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प्रश्न
Find `bb(dy/dx)` in the following:
y = `cos^(-1) ((2x)/(1+x^2))`, −1 < x < 1
योग
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उत्तर
y = `cos^(-1) ((2x)/(1+x^2))`
Let, x = tan θ
⇒ θ = tan−1 x
∴ y = `cos^-1 ((2 tan theta)/(1 + tan^2 theta))`
= cos−1 (sin 2 θ)
= `cos^-1 {cos (pi/2 - 2 theta)}`
= `pi/2 - 2 theta`
= `pi/2 - 2 tan^-1 x`
On differentiating with respect to x,
`dy/dx = pi/2 d/dx (1) - 2 d/dx tan^-1 x`
`dy/dx = 0 - 2 xx 1/(1 + x^2)`
`dy/dx = -2/(1 + x^2)`
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