If y = sin - 1 ( 2 x 1 + x 2 ) , then d y d x is equal to:

Question

If $y = \sin^{- 1} (\frac{2 x}{1 + x^{2}}),$ then $\frac{d y}{d x}$ is equal to:

Smart Achievers · Accepted Answer

(2)

Ans. $\frac{2}{1 + x^{2}}$

Explanation:

Let $x = \tan θ$

Then $y = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ})$

Using the identity $\frac{2 \tan θ}{1 + \tan^{2} θ} = \sin 2 θ,$

we get $y = \sin^{- 1} (\sin 2 θ)$

Since $2 θ$ lies in the principal value range, $y = 2 θ = 2 \tan^{- 1} x$

Differentiating,

$\frac{d y}{d x} = 2 \cdot \frac{1}{1 + x^{2}} = \frac{2}{1 + x^{2}}$

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