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Find dy/dx in the following: y = sec−1⁡(1/2⁢𝑥2−1), 0 < 𝑥 < 1√2 - Mathematics

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Question

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

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

Sum
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Solution

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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Chapter 5: Continuity and Differentiability - Exercise 5.3 [Page 169]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 5 Continuity and Differentiability
Exercise 5.3 | Q 15 | Page 169

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