Question

Find `bb(dy/dx)` in the following:

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

Answer 1

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))`