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Qn #849
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If $y=\sin ^{-1}(\frac{{x}^2+1}{\sqrt[]{1+3{x}^2+{x}^4}}),\, (x>0),$ then $\frac{dy}{dx}$=
A.
$\frac{{x}^2-1}{{x}^4+3{x}^2+1}$
B.
$\frac{{x}^2+1}{{x}^4+3{x}^2+1}$
C.
$\frac{{x}^2-1}{{x}^4-3{x}^2+1}$
D.
$\frac{{x}^2+1}{{x}^4-3{x}^2+1}$
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❓ Question: If $y=\sin ^{-1}(\frac{{x}^2+1}{\sqrt[]{1+3{x}^2+{... | TANCET Group Studies | Tancet Group Studies