}[\sqrt{x-a x}-\sqrt{a-a x}]$
$y =\sin ^{-1}[\sqrt{x(1-a)}-\sqrt{a(1-x)}]$
$=\sin ^{-1}\left[\sqrt{x} \sqrt{1-(\sqrt{a})^{2}}-\sqrt{a} \sqrt{1-(\sqrt{x})^{2}}\right]$ Put $\sqrt{x}=\sin \theta, \sqrt{a}=\sin \alpha$
$=\sin ^{-1}(\sin \theta \cos \alpha-\sin \alpha \cos \theta)$
$=\sin ^{-1}(\sin (\theta-\alpha))=\theta-\alpha$
$y =\sin ^{-1} \sqrt{x}-\sin ^{-1} a$
$\frac{d y}{d x} =\frac{1}{\sqrt{1-x}} \frac{1}{2 \sqrt{x}}-0=\frac{1}{2 \sqrt{x} \sqrt{1-x}}$

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