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Question
Find `dy/dx` in the following:
`y = sec^(-1) (1/(2x^2 - 1)), 0 < x < 1/sqrt2`
Solution 1
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`
Differentiating 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)`
Solution 2
y = `sec^-1 (1/(2x^2 - 1))`
Let, x = `cos theta => theta = cos^-1 x`
`therefore y = sec^-1 (1/(2 cos^2 theta - 1))`
`= sec^-1 (1/(cos 2 theta))`
`= sec^-1 (sec 2 theta) = 2 theta`
`y = 2 cos^-1 x`
`therefore "dy"/"dx" = - 2/sqrt(1 - "x"^2)`
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