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प्रश्न
Find `bb(dy/dx)` in the following:
y = `sec^(-1) (1/(2x^2 - 1)), 0 < x < 1/sqrt2`
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उत्तर
y = `sec^-1 (1/(2x^2 - 1))`
Let, x = cos θ
⇒ θ = cos−1 x
∴ y = `sec^-1 (1/(2 cos^2 theta - 1))`
= `sec^-1 (1/(cos 2 theta))`
= sec−1 (sec 2 θ)
= 2 θ
= 2 cos−1 x
On differentiating with respect to x,
`dy/dx = 2 d/dx cos^-1 x`
`dy/dx = 2 xx -1/(sqrt(1 - x^2))`
`dy/dx = -2/(sqrt(1 - x^2))`
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