मराठी

Find dy/dx in the following: y = cos-1 (1 - x2/1 + x2), 0 < x < 1 - Mathematics

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प्रश्न

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

y = `cos^(-1) ((1-x^2)/(1+x^2))`, 0 < x < 1

बेरीज
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उत्तर

y = `cos^-1 ((1 - x^2)/(1 + x^2))`

Let, x = tan θ

⇒ θ = tan−1 x

∴ y = `cos^-1 ((1 - tan^2 theta)/(1 + tan^2 theta))`

= cos−1 (cos 2 θ)

= 2 θ

= 2 tan−1 x

On differentiating with respect to x,

`dy/dx = 2 d/dx tan^-1 x`

`dy/dx = 2 xx 1/(1 + x^2)`

`dy/dx = 2/(1 + x^2)`

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पाठ 5: Continuity and Differentiability - Exercise 5.3 [पृष्ठ १६९]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 5 Continuity and Differentiability
Exercise 5.3 | Q 11 | पृष्ठ १६९

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