Question

Question 14:Find dy/dx : y=sin¯¹(2x √1-x²), -1/2 <√x < 1/√2

Class 12 - Math - Continuity and Differentiability

Answer 1

y=sin¯¹(2x √1-x²)
Put x = sin∅
y = sin-¹(2sin∅√1-sin²∅)
y = sin-¹(2sin∅cos∅ { 1-sin²∅ = cos²∅
y = sin-¹(sin2∅) { 2sin∅cos∅ = sin2∅
y = 2∅
y = 2sin-¹x
dydx = 2/√1-x² { d/dx(sin-¹x = 1/√1-x²

Answer 2

★ DIFFERENTIATION ★

Given function :

y = Sin^-1 ( 2x √1-x² ) where x has a range defined as -1/2 < √x < 1/ √2

Y = Sin ^-1 ( 2x√1 - x² )
substantially , x = Sinθ

y = Sin ^ -1 ( 2 sin θ √ 1 - Sin² θ )

y = Sin^-1 ( sin2 θ ) as 1 - sin²θ = Cos² θ

y = 2θ

y = 2 sin ^-1

Hence , dy/ dx = 2 / √1 - x² Aslike , dy/dx = 1 / √1 - x²

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