If y - sin2 tan-1-1 - x21 - x2 - then dydx -

The correct option is D x y = sin^2 [tan^ 1√(1 x^21 + x^2) ] Put x^2 = cosθ y = sin^2 [tan^ 1 (tanθ2)] = sin^2θ2 = 1 cosθ2 = 1 x^22 ...

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Algebra of Derivatives

If y = sin2...

Question

If $y = {sin}^{2} ({tan}^{- 1} \sqrt{\frac{1 - x^{2}}{1 + x^{2}}})$ , then $\frac{d y}{d x} =$

x

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Similar questions

y = {tan}^{- 1} \frac{x}{1 + \sqrt{1 - x^{2}}} + sin (2 {tan}^{- 1} \sqrt{\frac{1 - x}{1 + x}})

\frac{d y}{d x}

x \in (- 1, 1)

y = x log ∣ ∣ x + \sqrt{(1 + x^{2})} ∣ ∣ - \sqrt{1 + x^{2}}

\frac{d y}{d x} =

y = {sin}^{- 1} \frac{1}{\sqrt{1 + x^{2}}} + {tan}^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})

\frac{d y}{d x}

(x \neq 0)

y = {tan}^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x}

\frac{d y}{d x} =

y = {tan}^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}]

\frac{d y}{d x}

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