Question

What is the derivative of `arcsin(x-1)`?

Answer 1

`1/sqrt(1-(x-1)^2`

Explanation:

derivative of inverse trigonometric functions

the general formula to differentiate the arcsin functions is

`intsin^-1u=1/sqrt(1-u^2)(du)/dx`

`d/dxsin^-1(x-1)=1/sqrt(1-(x-1)^2)*(d(x-1))/dx` `"rarr` chain rule

`d/dxsin^-1(x-1)=1/sqrt(1-(x-1)^2)*1`

Answer 2

`1/(sqrt(1-(x-1)^2)`

Explanation:

We got:

`d/dx(arcsin(x-1))`

Let `y=arcsin(x-1)`

Let's use the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

Let `u=x-1,:.(du)/dx=1`

Then `y=arcsinu,:.dy/(du)=1/(sqrt(1-u^2))`.

Combining,

`dy/dx=1/(sqrt(1-u^2))*1`

`=1/(sqrt(1-u^2))`

Substitute back `u=x-1` to get the final answer:

`=1/(sqrt(1-(x-1)^2)`