Question

Please answer this question
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Answer 1

Answer---> (4) -2 /√( 1-x² ) , is correct option

To find --->

d/dx { Cos⁻¹x + Sin⁻¹ √(1 - x² ) }

Solution---> First we find value of Sin⁻¹√(1 - x²)

y = Sin⁻¹√(1 - x²)

Let x = Cosθ => θ = Cos⁻¹x

y = Sin⁻¹√(1 - Cos²θ )

We know that , 1 - Cos²θ = Sin²θ

y = Sin⁻¹√Sin²θ

= Sin⁻¹ ( Sinθ )

= θ

= Cos⁻¹x

So, Sin⁻¹√(1 - x²) = Cos⁻¹x

Now returning to the original problem

d/dx { Cos⁻¹x + Sin⁻¹√1 - x² }

Putting , Sin⁻¹√( 1 - x² ) = Cos⁻¹ x , we get,

= d/dx ( Cos⁻¹x + Cos⁻¹x )

= d/dx ( 2 Cos⁻¹ x )

= 2 d / dx ( Cos⁻¹x )

We know that

d/dx ( Cos⁻¹x ) = - 1 /√(1 - x²) , applying it here , we get

= 2 { - 1 / √(1 - x²) }

= -2 / √( 1 - x² )