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Question Number 190660 by vishal1234 last updated on 08/Apr/23

i f y = {s i n}^{- 1} (2 x \sqrt{1 - x^{2}}) w h e r e x \in [- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}], t h e n f i n d \frac{d y}{d x} .

Answered by BaliramKumar last updated on 08/Apr/23

let, x = s i n θ \Rightarrow θ = {s i n}^{- 1} x

y = {s i n}^{- 1} (2 s i n θ \sqrt{1 - {s i n}^{2} θ})

y = {s i n}^{- 1} (2 s i n θ \sqrt{{c o s}^{2} θ})

y = {s i n}^{- 1} (2 s i n θ \cdot c o s θ)

y = {s i n}^{- 1} (s i n 2 θ)

y = 2 θ = 2 {s i n}^{- 1} x

\frac{d y}{d x} = 2 (\frac{1}{\sqrt{1 - x^{2}}}) = \frac{2}{\sqrt{1 - x^{2}}}

Commented by vishal1234 last updated on 14/Apr/23

w h a t w i l l h a p p e n, ? i f w e t a k e x = c o s θ ?

Answered by horsebrand11 last updated on 09/Apr/23

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

2 x \sqrt{1 - x^{2}} = \sin y

2 \sqrt{1 - x^{2}} + 2 x (\frac{- 2 x}{2 \sqrt{1 - x^{2}}}) = \cos y . \frac{d y}{d x}

2 \sqrt{1 - x^{2}} - \frac{2 x^{2}}{\sqrt{1 - x^{2}}} = \cos y . \frac{d y}{d x}

\frac{2 - 4 x^{2}}{\sqrt{1 - x^{2}}} . \frac{1}{\sqrt{1 - 4 x^{2} (1 - x^{2})}} = \frac{d y}{d x}

\frac{d y}{d x} = \frac{2 - 4 x^{2}}{\sqrt{1 - x^{2}} \sqrt{4 x^{4} - 4 x^{2} + 1}}

\frac{d y}{d x} = \frac{2 - 4 x^{2}}{(2 x - 1) \sqrt{1 - x^{2}}}

Commented by som(math1967) last updated on 09/Apr/23

4 x^{4} - 4 x^{2} + 1 = {(2 x^{2} - 1)}^{2} o r {(1 - 2 x^{2})}^{2}