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Find `dy/dx`, if y = `sin^-1 x + sin^-1 sqrt (1 - x^2)`, 0 < x < 1.

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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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рдкрд╛рда 5: Continuity and Differentiability - Exercise 5.9 [рдкреГрд╖реНрда резрепрез]

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рдПрдирд╕реАрдИрдЖрд░рдЯреА Mathematics Part 1 and 2 [English] Class 12
рдкрд╛рда 5 Continuity and Differentiability
Exercise 5.9 | Q 13 | рдкреГрд╖реНрда резрепрез

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