If y=sin-1x2x4+a4 then dydx=

y=sin-1x2x4+a4 x2 =a2tanθ y=sin-1tanθsecθ =sin-1sinθ =θ=tan-1x2a2 dydx=2a2xa4+x4

If y=sin-1x2x4+a4 then dydx=

y=sin-1x2x4+a4 x2 =a2tanθ y=sin-1tanθsecθ =sin-1sinθ =θ=tan-1x2a2 dydx=2a2xa4+x4

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If $y = \sin^{- 1} (\frac{x^{2}}{\sqrt{x^{4} + a^{4}}})$ then $\frac{dy}{dx} =$

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$\frac{2 a^{2} x^{2}}{a^{4} + x^{4}}$

$\frac{2 a^{2} x}{a^{4} + x^{4}}$

$\frac{ax}{a^{4} + x^{4}}$

$\frac{- ax}{a^{4} + x^{4}}$

answer is B.

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Detailed Solution

$y = \sin^{- 1} (\frac{x^{2}}{\sqrt{x^{4} + a^{4}}}) x^{2} = a^{2} tanθ y = \sin^{- 1} (\frac{tanθ}{secθ}) = \sin^{- 1} (s i nθ) = θ = \tan^{- 1} (\frac{x^{2}}{a^{2}}) \frac{dy}{dx} = \frac{2 a^{2} x}{a^{4} + x^{4}}$

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If $y = \sin^{- 1} (\frac{x^{2}}{\sqrt{x^{4} + a^{4}}})$ then $\frac{dy}{dx} =$