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

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Answer 1 · 2018-04-12T10:17:21

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

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 · 2018-04-13T12:39:26

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

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:

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

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