Question

Second order differentiation
of y= sin inverse of x​

Answer 1

Answer:

x / ( 1 - x² )³/²

Step-by-step explanation:

To find---> Second order derivative of Sin⁻¹x

Solution---> We have some formulee of differentiation as follows

1) d / dx ( Sin⁻¹x ) = 1 / √(1 - x²)

2) d / dx ( √x ) = 1 / 2√x

3) d / dx ( x² ) = 2x

Now returning to original problem

y = Sin⁻¹ x

Differentiating with respect to x , we get,

dy / dx = d / dx ( Sin⁻¹x )

=> dy / dx = 1 / √(1 - x²)

= 1 / ( 1 - x² )¹/²

=> dy / dx = ( 1 - x² )⁻¹/²

Differentiating with respect to x again , we get

=> d/dx ( dy/dx ) = d / dx { ( 1 - x² )⁻¹/² }

=> d²y / dx² = ( -1 /2 ) ( 1 - x² )⁻¹/² ⁻¹ d/dx ( 1 - x² )

=> d²y / dx² = (- 1 / 2 ) ( 1 - x² )⁻³/² ( 0 - 2x )

=> d²y / dx² = x / ( 1 - x² )³/²

Answer 2

Answer:

x / ( 1 - x² )³/²

Step-by-step explanation:

Given : f y = sin inverse of x​

To find : Second order derivative of Sin⁻¹x

Solution :

We have some formula of differentiation as follows

1) d / dx ( Sin⁻¹x ) = 1 / √(1 - x²)

2) d / dx ( √x ) = 1 / 2√x

3) d / dx ( x² ) = 2x

Now returning to original problem

y = Sin⁻¹ x

Differentiating with respect to x , we get,

dy / dx = d / dx ( Sin⁻¹x )

=> dy / dx = 1 / √(1 - x²)

= 1 / ( 1 - x² )¹/²

=> dy / dx = ( 1 - x² )⁻¹/²

Differentiating with respect to x again , we get

=> d/dx ( dy/dx ) = d / dx { ( 1 - x² )⁻¹/² }

=> d²y / dx² = ( -1 /2 ) ( 1 - x² )⁻¹/² ⁻¹ d/dx ( 1 - x² )

=> d²y / dx² = (- 1 / 2 ) ( 1 - x² )⁻³/² ( 0 - 2x )

=> d²y / dx² = x / ( 1 - x² )³/²

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