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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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