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рдкреНрд░рд╢реНрди
Find `dy/dx`, if y = `sin^-1 x + sin^-1 sqrt (1 - x^2)`, 0 < x < 1.
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рдЙрддреНрддрд░
Here, y = `sin^-1 x + sin^-1 sqrt (1- x^2)`
Let u = sin−1 x and v = `sin^-1 sqrt (1-x^2)`
`(du)/dx = 1/sqrt(1 - x^2)`
Now v = `sin^-1 sqrt(1 - x^2)`
Put x = cos θ
∴ v = `sin^-1 sqrt (1 - cos^2 θ)`
= `sin^-1 sqrt (sin^2 θ)`
= sin−1 (sin θ)
= θ
= cos−1 x
∴ `(dv)/dx = -1/sqrt(1 - x^2)`
As `dy/dx = (du)/dx + (dv)/dx`
`= 1/ sqrt(1-x^2) + -1/ sqrt (1 - x^2)`
= 0
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