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प्रश्न
Find `dy/dx` in the following:
`y = sin^(-1) ((1-x^2)/(1+x^2)), 0 < x < 1`
उत्तर
y = `sin^-1 ((1 - x^2)/(1 + x^2))`
Putting x`= tan theta`,
`therefore y = sin^-1 ((1 - tan^2 theta)/(1 + tan^2 theta)) = sin^-1 (cos 2 theta)`
`y = sin^-1 sin (pi/2 - 2 theta)`
`y = pi/2 - 2theta`
`= 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`
`= 0 - 2 xx 1/(1 + x^2) = - 2/(1 + x^2)`
`= dy/dx = -2/(1 + x^2)`
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