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

Hint: In this question we have to find the derivative of the given function. We should that the given function is a trigonometric ratio which is in the form of inverse trigonometric function. So we will apply the formula of chain rule to solve this question. After that we will assume that $u = x - 1$ , then we will simplify the function according to the formula.
We will here us the general formula of the derivative of \[arcsinx\] which is written as $ \Rightarrow \dfrac{d}{{dx}}\arcsin x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$

Complete step by step solution:
We have been given to find the derivative of $\arcsin (x - 1)$.
We can write this function as $\dfrac{d}{{dx}}(\arcsin (x - 1))$ .
Let us use the chain rule, which says that
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ .
Let us take $u = \arcsin (x - 1)$
Therefore we can say that
$\dfrac{{du}}{{dx}} = 1$
Then again we have $y = \arcsin u$, so it gives us the formula which says that
$\dfrac{{dy}}{{du}} = \dfrac{1}{{\sqrt {1 - {u^2}} }}$
Now by combing the equation, we can write it as:
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {u^2}} }} \times 1$
Here we will substitute the value back in the expression i.e. $u = x - 1$
It gives us:
$ = \dfrac{1}{{1 - {{\left( {x - 1} \right)}^2}}}$
$ = \dfrac{1}{1-(x^2-2x+1)}$
$=\dfrac{1}{1-x^2+2x-1}$
$=\dfrac{1}{2x-x^2}$
Hence this is the required answer of the given function.
Therefore, the derivative of $\arcsin (x - 1)$ is $\dfrac{1}{2x-x^2}$.

Note:
Before solving this kind of questions, we should always know all the formulas of inverse trigonometric functions and the derivative and the formulas of differentiation of trigonometric ratios.
In this question we have the inverse trigonometric function $\arcsin x$ which is called the inverse sine function or the sine inverse function. We can also represent the given function as ${\sin ^{ - 1}}x$. We use the chain rule whenever there are two functions in the expressions or there can be more than $2$ functions in the expression too.