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Find Dy/Dx for Y = Sec^(-1) (1/(2x^2 - 1)), 0 < X < 1/Sqrt2 - Mathematics

Sum

Find `dy/dx`

`y = sec^(-1) (1/(2x^2 - 1)), 0 < x < 1/sqrt2`

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Solution

The given relationship is `y = sec^(-1) (1/(2x^2 - 1))`

`y = sec^(-1) (1/(2x^2 - 1))`

⇒ sec y = `1/(2x^2 - 1)`

⇒ cos y = `2x^2 - 1`

⇒ `2x^2 = 1 + cosy`

⇒ `2x^2 = 2cos^2  y/2`

⇒ `x = cos  y/2`

Differebtiating this relationship with respect to x  , we obtain

`d/dx (x) = d/dx (cos  y/2)`

⇒ `1 = -sin  y/2 . d/dx(y/2)`

⇒ `(-1)/(sin  y/2) = 1/2 dy/dx`

⇒ `dy/dx = (-2)/(sin  y/2) = (-2)/sqrt(1 - cos^2  y/2)`

⇒ `dy/dx = (-2)/sqrt(1 - x^2)`

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APPEARS IN

NCERT Class 12 Maths
Chapter 5 Continuity and Differentiability
Q 15 | Page 169
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